Branislav M. Notaros
Electrostatic Field
in
Free Space
Page
1
Dielectrics, Capacitance,
and Electric Energy
Page 61
Steady Electric
Currents
Page 124
Magnetostatic Field in
Free Space
Page 173
Magnetostatic Field in
Material Media
Page 221
Slowly Time-Varying
Electromagnetic Field
Page 263
Inductance and
Magnetic Energy
Page 311
8
Rapidly Time- Varying
Electromagnetic Field
Page 351
10
6
3
1
0
10
9
10
12
s
r adio
microwaves
waves
i
/
\
15
10 18
1C
>
#
9
j
0
/[Hz]
o
infrared
i
6
i
i
10 3
Page 408
x-rays
ultraviolet
cosmic rays
7 -rays
K
>
Uniform Plane
Electromagnetic Waves
2l
A[m]
i
i
10 3
1C
-*
-9
io
io~
12
icr
15
10
Reflection and Transmission
of Plane
Waves
Page 471
integrated circuits
11
Field Analysis of
Transmission Lines
Page 533
12
Circuit Analysis of
Transmission Lines
Page 576
13
Waveguides and
Cavity Resonators
short circuit
Page 662
14
Antennas and Wireless
Communication Systems
Page 713
ELECTROMAGNETICS
Branislav M. Notaros
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PEARSON
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River,
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Electromagnetics
Electromagnetics
Branislav
M. Notaros
Department of Electrical and Computer Engineering
Colorado State University
Prentice Hall
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These
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include the development, research, and testing of the theories and programs to determine their effective-
The author and publisher make no warranty of any kind, expressed or implied, with regard to these
programs or the documentation contained in this book. The author and publisher shall not be liable in
any event for incidental or consequential damages in connection with, or arising out of, the furnishing,
ness.
performance, or use of these programs.
Library of Congress Cataloging-in-Publication Data
Notaros, Branislav M.
Electromagnetics
/
Branislav
M. Notaros,
cm.
ISBN 0-13-243384-2
p.
Electromagnetism
1.
—
Textbooks.
I.
Title.
QC760.N68 2010
537-dc22
2010002214
Prentice Hall
an imprint of
is
PEARSON
www.pearsonhighered.com
ISBN-13: 978-0-13-243384-6
0-13-243384-2
ISBN-10:
To the pioneering
giants of electromagnetics
Michael Faraday, James Clerk Maxwell, and others (please see the inside back cover)
for providing the foundation of this book.
To
my professors and colleagues
Branko Popovic
(late),
for making
To
Antonije Djordjevic, and others
me nearly
all
understand and fully love
my students
for teaching
in all
my classes
me to
over
this stuff
all
these years
teach.
To Olivera,
Jelena,
and Milica
for everything
else.
Contents
Preface
xi
2.2
Polarization Vector
2.3
Bound Volume and
64
Densities
1
Evaluation of the Electric Field and
2.4
Electrostatic Field in Free
Coulomb’s Law
1.1
Space
63
Surface Charge
1
Potential due to Polarized Dielectrics
2
Generalized Gauss’
2.5
Law
68
70
1.2
Definition of the Electric Field Intensity
2.6
Characterization of Dielectric Materials
2.7
Maxwell’s Equations for the Electrostatic
1.3
Vector 7
Continuous Charge Distributions
1.4
On
1.5
Electric Field Intensity Vector
1.6
Definition of the Electric Scalar
1.7
Electric Potential due to
the
Volume and Surface
Charge Distributions
Distributions
Differential Relationship
Gradient
1.11
3-D and 2-D
1.14
1.15
1.16
1.17
1.19
22
Poisson’s and Laplace’s Equations
Finite-Difference
Charged Conductors
2.13
Analysis of Capacitors with
Capacitor
Homogeneous
88
95
2.16
Energy of an Electrostatic
System 102
Electric Energy Density
104
2.17
Dielectric
Breakdown
Systems
108
2.15
in Electrostatic
43
46
3
48
1.20
Method
1.21
Charged Metallic Bodies
Image Theory 51
Moments for Numerical
of
Analysis
3.1
49
Electric
3.2
3.3
2
and
Electric
61
Polarization of Dielectrics
62
Currents
124
Current Density Vector and Current
Intensity
125
Conductivity and
Form
2.1
86
Dielectrics
Steady
Energy
for
Analysis of Capacitors with Inhomogeneous
2.14
Charge Distribution on Metallic Bodies of
Dielectrics, Capacitance,
82
2.12
Arbitrary Shapes
of
Method
Numerical
Solution of Laplace’s Equation 84
Definition of the Capacitance of a
Dielectrics
Electrostatic Shielding
Boundary
79
2.11
23
Electric Dipoles
26
Formulation and Proof of Gauss’ Law 28
Applications of Gauss’ Law 31
Differential Form of Gauss’ Law 35
Divergence 36
Conductors in the Electrostatic Field 39
Evaluation of the Electric Field and
and
75
2.10
between the Field
in Electrostatics
Potential due to
1.18
Conditions
18
1.9
1.10
Dielectric-Dielectric
2.9
21
and Potential
Homogeneous Media
due to Given
Given Charge
Voltage
75
Electrostatic Field in Linear, Isotropic,
2.8
10
1.8
1.13
9
16
Potential
1.12
Field
8
Integration
71
Ohm’s Law
in
Local
128
Losses in Conductors and Joule’s
Local Form 132
3.4
Continuity Equation
3.5
Boundary Conditions
Currents
Law
in
133
for Steady
137
vii
Contents
viii
3.6
Distribution of Charge in a Steady Current
3.7
Relaxation Time
3.8
Resistance,
Law
3.9
3.10
1
Maxwell’s Equations for the Time-Invariant
Electromagnetic Field 258
6
Duality between Conductance and
Slowly Time-Varyinq Electromaqnetic
Capacitance
Field
146
External Electric Energy Volume Sources
6.1
149
6.2
Analysis of Capacitors with Imperfect
Dielectrics
152
6.3
Analysis of Lossy Transmission Lines with
Steady Currents
3.13
1
Joule’s
140
Inhomogeneous
3.12
1
139
Ohm's Law, and
and Generators
3.1
5.
138
Field
Induced Electric Field Intensity Vector 264
Slowly Time-Varying Electric and Magnetic
Fields
269
Faraday’s Law of Electromagnetic
Induction
156
Grounding Electrodes
263
Maxwell’s Equations for the Slowly
Time-Varying Electromagnetic Field
6.5
Computation of Transformer
6.6
Electromagnetic Induction due to
Motion 283
6.7
Total Electromagnetic Induction
6.8
Eddy Currents
162
4
Induction
Magnetostatic Field
4.1
in
Free Space
173
Magnetic Force and Magnetic Flux Density
Vector
174
Law
4.2
Biot-Savart
4.3
4.4
Magnetic Flux Density Vector due to Given
Current Distributions 179
Formulation of Ampere’s Law 185
4.5
Applications of Ampere’s
4.6
Differential
Form
of
Law
Ampere’s Law
4.9
Magnetic Vector Potential 201
Proof of Ampere’s Law 204
Magnetic Dipole 206
The Lorentz Force and Hall Effect 209
Evaluation of Magnetic Forces 211
7.1
195
of Conservation of Magnetic Flux
Magnetostatic Field
in
Material Media
7.3
198
7.4
7.5
7.6
5
221
5.2
5.3
Materials 223
Magnetization Volume and Surface Current
5.5
5.6
5.7
5.8
5.9
Field
227
Generalized Ampere’s Law 234
Permeability of Magnetic Materials 236
Maxwell’s Equations and Boundary
Conditions for the Magnetostatic Field 239
Image Theory for the Magnetic Field 241
Magnetization Curves and Hysteresis 243
Magnetic Circuits - Basic Assumptions for
the Analysis
5.10
Kirchhoff’s
247
Laws
for
Magnetic Circuits
312
318
Analysis of Magnetically Coupled
Circuits
324
Magnetic Energy of Current-Carrying
Conductors 331
Magnetic Energy Density 334
Internal and External Inductance in Terms
of Magnetic Energy 342
Self-Inductance
Mutual Inductance
8
8.1
Displacement Current
8.2
Maxwell’s Equations for the Rapidly
250
352
Time- Varying Electromagnetic Field
Densities
5.4
311
Rapidly Time-Varying Electromagnetic
351
Magnetization Vector 222
Behavior and Classification of Magnetic
5.1
Inductance and Magnetic Energy
7.2
Law
4.13
289
294
7
193
Curl
4.12
277
187
4.8
4.11
276
177
4.7
4.10
271
6.4
8.3
Electromagnetic Waves
8.4
Boundary Conditions
8.5
8.6
8.7
357
361
for the Rapidly
Time- Varying Electromagnetic Field 363
Different Forms of the Continuity Equation
364
for Rapidly Time-Varying Currents
Time-Harmonic Electromagnetics 366
Complex Representatives of
Time-Harmonic Field and Circuit
Quantities
369
Contents
8.8
Maxwell’s Equations in Complex
Domain
8.9
8.10
8.11
8.12
10.9
Lorenz Electromagnetic Potentials 376
Computation of High-Frequency Potentials
and Fields in Complex Domain 381
Theorem
389
Complex Poynting Vector
Poynting’s
Wave Propagation
Media
373
11
533
Transmission Lines
TEM Waves in Lossless Transmission Lines
with
11.2
9
in Multilayer
520
Field Analysis of
11.1
397
Homogeneous
Electrostatic
534
Dielectrics
and Magnetostatic Field
538
Distributions in Transversal Planes
Uniform Plane Electromagnetic Waves
408
ix
11.3
Currents and Charges of Line
Conductors
539
9.1
Wave Equations
9.2
Uniform-Plane-Wave Approximation 411
Time-Domain Analysis of Uniform Plane
11.4
Analysis of Two-Conductor Transmission
Waves 412
Time-Harmonic Uniform Plane
Waves and Complex-Domain Analysis 416
The Electromagnetic Spectrum 425
Arbitrarily Directed Uniform TEM
Waves 427
Theory of Time-Harmonic Waves in Lossy
Media 429
11.5
Transmission Lines with Small Losses
11.6
Attenuation Coefficients for Line
11.7
Conductors and Dielectric 550
High-Frequency Internal Inductance of
9.3
9.4
9.5
9.6
9.7
409
9.8
Explicit Expressions for Basic Propagation
9.9
9.10
Wave Propagation
Wave Propagation
Conductors
439
9.11
Skin Effect
441
Parameters
in
Good
Good
Dielectrics
436
Polarization of Electromagnetic
Circuit Parameters of Transmission
Lines
11.9
11.10
Waves
12.1
458
10.3
10.4
10.5
10.6
10.7
10.8
567
576
Telegrapher’s Equations and Their Solution
in
12.3
and Transmission of Plane
Waves 471
10.2
563
Multilayer Printed Circuit Board
Complex Domain
12.4
472
Normal Incidence on a Penetrable Planar
Interface 483
Surface Resistance of Good
Conductors 492
Perturbation Method for Evaluation
of Small Losses 497
Oblique Incidence on a Perfect
Conductor 499
Concept of a Rectangular Waveguide 505
Oblique Incidence on a Dielectric
Boundary 507
Total Internal Reflection and Brewster
Angle 513
581
Circuit Analysis of
Lines
a Perfectly Conducting
577
Circuit Analysis of Lossless Transmission
Lines
10
Plane
557
Transmission Lines with Inhomogeneous
Circuit Analysis of Transmission Lines
12.2
Normal Incidence on
556
12
9.14
10.1
547
Evaluation of Primary and Secondary
Dielectrics
in
9.13
Reflection
540
Transmission Lines
11.8
433
Wave Propagation in Plasmas 447
Dispersion and Group Velocity 452
9.12
Lines
Low-Loss Transmission
581
Reflection Coefficient for Transmission
Lines
583
12.5
Power Computations of Transmission
12.6
Transmission-Line Impedance
12.7
Complete Solution
Lines
589
Current
12.8
and
597
Short-Circuited, Open-Circuited, and
Matched Transmission Lines
12.9
592
for Line Voltage
Transmission-Line Resonators
601
608
12.10
Quality Factor of Resonators with Small
12.11
The Smith Chart - Construction and Basic
Losses
610
Properties
12.12
614
Circuit Analysis of Transmission Lines
Using the Smith Chart
618
x
Contents
12.13
Transient Analysis of Transmission
14.1
628
Thevenin Equivalent Generator Pair and
Reflection Coefficients for Line
Transients 630
Step Response of Transmission Lines with
Purely Resistive Terminations 634
Analysis of Transmission Lines with Pulse
Excitations 640
Bounce Diagrams 646
Transient Response for Reactive or
Nonlinear Terminations 649
Lines
12.14
12. 15
12.16
12.17
12.18
14.2
14.3
14.4
14.5
14.6
14.7
14.8
14.9
13
14.10
Waveguides and Cavity Resonators
13.1
662
Analysis of Rectangular Waveguides Based
on Multiple Reflections of Plane
13.2
663
Propagating and Evanescent Waves
13.4
Dominant Waveguide Mode
General TE Modal Analysis
13.5
TM Modes in a Rectangular
13.3
Waveguides
Waveguide
666
of Rectangular
13.7
13.8
Waves 680
Power Flow along
of
TE and TM
a Waveguide
681
Waveguides with Small Losses 684
Waveguide Dispersion and Wave
688
Waveguide Couplers 692
Rectangular Cavity Resonators 696
Electromagnetic Energy Stored in a Cavity
Resonator 700
Quality Factor of Rectangular Cavities with
Small Losses 703
Velocities
13.12
13.13
13.14
14.14
14.15
APPENDICES
676
Modes 677
Wave Impedances
13.11
14.12
715
720
Steps in Far-Fieid Evaluation of an
Arbitrary Antenna 722
Radiated Power, Radiation Resistance,
Antenna Losses, and Input Impedance 730
Antenna Characteristic Radiation Function
and Radiation Patterns 736
Antenna Directivity and Gain 740
Antenna Polarization 745
Wire Dipole Antennas 745
Image Theory for Antennas above a
Perfectly Conducting Ground Plane 751
Monopole Antennas 754
Magnetic Dipole (Small Loop)
Antenna 758
Theory of Receiving Antennas 760
Antenna Effective Aperture 766
Friis Transmission Formula for a Wireless
Link 768
Antenna Arrays 772
Far Field and Near Field
671
Cutoff Frequencies of Arbitrary Waveguide
13.9
1
668
13.6
13.10
14.1
14.13
Waves
Electromagnetic Potentials and Field
Vectors of a Hertzian Dipole
1
Quantities, Symbols, Units,
Constants
and
791
2
Mathematical Facts and Identities
3
Vector Algebra and Calculus Index
4
Answers to Selected Problems
14
Antennas and Wireless Communication
Systems 713
796
Bibliography
Index
809
806
802
801
)
Preface
E
lectromagnetic theory
is
a fundamental under-
pinning of technical education, but, at the same
one of the most
time,
difficult subjects for
students
to master. In order to help address this difficulty
contribute to overcoming
on electromagnetic
fields
it,
here
is
and
another textbook
and waves
for undergrad-
uates, entitled, simply, Electromagnetics. This text
provides engineering and physics students and other
users with a comprehensive
knowledge and firm
and wave computation and
most importantly, outstanding (by the judgment of students so far) workedfor electromagnetic field
problem
solving, and,
out examples,
and
tions,
homework problems, conceptual ques-
MATLAB
exercises.
The goal
is
to sig-
improve students’ understanding of electromagnetics and their attitude toward it. Overall,
the book is meant as an “ultimate resource ” for
undergraduate electromagnetics.
nificantly
grasp of electromagnetic fundamentals by emphasizing
both mathematical rigor and physical under-
standing of electromagnetic theory, aimed toward
is
distinguishing features of the
371
practical engineering applications.
The book
The
designed primarily (but by no means
realistic
book
examples with very detailed and instrucoupled to the theory, includ-
ctive solutions, tightly
exclusively) for junior-level undergraduate university
ing strategies for problem solving
and college students in electrical and computer engineering, physics, and similar departments, for both
fully supported
two-semester
(or
two-quarter)
course
sequences
and one-semester (one-quarter) courses. It includes
14 chapters on electrostatic fields, steady electric
currents, magnetostatic fields, slowly time-varying
(low-frequency) electromagnetic
fields,
rapidly time-
varying (high-frequency) electromagnetic
fields,
all
of them.
It
also introduces
many new pedagogical
features not present in any of the existing texts.
This text provides
cally
and
new
style
many nonstandard
practically important sections
650
realistic
theoreti-
and chapters,
and approaches to presenting challenging topics and abstract electromagnetic phenomena, innovative strategies and pedagogical guides
end-of-chapter problems, strongly and
by solved examples (there
example for every homework problem
Clear, rigorous, complete,
of material, balance
and
with
is
a
demo
logical presentation
of breadth and depth, balance
of static (one third) and dynamic ( two
uni-
form plane electromagnetic waves, transmission
lines, waveguides and cavity resonators, and antennas
and wireless communication systems.
Apparently, there are an extremely large number of quite different books for undergraduate electromagnetics available (perhaps more than for any
other discipline in science and engineering), which
are all very good and important. This book, however,
aims to combine the best features and advantages of
are:
thirds) fields,
no missing steps
emphaand ordering the material in a course or courses,
Flexibility for different options in coverage,
sis,
including the transmission-lines-first approach
Many nonstandard
derivations,
topics
and subtopics and new
explanations, proofs,
examples, pedagogical
style,
and
interpretations,
visualizations
500 multiple-choice conceptual questions (on the
Companion
Website), checking conceptual under-
standing of the book material
400
MATLAB computer exercises and projects
the
Companion
tions (tutorials)
Website),
and
many
(on
with detailed solu-
MATLAB codes (m files)
www.pearsonhighered.com/notaros
The following sections explain these and other
tures in more detail.
fea-
XI
Preface
xii
WORKED EXAMPLES AND
HOMEWORK PROBLEMS
a strong appreciation for both
mentals and
its
theoretical funda-
its
practical applications.
“Physical” nontrivial examples are good also
-
and
- as
The most important feature of the book is an
extremely large number of realistic examples, with
for instructors
detailed and pedagogically instructive solutions, and
and discussion in the class
than the “plug-and-chug” or purely “mathematical”
end-of-chapter (homework) problems, strongly and
fully
supported by solved examples. There are a
worked examples, all tightly coupled
for lectures
much more
they are
interesting
recitations
and suitable
for
presentation
logical
examples.
total of 371
to the theory, strongly reinforcing theoretical con-
cepts and smoothly and systematically developing
the problem-solving skills of students,
and a
total of
CLARITY, RIGOR,
AND
COMPLETENESS
650 end-of-chapter problems, which are essentially
offered and meant as end-of-section problems (indications appear at the ends of sections as to which
problems correspond to that section).
Along with the number and type of examples and
problems (and questions and exercises), the most
homework problem
attention to clarity, completeness, and pedagogical
always an example or
whose detailed solution
soundness of presentation of the material throughout the entire text, aiming for an optimal balance of
breadth and depth. Electromagnetics, as a fundamen-
Most importantly,
for each
or set of problems, there
set of
examples
is
in the text
provides the students and other readers with
all
nec-
characteristic feature of the
book
is
its
consistent
essary instruction and guidance to be able to solve
tal
problem on their own, and to complete all homework assignments and practice for tests and exams.
The abundance and quality of examples and problems are enormously important for the success of
the course and class: students always ask for more
and more solved examples, which must be relevant
for the many problems that follow (for homework
and exam preparation) - and this is exactly what this
book attempts to offer.
Examples and problems in the book emphasize physical conceptual reasoning and mathematical
synthesis of solutions, and not pure formulaic (plugand-chug) solving. They also do not carry dry and too
complicated pure mathematical formalisms. The primary goal is to teach the readers to reason through
different (more or less challenging) situations and to
help them gain confidence and really understand and
like the material. Many examples and problems have
plete physical explanations for (almost) everything
the
a strong practical engineering context.
show and explain every
ample discussions of approaches, strategies, and alternatives. Very often, solutions are presented in more than one way to aid understanding
and development of true electromagnetic problemSolutions to examples
step, with
By acquiring such
skills, which are
browsing through
the book pages in a quest for a suitable “black-box”
formula or set of formulas nor a skillful use of pocket
calculators to plug-and-chug, the reader also acquires
true confidence and pride in electromagnetics, and
solving
skills.
definitely not limited to a skillful
science and engineering discipline, provides
within
com-
scope and rigorous mathematical models
its
for everything
it
covers. Thus, besides a couple of
Coulomb’s law)
model to
building the most impres-
experimental fundamental laws
(like
that have to be taken for granted for the
build on,
all
other steps in
and exciting structure called the electromagnetic
theory can be readily presented to the reader in a
consistent and meaningful manner and with enough
detail to be understandable and appreciable. This is
exactly what this book attempts to do.
Simply speaking, literally everything is derived,
proved, and explained (except for a couple of expersive
imental
facts),
with
many new
derivations, expla-
and visualizations.
and important concepts and derivations are
regularly presented in more than one way to help
students understand and master the subject at hand.
Maximum effort has been devoted to a continuous
logical flow of topics, concepts, equations, and ideas,
with practically no “intentionally skipped” steps and
parts. This, however, is done in a structural and modular manner, so that the reader who feels that some
steps, derivations, and proofs can be bypassed at the
time (with an opportunity of redoing it later) can do
nations, proofs, interpretations,
Difficult
so,
but this
discretion
is left
to the reader’s discretion (or to the
and advice of the course
instructor), not
the author’s.
Overall,
(or
all
my
approach
is
to provide
all
possible
necessary) explanations, guidance, and detail
Preface
transmission
lines,
waveguides, and antennas). In
whereas students’ actual understanding of the mate-
addition, the
book
features a favorable balance of
“on their own feet,” and ability
independent
work are tested and challenged
to do
through numerous and relevant end-of-chapter problems and conceptual questions, and not through
filling the missing gaps in the text.
On the other hand, I am fully aware that
static
in the theoretical parts
rial,
and examples
xiii
in the text,
their thinking
brevity
may seem attractive
to students at
first
glance
(one third) and dynamic (two thirds) fields.
or a sequence of courses using
cover the book material,
would
completely
text
Ideally, a course
this
with a likelihood that some portions would be given
to students as a reading assignment only.
book allows
the
a lot of flexibility and
However,
many
dif-
ferent options in actually covering the material, or
means fewer pages for readHowever, most students will readily
acknowledge that it is indeed much easier and faster
to read, grasp, and use several pages of thoroughly
explained and presented material as opposed to a
single page of condensed material with too many
missing parts. During my dealings with students over
so many years, I have been constantly told that
they in fact prefer having everything derived and
explained, and host of sample problems solved, to
a lower page count and too many important parts,
steps, and explanations missing, and too few detailed
solutions, and this was the principal motivation for
parts of
my writing this book.
able for different areas of emphasis and specialized
because
it
typically
ing assignments.
This approach, in
my
opinion,
is
also
good
for
have a self-contained, ready-touse continuous “story” for each of their lectures,
instead of a set of discrete formulas and sample
facts with little or no explanations and detail. On
the other hand, the instructor may choose to present
only main facts for a given topic in class and rely on
students for the rest, as they will be able to quickly
and readily understand all reading assignments from
the book. Indeed, I expect that every instructor
using this text will have different “favorite” topics presented in class with all details and in great
instructors, as they
depth, including a
number
of examples, while “giv-
away” some other topics to students to cover on
their own, with more or less depth, including worked
ing
examples.
it,
and ordering the topics
in a course (or
courses).
One
1-7,
do
scenario
is
to quickly go through Chapters
just basic concepts
and equations, and a
couple of examples in each chapter, quickly reach
Chapter 8 (general Maxwell’s equations,
etc.),
and
then do everything else as applications of general
Maxwell’s equations, including selected topics from
Chapters 1-7 and more or less complete coverage
of
all
other chapters. This scenario would essenreflect the inverse
tially
(nonchronological) order
of topics in teaching/learning electromagnetics. In
fact,
there
may be many
different scenarios suit-
outcomes of the course and the available time,
of
them advancing
in chronological order,
all
through
Chapters 1-14 of the book, just with different
speeds and different levels of coverage of individual
chapters.
To help the instructors create a plan for using the
book material in their courses and students and other
readers prioritize the book contents in accordance
with their learning objectives and needs, Tables 1
and 2 provide classifications of all book chapters
and sections, respectively, in two levels, indicating
which chapters and sections within chapters are suggested as more likely candidates to be skipped or
skimmed (covered lightly). This is just a guideline,
and I expect that there will be numerous extremely
creative, effective, and diverse combinations of book
and subtopics constituting course outlines and
meet the
preferences, interests, and needs of instructors, students, and other book users.
Most importantly, if chapters and sections are
topics
learning/training plans, customized to best
OPTIONS IN COVERAGE OF THE
MATERIAL
This book promotes and implements the direct or
chronological and not inverse order of topics in
teaching/learning electromagnetics, which can briefly
and then dynamic
topics, or first fields (static, quasistatic, and rapidly
time-varying) and then waves (uniform plane waves.
be characterized
as: first static
skipped or skimmed in the class, they are not skipped
nor skimmed in the book, and the student will always
be able to quickly find and apprehend additional
information and fill any missing gaps using pieces of
the book material from chapters and sections that are
not planned to be covered in detail.
XIV
Preface
Table
would
Classification of
1.
book chapters
in
two groups, where “mandatory” chapters
are those that
be covered in most courses, while some of the “elective” chapters could be skipped (or skimmed)
based on specific areas of emphasis and desired outcomes of the course or sequence of courses and the available time. In selecting the material for the course(s), this classification at the chapter level could be combined
with the classification at the section level given in Table 2.
likely
“Mandatory” Chapters:
1, 3, 4, 6, 8, 9,
“Elective” Chapters:
12
14
2, 5, 7, 10, 11, 13,
1.
Electrostatic Field in Free Space
2.
Dielectrics, Capacitance,
3.
Steady Electric Currents
5.
Magnetostatic Field in Material Media
4.
Magnetostatic Field
7.
Inductance and Magnetic Energy
6.
Slowly Time-Varying Electromagnetic Field
10.
8.
Rapidly Time-Varying Electromagnetic Field
11. Field
9.
Uniform Plane Electromagnetic Waves
13.
Waveguides and Cavity Resonators
14.
Antennas and Wireless Communication Systems
12. Circuit
in
Free Space
Analysis of Transmission Lines
and Electric Energy
Reflection and Transmission of Plane
Waves
Analysis of Transmission Lines
2. Classification of book sections in two “tiers” in terms of the suggested priority for coverage; if
one or more sections in any of the chapters are to be skipped (or skimmed) given the areas of emphasis and
specialized outcomes of the course or courses and the available time, then it is suggested that they be selected
from the “tier two” sections, which certainly does not rule out possible omission (or lighter coverage) of some
of the “tier one” sections as well.
Table
Chapter
1.
“Tier
Electrostatic Field in Free Space
2. Dielectrics,
1.
Capacitance, and Electric Energy
One”
Sections
1-1.4, 1.6, 1.8-1.10, 1.13-1.16
2.1,2.6,2.7,2.9,2.10,2.12,
“Tier
Two”
Sections
1.5, 1.7, 1.11, 1.12,
1.17-1.21
2.2-2.5,2.8,2.11,2.14, 2.17
2.13,2.15,2.16
3.
Steady Electric Currents
3.1-3.4,3.8,3.10,3.12
3.5-3.7,3.9,3.11,3.13
4.
Magnetostatic Field in Free Space
4.1, 4.2, 4.4-4.7, 4.9
4.3, 4.8,
5.
Magnetostatic Field
5.1,5.5,5.6,5.8,5.11
5.2-5.4, 5.7, 5.9, 5.10
6.
Slowly Time-Varying Electromagnetic Field
6.2-6.5
6.1, 6.6-6.8
7.
Inductance and Magnetic Energy
7.1, 7.4, 7.5
7.2, 7.3, 7.6
8.
Rapidly Time-Varying Electromagnetic Field
8.2, 8.4, 8.6-8. 8,
9.
Uniform Plane Electromagnetic Waves
9. 3-9.7,
in
Material Media
10.
Reflection and Transmission of Plane
1 1
Field Analysis of Transmission Lines
.
12. Circuit
Waves
Analysis of Transmission Lines
8.11,8.12
9.11,9.14
4.10-4.13
8.1,8.3,8.5,8.9,8.10
9.1,9.2,9.8-9.10, 9.12,9.13
10.1, 10.2, 10.4-10.7
10.3, 10.8, 10.9
11.4-11.6, 11.8
11.1-11.3, 11.7, 11.9, 11.10
12.1-12.6, 12.11, 12.12, 12.15
12.7-12.10, 12.13, 12.14,
12.16-12.18
13.
14.
Waveguides and Cavity Resonators
13.1-13.3, 13.6, 13.8, 13.9,
13.4, 13.5, 13.7, 13.10, 13.11,
13.12
13.13, 13.14
Antennas and Wireless Communication
14.1, 14.2, 14.4-14.6, 14.8,
14.3, 14.7, 14.9-14.13
Systems
14.14,14.15
Preface
Table
xv
Ordering the book material for the transmission-lines-first approach; Chapter 12 (Circuit
3.
is written using only pure circuit-theory concepts (all field-theory aspects of
transmission lines are placed in Chapter 11 - Field Analysis of Transmission Lines), so it can be taken at the very
Analysis of Transmission Lines)
beginning of the course (or at any other time in the course). Note that two sections introducing (or reviewing)
complex representatives of time-harmonic voltages and currents (Sections 8.6 and 8.7) must be done before
Chapter
12.
Section
8.6:
Time-Harmonic Electromagnetics
Section
8.7:
Complex Representatives of Time-Harmonic
Chapter
12: Circuit Analysis of
Transmission Lines (or a selection of sections from Chapter 12 - see Table 2)
Chapters 1-11, 13, 14 or a selection of chapters (see Table
TRANSMISSION-LINES-FIRST
APPROACH
One
possible
Field and Circuit Quantities
1)
and sections (see Table 2)
an invaluable resource. They are also ideal for
and discussions (so-called active
teaching and learning) to be combined with traditional lecturing - if so desired.
in-class questions
exception
from the chronological
sequence of chapters (topics) in using this text
implies a different placement of Chapter 12 (Circuit
Analysis of Transmission Lines), which is written in
such a manner that it can be taken at any time,
even at the very beginning of the course, hence
constituting the transmission-lines-first approach to
the course and learning the material.
Namely, the field and circuit analyses of transmission
lines are completely decoupled in the book, so that
any field-theory aspects are placed in Chapter 11
(Field Analysis of Transmission Lines) and only pure
circuit-theory concepts are used in Chapter 12 with
per-unit-length characteristics (distributed parameters) of the lines being taken for granted (are
assumed to be known) from the field analysis if
the circuit analysis is done first. Table 3 shows the
transmission-lines-first scenario using this book.
teaching
In addition, conceptual questions are perfectly
suited for class assessments, namely, to assess stu-
and evaluate the effectiveness
between the
course “pretest” and “posttest” scores, and especially in light of ABET and similar accreditation
criteria (the key word in these criteria is “assessment”). Selected conceptual questions from the large
collection provided in the book can readily be
used by instructors as partial and final assessment
dents’ performance
of instruction, usually as the “gain”
instruments for individual topics at different points
in the course
on the Companion Website,
and comprehensive collection of
MATLAB computer exercises, strongly coupled to
the book material, both the theory and the worked
examples, and designed to help students develop a
stronger intuition and a deeper understanding of
electromagnetics, and find it more attractive and lika
The book
provides,
on the Companion Website, a
500 conceptual questions. These are multiplechoice questions that focus on the core concepts
of the material, requiring conceptual reasoning and
understanding rather than calculations. They serve
as checkpoints for readers following the theoretical
parts and worked examples (like homework problems, conceptual questions are referred to at the ends
of sections). Generally, conceptual questions may
appear simple, but students often find them harder
than the standard problems. Pedagogically, they are
total of
for the entire class.
MATLAB EXERCISES, TUTORIALS,
AND PROJECTS
The book
MULTIPLE-CHOICE CONCEPTUAL
QUESTIONS
and
very
able.
provides,
large
MATLAB
is
chosen principally because
it is
a
generally accepted standard in science and engineering education worldwide.
There are a total of 400 MATLAB exercises,
which are referred to regularly within all book chapters, at the ends of sections, to supplement problems
and conceptual questions. Each section of this collection starts with a comparatively very large
num-
ber of tutorial exercises with detailed completely
XVI
Preface
worked out
files).
solutions, as well as
MATLAB codes (m
This resource provides abundant opportunities
and homework
for instructors for assigning in-class
projects
-
if
fascinating biographies of
famous
scientists
and pio-
neers in the field of electricity and magnetism. There
my
are a total of 40 biographies, which are, in
view,
not only very interesting historically and informa-
so desired.
terms of providing the factual chronological
review of the development of one of the most imprestive in
VECTOR ALGEBRA AND CALCULUS
and complete theories of the entire
and technological world - the electromagnetic theory - but they also often provide additional
technical facts and explanations that complement the
sive, consistent,
scientific
Elements of vector algebra and vector calculus
are presented and used gradually across the book
sections with an emphasis on physical insight and
immediate links to electromagnetic field theory concepts, instead of having a purely mathematical review
in
a
separate chapter.
They are
fully
integrated
with the development of the electromagnetic theory,
where they actually belong and really come to life.
The mathematical concepts of gradient, divergence, curl, and Laplacian, as well as line (circulation), surface (flux), and volume integrals, are
literally derived from physics (electromagnetics),
where they naturally emanate as integral parts of
electromagnetic equations and laws and where their
physical meaning is almost obvious and can readbe made very visual. Furthermore, the text is
written in such a way that even a reader with litily
tle
background in vector algebra and vector calculus
indeed be able to learn or refresh vector analy-
will
sis
concepts directly through the
Appendix
ters (please see
3
first
several chap-
- Vector Algebra and
Calculus Index).
LINKS
I also feel that some basic knowledge about the discoverers - who made such epochal
scientific achievements and far-reaching contributions to humanity - like Faraday, Maxwell, Henry,
material in the text.
Hertz, Coulomb, Tesla, Heaviside, Oersted, Ampere,
Ohm, Weber, and
throughout
all
of
explanations for
chapters.
its
all
It
elements of
made an
irre-
engineering and physics students.
SUPPLEMENTS
The book
(for
accompanied by the Solutions Manual
is
instructors)
with
detailed
solutions
to
all
end-of-chapter problems (written in the same manin the examples in the book),
answers to all conceptual questions, and MATLAB
codes (m files) for all MATLAB computer exercises
and projects, as well as by PowerPoint slides with all
illustrations from the text and by other supplements.
Pearson eText of the book is also available.
ner as the solutions
www.pearsonhighered.com/notaros
TO CIRCUIT THEORY
The book provides detailed discussions of the links
between electromagnetic theory and circuit theory
others should be
placeable part of a sort of “general education” of our
ACKNOWLEDGMENTS
contains physical
circuit theory, for
both dc and ac regimes. All basic circuit-theory equations (circuit laws, element laws, etc.) are derived
from electromagnetic theory. The goal is for the
reader to develop both an appreciation of electromagnetic theory as a foundation of circuit theory
and electrical engineering as a whole, as well as an
understanding of limitations of circuit theory as an
approximation of field theory.
is based on my electromagnetics teachand research over more than 20 years at
This text
ing
the University of Belgrade, Yugoslavia
(Serbia),
Colorado at Boulder, University
of Massachusetts Dartmouth, and Colorado State
University
of
University,
in
students
advice,
at
Fort
my
acknowledge
these
ideas,
Collins,
U.S.A.
I
gratefully
colleagues and/or former Ph.D.
institutions
enthusiasm,
whose
initiatives,
discussions,
co-teaching,
and co-authorships have shaped my knowledge,
teaching style, pedagogy, and writing in electro-
HISTORICAL ASIDES
magnetics, including: Prof. Branko Popovic (late).
Throughout almost
Prof. Milan Ilic, Prof. Miroslav Djordjevic, Prof.
Antonije Djordjevic, Prof. Zoya Popovic, Gradimir
Bozilovic, Prof. Momcilo Dragovic (late), Prof.
all
chapters of the book, dozens
of Historical Asides appear with quite detailed and
Preface
Branko Kolundzija,
Prof.
Vladimir Petrovic, and
Jovan Surutka (late). All I know in electromagnetics and about its teaching I learned from them or
with them or because of them, and I am enormously
Prof.
thankful for that.
am
I
over
ing
all
grateful to
all
these years for
my
all
students in
the joy
them electromagnetics and
I
all
my
have had
classes
in teach-
for teaching
me
to
teach better.
I
Nada
Ana
Sekeljic,
my
current Ph.D. students
Manic, and Sanja Manic for
MATLAB computer
and codes, checking the derivaand examples in the book, and solving selected
their invaluable help in writing
exercises, tutorials,
tions
end-of-chapter problems.
gratitude to
Prof.
Milan
with the
manager Scott Disanno, for expertly leading the
book production, Marcia Horton, Vice President
and Editorial Director with Prentice Hall, for great
conversations and support in the initial phases of
the project, and Tom Robbins, former Publisher at
Prentice Hall, for the first encouragements. I hope
they enjoyed our dealings and discussions as extensively as
my
Ilic,
initial
I
owe
a particular debt of
colleague and former Ph.D. student
for his outstanding
work and help
electronic artwork in the book.
colleagues and former students Andjelija
Ilic
My
and
Prof. Miroslav Djordjevic, as well as Olivera Notaros,
also contributed very significantly to the artwork, for
which I am sincerely indebted.
I would like to express my gratitude to the reviewers of the manuscript for their extremely detailed,
useful, positive, and competent comments that I feel
helped me to significantly improve the quality of
the book, including: Professors Indira Chatterjee,
Robert J. Coleman, Cindy Harnett, Jianming Jin, Leo
Kempel, Edward F. Kuester, Yifei Li, Krzysztof A.
Michalski, Michael A. Parker, Andrew F. Peterson,
Costas D. Sarris, and Fernando L. Teixeira.
Special thanks to all members of the Pearson
Prentice Hall team, who all have been excellent, and
particularly to
my
editor
Andrew
Gilfillan,
who
has
been extremely helpful and supportive, and whose
input was essential at many stages in the development of the manuscript and book, my production
I
did.
my
thank
I
especially thank
xvii
wife
ECE
Olivera Notaros,
Department
who
also
Colorado
State University, not only for her great and constant support and understanding but also for her
direct involvement and absolutely phenomenal ideas,
advice, and help in all phases of writing the
manuscript and production of the book. Without her,
this book would not be possible or would, at least, be
very different. I also acknowledge extraordinary support by my wonderful daughters Jelena and Milica,
and I hope that I will be able to keep my promise
to them that I will now take a long break from writing. I am very sad that the writing of this book took
me so long that my beloved parents Smilja and Mile
did not live to receive the first dedicated copy of the
book from me, as had been the case with my previous
teaches
in
the
at
books.
Finally,
on
a very personal note as well,
I
really
hope
convey at least a portion of my
admiration and enthusiasm to the readers and help
more and more students start liking and appreciatlove electromagnetics and teaching
that this
book
it,
and
I
will
ing this fascinating discipline with endless impacts.
I
am proud of being able to do that in my classes, and
am now excited and eager to try to spread that mesmuch larger audience using this text. Please
your comments, suggestions, questions, and
corrections (I hope there will not be many of these)
sage to a
send
me
regarding the book to notaros@colostate.edu.
Branislav M. Notaros
Fort Collins, Colorado
“I believe but
cannot explain that the author’s confidence
is
somehow
student as a trust that the text they are reading and learning from
—Anonymous reviewer of the book manuscript
is
transferred to the
worth
their time.”
'
Electrostatic Field in
Free Space
[
Introduction:
E
lectrostatics
ics
is
the branch of electromagnet-
that deals with
phenomena
associated with
which are essentially the consequence of a simple experimental fact - that charges
exert forces on one another. These forces are called
electric forces, and the special state in space due
to one charge in which the other charge is situated and which causes the force on it is called
static electricity,
the electric
field.
Any
charge distribution in space
with any time variation
tric field.
The
charges at rest
due to time-invariant
(charges that do not change in time
and do not move)
or electrostatic
a source of the elec-
is
electric field
is
field.
called the static electric field
This
is
the general electromagnetic
the simplest form of
field,
and
its
physics
and mathematics represent the foundation of the
entire electromagnetic theory.
On
the other hand,
a clear understanding of electrostatics
for
many
and forces
and systems.
electric fields, charges,
electronic devices
We
is
essential
practical applications that involve static
shall begin
in electrical
our study of electrostatics by
investigating the electrostatic field in a
air (free space),
and
which
will
vacuum
or
then be extended to
the analysis of electrostatic structures
composed
of charged conductors in free space (also in this
chapter). In the next chapter,
we
shall evaluate the
electrostatic field in the presence of dielectric
rials,
and include such materials
in
mateour discussion of
general electrostatic systems.
1
2
Chapter
1
Electrostatic Field in Free
Space
HISTORICAL ASIDE
The
first
dates
city
with
B.C.,
of Miletus
PXeKxpov
an
(624 B.C.-546 B.C.),
(Mayvrjala),
wrote that
amber rubbed in wool
attracts pieces of straw or feathers - which we now
Charles
Augustin
Coulomb
was
a
and
in
between like
and unlike charges (i.e., between two charges of
the same or opposite polarity) using his genuine
the
torsion balance apparatus, in a course of experi-
ments
originally
compass.
experimentalist in
electricity
word “magnet”
Greece named Magnesia
which ancient Greeks first noticed
origin of the
basic law for the electrostatic force
Engineering Corps of the
French Army and a brilliant
The
800 B.C.) that pieces of the black rock they
were standing on, now known as the iron mineral
magnetite (Fe 304 ), attracted one another.
de
in
Our experiences
(ca.
(1736-1806)
colonel
elektron ) for amber.
relates to the region in
time,
all
(
to ancient times.
pher and mathematician,
one of the greatest minds
of
fric-
“electron” for the
with magnetism, on the other side, also trace back
Greek philoso-
ancient
name
subatomic particle carrying the smallest amount
of (negative) charge comes from the Greek word
back to the
when Thales
a manifestation of electrification by
is
tion. In relation to this, the
electri-
century
sixth
know
record of our ex-
periences
magne-
aimed
He measured
tion or repulsion that
He
at
improving the mariner’s
the electric force of attrac-
two charged small
pith balls
graduated in
1761 from the School of
the Engineering Corps
( Ecole
du Genie), and
exerted on one another by the amount of twist pro-
was in charge of building the Fort Bourbon
on Martinique, in the West Indies, where he
showed his engineering and organizational skills.
each of the balls and inversely proportional to
tism.
In
1772,
Coulomb returned
to
France
duced on the torsion balance, and demonstrated
an inverse square law for such forces - the force
is
the square of the distance between their centers.
came out to be an underpinning of the
whole area of science and engineering now known
as electromagnetics, and of all of its applications.
Upon the outbreak of the French Revolution in
This result
with
impaired health, and began his research in applied
mechanics. In 1777, he invented a torsion balance
measure small forces, and as a result of his 1781
memoir on friction, he was elected to the French
Academy ( Academie des Sciences ). Between 1785
and 1791, he wrote a series of seven papers on
electricity and magnetism, out of which by far the
most important and famous is his work on the theory of attraction and repulsion between charged
bodies. Namely, Coulomb formulated in 1785 the
to
1.1
The
COULOMB
S
1789,
to
Coulomb
work
in
retired to a small estate near Blois,
peace on
his scientific
memoirs. His
last
post was that of the inspector general of public
instruction,
under Napoleon, from 1802 to 1806.
The law of electric forces on charges now bears his
name - Coulomb’s law - and his name is further
immortalized by the use of coulomb (C) as the unit
of charge.
LAW
basis of electrostatics
that the electric force
proportional to the product of the charges of
Fe
i
is
2
an experimental result called Coulomb’s law. It states
to a point charge Q\ in a
on a point charge Qi due
F
Section 1.1
vacuum
(or air)
is
given by 1 (Fig.
Coulomb's Law
3
1 1)
.
Fel2
1
Q1Q2
Attsq
R2
=
R 12
(1
-
1 )
.
Coulomb's law
With R12 denoting the position vector of Q2 relative to Qi, R = IR12I is the distance
between the two charges, R12 = R12/R is the unit vector 2 of the vector R12, and £0
is
the permittivity of a
vacuum
(free space),
£0
=
(p
10“ 12 and F
=
pF/m
8.8542
farad, the unit for capacitance,
is
(1
which
be studied
will
2)
.
permittivity of a
vacuum
in the
By point charges we mean charged bodies of arbitrary shapes whose
dimensions are much smaller than the distance between them. The SI (International
System of Units 3 ) unit for charge is the coulomb (abbreviated C), named in honor
next chapter).
of Charles Coulomb. This
which
is
is
a very large unit of charge.
c
in
magnitude
((^electron
QiQ 2 /( 47teor2 )
sion
The charge
of an electron,
negative, turns out to be
1.602 x
1
(T 19
C
(1.3)
Eq.
(
1 1)
.
Fe i2
represents the algebraic intensity (can be of
with respect to the unit vector R12.
sign or polarity (like charges), this intensity
Q2
are of the
the
same orientation
as R12,
and the force between charges
the electric force between unlike charges
(Q1Q2 < 0 )
is
is
is
If
positive,
Q\ and
Fe i2 has
repulsive. Conversely,
attractive.
and noting that R21 = — R12, we
obtain that F e 2i
e i2; he., the force on Q\ due to Q2 is equal in magnitude
and opposite in direction to the force on Q2 due to Q\. This result is essentially
an expression of Newton’s third law - to every action (force) in nature, there is an
By
reversing the indices
1
and 2
in
Eq.
(
1 1)
.
=—
opposed equal
If
sition,
charge of electron, magnitude
= — e). The unit for force (F) is the newton (N). The expres-
in
arbitrary sign) of the vector
same
=
Figured
Notation
Coulomb's
law, given
in
by
Eq- (1-1).
reaction.
we have more than two
which also
particular charge
is
point charges,
we can use
the principle of superpo-
on a
by each
a result of experiments, to determine the resultant force
- by adding up vectorially the
partial forces exerted
on
it
of the remaining charges individually.
is carried out component by component
an arbitrary number of vectors), most frequently in the Cartesian coordinate
system. Cartesian (or rectangular) coordinates, x, y, and z, and coordinate unit vectors, x, y, and z (unit vectors along the x, y, and z directions), are shown in Fig. 1 2
The unit vectors are mutually perpendicular, and an arbitrary vector a in Cartesian
coordinates can be represented as
In the general case, vector addition
(for
.
a
Tn
typewritten work, vectors are
= ax x + a y y + a z z.
e.g.,
F,
whereas
in
handwritten work, they are denoted by placing a right-handed arrow over the symbol, as F.
2
All unit vectors in this text will be represented using the “hat” notation, so the unit vector in the x direc-
tion (in the rectangular coordinate system), for example,
widely used notations for unit vectors would represent
3
SI
is
the modernized version of the metric system.
International d’Unites.
is
this
given as x (note that
vector as a v
The abbreviation
,
is
i*,
and
some
y, z)
unit vectors
.
(1.4)
commonly represented by boldface symbols,
Figure 1.2 Point M(x,
and coordinate
of the alternative
u*, respectively).
from the French name Systeme
in
the Cartesian coordinate
system.
Cartesian vector components
6
.
4
Chapter
1
Space
Electrostatic Field in Free
Here, a x ay and a z are the components of vector a
,
,
system, and
its
magnitude
a
The
is
unit vector of a
unity,
=
|a|
|a|
/a
a
Shown in Fig.
1.2
is
is
=
a
1.
Cartesian coordinate
in the
is
=
|a|
(1.5)
=
a /a. Of course, the magnitude of a, and
The sum of two vectors is given by
+b=
(a x
+ bx
)
x
+
+ by )
(a y
+
y
(a z
+ bz
)
of any unit vector,
z.
also the position vector r of an arbitrary point
(1 .6)
M(x,
y, z ) in space,
with respect to the coordinate origin (O),
position vector of a point
r
where, using Eq.
points
O
(1.5), r
=
= xx + yy + zi,
=
|r|
yjx2
(1.7)
OM
+ y2 + z 2 —
is
the distance 4 between
and M.
Example
Three Equal Point Charges at Triangle Vertices
1.1
Three small charged bodies of charge Q are placed at three vertices of an equilateral triangle
with sides a in air. The bodies can be considered as point charges. Find the direction and
magnitude of the electric force on each of the charges.
,
Solution Even with no computation whatsoever, we can conclude from the symmetry of
problem that the resultant forces on the charges, Fe i, Fe 2 and Fe 3 all have the same
magnitude and are positioned in the plane of the triangle as indicated in Fig. 1.3(a). Let
us compute the resultant force on the lower right charge - charge 3. Using the principle of
superposition, this force represents the vector sum of partial forces due to charges 1 and 2,
this
respectively, that
is [Fig.
1.3(b)],
Fe3
From Coulomb's
=Fet T F 23
3
law, Eq. (1.1),
F e23
Fe3
and both forces are
We
of
an equilateral triangle
and
(b)
(1-8)
Q2
= Fe 23 =
(1.9)
'
4neoa 2
repulsive.
note that the vector
2, i.e.,
,
computation of the
on
Fe 3 =
resultant electric force
one of the charges;
Example 1 .1
(vector superposition).
Fe3 is positioned along the symmetry line between charges 1
between vectors Fe i 3 and Fe 23 and it makes the angle a = n/ with both vectors.
The magnitude of the resultant vector is therefore twice the projection of any of the partial
vectors on the symmetry line, which yields
and
Figure 1.3 (a) Three equal
point charges at the vertices
e
magnitudes of the individual partial forces are given by
Fel3
(b)
,
,
2 (Fel3 cos a)
= 2 Fe 3 ^- = Fe uV?> =
(1
i
.
10 )
for
4
While dealing with
a
wide variety of vector quantities
(draw) them as arrows
tion.
However, we
shall
in space, like the force
always have
in
mind
vector
in
Fe
i
electromagnetics,
2 in Fig. 1.1,
we
shall regularly visualize
and computaand some other
to aid the analysis
that only position vectors, like r in Fig. 1.2,
length vectors to be introduced later have this feature of their magnitude being the actual geometrical
distance in space. Magnitudes of
sizes (lengths) of
arrows
in
nitudes of quantities of the
which
is
all
other vectors are measured in units different from meter, and the
space that they are associated with can only be indicative of relative mag-
same nature (with the same
unit), like
two forces acting on the same body,
equally useful and will be utilized extensively in this text as well.
.
Section
Q
2
Figure 1.4 Three point charges
equal
magnitude but with
in
different polarities at the
an equilateral
- computation of the
resultant electric force on
charge 3; for Example 1 .2.
vertices of
triangle
Example
Three Unequal Point Charges at Triangle Vertices
1.2
Determine the resultant force on the lower
Assume that Q and a are given quantities.
The only
Solution
Fe
i
3 is
now
p
is
,
=
tt/3. Its
1.3
Point charges
Q\
1,
i
3
Compute
= 2Fe
cos£)
Three Point Charges
=
1
Q 2 = —2
/xC,
/zC,
by Cartesian coordinates
defined
The
and the angle
magnitude is hence
Fe3 = 2(Fe
Example
shown
difference with respect to the configuration in Fig. 1.3
attractive, as indicated in Fig. 1.4.
the line connecting charges 2 and
and F e 23
right charge in the configuration
(1
in
i3
it
is
is
.4.
parallel to
the partial forces,
^£
= Fel3 =
-
1
that the force
resultant force on charge 3
makes with any of
in Fig.
Fc
]
3
(1.11)
Cartesian Coordinate System
and Q 3 =2 /xC are situated in free space at points
m, 0, 0), (0,1m, 0), and (0,0,1m), respectively.
the resultant electric force on charge Q\.
Solution
From Coulomb’s law and
Fig. 1.5(a), the
magnitudes of the individual forces on
the charge Q\ are
-= fl=iS =9mNi
01 2)
f
where
R
is
Q
the distance of
Fe3 t we decompose them
,
from 02
1
(
or
Q3
)-
In order to
Cartesian coordinate system. Based on Figs. 1.5(b) and
so the resultant force
Fe2 i
= — Fe2
F e3 i
=
cos a x
i
„
Fe3i COS
/l
X
+ Fe2
—
i
sin
Fe3] Sin
= Fe21 + Fe31 =
1
The Cartesian components
^eix
its
i
/3
(c),
a
a
y,
„
Z,
=
(1
.
13 )
7T
£=4’
(1.14)
is
FC
and
add together vectors F e 2 and
- into components in the
into convenient components, in this case
magnitude [Eq.
—
(1.5)]
^ei
0)
of the vector
Feiy — —Fe
comes out
=
74
x
+
^e21
V2 „
"yfy ~
Fe amount
i
\7
„
Z).
(1.15)
to
— Fe 2i~2~ ~
^-26
mN,
(1.16)
to be
4
y
+ Fjlz = Fe21 =
9
mN.
(1.17)
1
Coulomb's Law
5
Chapter
1
Electrostatic Field in Free
Space
Figure 1.5 Summation of
electric forces in
the Cartesian
coordinate system:
point charges
in
(a) three
space, with
partial force vectors
Fe 21
and F e 3 i, (b) component
decomposition of Fe 2 i,
(c) decomposition of F e 3 i,
and (d) alternative addition
of
forces using the head-to-tail
and the cosine formula;
Example 1 .3.
rule
for
Note
that
Fig. 1.5(d ),
5
in
Fe \ can alternatively be obtained using the head-to-tail
combination with the cosine formula 6 which yields
Pel
Note
is
=
portrayed
in
yj
F\2 \
also that the vector
+ ^e 3 ~ 2 /re 2 lFe3
Fe
i
i
is
i
cos y
= F2 =
\
9
mN,
y
parallel to the line connecting charges
=
Q3
(1.18)
J.
and
Q2
,
and that
it
positioned at an angle of n/4 with respect to the plane xy.
Four Charges at Tetrahedron Vertices
Example 1.4
Four point charges
Q
are positioned in free space at four vertices of a regular (equilateral)
tetrahedron with the side length
5
rule, as
,
By
Find the electric force on one of the charges.
a.
the head-to-tail rule for vector addition, to obtain graphically the vector
arrange the two vectors (usually translate b from
(second vector)
“connected"
we draw
is
placed at the head of a
to the tail of the second,
(first
its
sum c = a + b, we
way that the tail
original position) in such a
vector). In other words, the
and hence the term “head-to-tail”
head of the
first
first
of b
vector
for this arrangement.
is
Then
extending from the tail of a to the head of b, as in Fig. 1.5(d).
add two vectors together is the parallelogram rule, where c = a + b
corresponds to a diagonal of the parallelogram formed by a and b, which can also be seen in Fig. 1.5(d).
To add more than two vectors, e.g., d = a + b + c, we simply apply the head-to-tail rule to add c to the
already found a + b, and so on - the resultant vector extends from the tail of the first vector to the head
of the last vector in the multiple head-to-tail chain, and a polygon is thus obtained, which is why this
procedure is often referred to as the polygon rule.
An
ft
In
c (resultant vector) as a vector
equivalent graphical
method
to
an arbitrary triangle of side lengths
a,
/;,
and c and angles a
,
fi ,
and
y, the
square of the length c of the
2
2
side opposite to the angle y equals c
a + b 2 — lab cos y (and analogously for a and b
and cos /l, respectively), and this is known as the cosine formula (rule) or law of cosines.
=
2
2
using cos a
Section
Note
Solution
1
7
Definition of the Electric Field Intensity Vector
.2
that this configuration actually represents a spatial version of the planar
configuration of Fig.
1.3.
Referring to Fig.
of the tetrahedron - charge
4.
This force
Fe4
=
1.6, let
us find the force on the charge on the top
given by
is
Fel4
+ Fe24 + Fe34,
(1-19)
same magnitude, equal to Fe 14 = Q / (4n eqo 2 )
The horizontal components of the force vectors all lie in one plane and the angle between
where
all
each two
2
the three partial forces are of the
add up to
120°, so that they vectorially
is
.
zero. Thus, the resultant vector
component only, whose magnitude amounts
component of each partial force,
Fe 4
To determine
coscr (as
=
3 (Fe i 4 COS a)
Fe 4
has
to three times that of the vertical
a vertical
(1-20)
.
H/a) from the right-angled triangle A014 in Fig. 1.6, we first find
1 and point O) from the equilateral triangle A 123 (the base
Figure 1.6 Four point
charges at tetrahedron
vertices; for
Example
definition of
E
electric field
due
1
.4.
the distance b (between charge
of the tetrahedron), as 2/3 of the height of this triangle
2
in
a
Eq. (1.20) results
1. 1-1.7;
b2
(1.21)
a
in
Fe 4 = 3Fe i4
:
we have
V a2 —
COSO!
3
Problems
so
//
b
which substituted
7
,
a/6
V6Q 2
T"
4nsoa 2
(1
Conceptual Questions (on Companion Website):
and
1.1
.22)
1.2;
MATLAB Exercises (on Companion Website).
DEFINITION OF THE ELECTRIC FIELD INTENSITY
1.2
VECTOR
The
is a special physical state existing in a space around charged
fundamental property is that there is a force (Coulomb force) acting
on any stationary charge placed in the space. To quantitatively describe this field,
electric field
objects. Its
we
introduce a vector quantity called the electric field intensity vector, E.
nition,
it is
equal to the electric force
the electric field, divided by
Qp
,
that
e=
The probe charge has
Fe on
a probe (test) point charge
By
defi-
Q p placed in
is,
E
(Gp
-
(1.23)
o).
(unit:
V/m)
be small enough in magnitude in order to practically not
which are the sources of E. The unit for the electric
field intensity we use is volt per meter (V/m).
From the definition in Eq. (1.23) and Coulomb’s law, Eq. (1.1), we obtain the
expression for the electric field intensity vector of a point charge Q at a distance R
from the charge (Fig. 1.7)
to
affect the distribution of charges
Q
E=
4tt 8q
7
R2
R,
Note that the orthocenter (point O in Fig. 1.6) of an equilateral
1, so into segments 2/z/3 and /i/3 long. Note also that h
ratio 2
:
being the side length of the triangle.
(1.24)
charge
triangle partitions
=
C3a/2
(in
its
heights ( h ) in the
an equilateral triangle), a
in free
to
a point
space
—
8
Chapter
Electrostatic Field in Free
1
R
where
Space
is
the unit vector along
R
directed from the center of the charge (source
point) toward the point at which the field
is
(to be)
determined
observation
(field or
point).
By superposition, the electric field
Qn) at a point that is at
(Q i, Q2,
produced by N point charges
R2
Rn, respectively, from
intensity vector
distances R\.
the charges can be obtained as
Figure 1.7 Electric
intensity vector
point charge
field
due to
in free
N
a
e = e + e2 +
1
space.
where
R
/, /
Problems
1 .3
:
=
1,2, ...
1.8;
...
+ e^ =
-
1
O
0-25)
R<’
4ne 0 *ri Rf
,N, are the corresponding unit vectors.
MATLAB
Exercises (on
Companion
Website).
CONTINUOUS CHARGE DISTRIBUTIONS
A point charge
the simplest case of a charge distribution, which, mathematically,
is
corresponds to the space (three-dimensional) delta function. In the general case,
however, charge can be distributed throughout a volume, on a surface, or along a
line. Each of these three characteristic continuous charge distributions is described
by a suitable charge density function. The volume charge density
defined as
volume charge density
(in a
volume
v)
is
[Fig. 1.8(a)]
(1.26)
(unit:
C/m 3 )
the surface charge density (on a surface S)
is
given by
[Fig. 1.8(b)]
surface charge density (unit:
(1
.27)
C/m 2 )
and the
line
charge density
charge density (along a line
line
/) is
[Fig. 1.8(c)]
(1.28)
(unit:
C/m)
symbol pv is sometimes used instead of p, a instead of ps and p\
In the above equations, dQ represents the elemental charge in a
volume element dv, on a surface element dS, and along a line element dl, respec3
C/m 2 and C/m.
tively, and the corresponding units for charge densities are C/m
Note
that the
instead of Q'
,
.
,
,
charge Q for the three characteristic charge distributions
obtained as / d Q (adding charge elements d<2), which leads to
The
total
Q
=
'm v
P dv,
Qon
JV
d2
(
dv
Figure 1.8 Three characteristic
,
/
continuous charge distributions
and charge elements:
(a) volume charge, (b) surface
charge, and (c) line charge.
v
(a)
/
\
J
S
and
C?along
/
—
JQ
d/.
in Fig. 1.8 is
(1.29)
r
,
Section
1
.4
On
the
Volume and Surface
Integration
9
respectively. Special, but important, cases of continuous charge distributions are
uniform volume, surface, and line charge distributions. A charge distribution is said
to be uniform if the associated charge density is constant over the entire region (v, S,
or /) with charges. The expressions in Eqs. (1.29) then become much simpler,
Qinv
= pv
(/0
= const),
Qalong
Note
that Q' (Q'
length
(p.u.l.)
=
const)
is
/
=
= PsS (p = const),
Q = const).
QonS
s
'
Q'l
(
(1
also used to represent the so-called charge per unit
of a long uniformly charged structure
thin or thick cylinder),
(e.g.,
defined as the charge on one meter (unit of length) of the structure divided by
Q =
„
Glp.u.i.
y
—
Qalong
=
Qforlm
;
/
1
l
1.5
A volume charge
Nonuniform Volume Charge Distribution
is
where
total
r
=
Po
-
(0
<
r
<
m,
(|-31)
charge per unit length,
in
in a
distributed in free space inside a sphere of radius
P(r)
1
length
m
and hence Q' numerically equals the charge on each meter of the
Example
.30)
C/m
structure.
Sphere
a.
The charge density
is
(1.32)
a).
stands for the radial distance from the sphere center, and po
is
a constant. Find the
charge of the sphere.
we need to integrate in the first expreswe adopt dv in the form of a thin
shown in Fig. 1.9. The volume of the shell is
Solution Since the charge density depends on
r only,
sion in Eqs. (1.29) only with respect to that coordinate, and
spherical shell of radius r and thickness dr, as
dv
= 4nr1 dr,
0-33)
which can be visualized as the volume of a thin flat slab of the same thickness ( dr) and the
same surface area (S = 4nr2 ), dv = S dr. The total charge of the sphere comes out to be
Q=
[ p dv
Jv
=
[
Jr=0
po
- 4n
2
dr
=
pqtt a
3
.
0-34)
a
Figure 1.9 Integration of
nonuniform volume charge
density
in
Example
Problems
1
.4
:
1.9-1.12.
ON THE VOLUME AND SURFACE INTEGRATION
A few additional comments about the volume integration performed in Eq.
(1.34),
and similar multiple integrations, may be useful at this point. In general, our strategy for solving volume integrals, fv f dv, is to adopt as large volume elements dv as
possible, the only restriction being the condition that / = const in dv. For instance,
for the function p = p(r) in Eq. (1.32) and a sphere as the domain of integration,
the largest volume element over which p = const is a thin spherical shell, with dv
given in Eq. (1.33). This adoption enables us to perform the volume integration in
Eq. (1.34) by integrating along r only, from 0 to a, whereas the adoption of the standard differential volume element (elementary curvilinear cuboid) in the spherical
coordinate system (Fig. 1.10),
dv'
=
dr(rd6) (rsin0d0)
=
2
r
sin 6 dr
d6 d0
(1.35)
1
a sphere; for
.5.
10
Chapter
1
Electrostatic Field in Free
Space
z
r sin0
Figure 1.10 Standard
differential
(
volume element
dv') in the spherical
x
coordinate system.
would require two additional integrations, i.e., the integration in 6 from 0 to n and
the integration in <p from 0 to 2 n. Note, however, that these two integrations are
implicitly contained in the expression for
dv
in
Eq. (1.33), as
(1.36)
Note
also that the elementary cuboid dv'
p(r 9
,
,
(p ).
cylindrical
A
dv
the
for a charge den-
all
and Cartesian coordinate systems.
example for our integration strategy
trivial
in cases
[e.g.,
would be necessary
three coordinates in the spherical coordinate system, p =
Similar considerations apply to volume integrations associated with the
depending on
sity
when /
first
=
is
the adoption of v in place of
const in the entire integration domain
v,
yielding
fv f dv — fv
expression in Eqs. (1.30)].
The same
principle
is
adopting surface elements dS for solving
utilized for
surface integrals.
1 .5
ELECTRIC FIELD INTENSITY VECTOR DUE TO GIVEN
CHARGE DISTRIBUTIONS
Using the superposition principle, the
electric field intensity vector
due to each of
the (uniform or nonuniform) charge distributions p, p s and Q' can be regarded
as the vector summation of the field intensities contributed by the numerous
,
equivalent point charges making up the charge distribution. Thus, by replacing
in
electric field
Eq. (1.24) with charge element d Q
=
Q
p dv, ps d S, or Q' d / and integrating, we get
due
to
volume
(1.37)
due
to surface
(1.38)
due
to line
charge
electric field
charge
electric field
charge
(1.39)
Section
Note
R
that, in the general case,
R
and
1
Electric Field Intensity
.5
Vector due to Given Charge Distributions
vary as the integrals in Eqs. (1.37)-(1.39)
We shall now apply these
are evaluated, along with the functions p, ps and
expressions to some specific charge distributions and field points, for which the inte-
Q
,
grals
'
.
can be evaluated analytically. It is very important that we develop analytical
solving these (and similar), true three-dimensional (3-D), vector problems.
skills for
There is not a unique recipe for an optimal solution algorithm. A general advice,
however, is to use superposition whenever possible - to break up a complex problem into simpler ones, and then add up (integrate) their solutions to get the solution
to the original problem. In doing this, sometimes we use directly the expressions in
Eqs. (1.37)-(1.39), which essentially imply breaking up the structure into equivalent
point charges. Often, on the other hand, we do not go directly all the way down to
point charges. Instead, we decompose the structure and apply superposition “layer
by layer,” modularly, going down toward simpler structures level by level. In this,
we always try to incorporate already known solutions to relevant simpler problems
into the solution to the problem under consideration.
Example
1.6
Charged Ring
A line charge of uniform charge density Q
of radius a in
its
air.
Find the electric
is
distributed around the circumference of a ring
vector along the axis of the ring normal to
field intensity
plane.
We subdivide the ring (contour C) into elemental segments of length
Solution
and apply Eq. (1.39). The contribution to the
(z-axis) by each charged segment is
in Fig. 1.11,
axis
dE =
with the position of
P along the
4neoR z
R=
R,
axis being defined
total electric field intensity vector
is
\Jz
electric field at a point
2
+ a2
d /, as shown
P on
the ring
(1
,
by the coordinate
.40)
z(-oo<z<oo). The
obtained as
(1.41)
Due
to symmetry, the horizontal (radial)
element d / there
is
opposite horizontal electric
and hence
component of the vector E
is
zero (for every
a corresponding diametrically opposite element that gives
E has a vertical
an equal but
field component, so that the two contributions cancel each other),
(axial)
component only
E = EZ
Substituting the expression for
Ez
d E7
dE from Eq.
(Fig. 1.11),
= dE cos a = dE
(1.40) into Eq. (1.42),
d1
2eoR 3
47reoR 3 Jc
axis of a
along the
its
plane; for
.6.
(1.42)
R
we
field
charged ring
normal to
Example 1
= = 2na
1
Figure 1.11 Evaluation of
the electric
get
(1
.43)
(1
.44)
or
47T£o (z
2
+ a2
where Q = Q'2na is the total charge of the ring.
2
2
2
Note that for |z|
a, z + a
and Eq.
z
,
E«
(\z\
» a).
electric field
of charge
(1.44) yields
Q
4neoZ 2
3/2
)
(1.45)
due
to
a ring
11
a
12
Chapter
1
Electrostatic Field in Free
.
Space
This means that far away from the ring,
at its center. In
other words,
when
its
charge
is
equivalent to a point charge
the distance of the field point from the ring
is
Q
located
much
larger
than the ring dimensions, the ring can be considered as a point charge and the actual shape of
the ring (or any other charged body) does not matter. This in fact
is
the definition of a point
charge or a small charged body.
Example
A
line
Semicircular Line Charge
1.7
charge in the form of a semicircle of radius a is situated in free space, as shown in
The line charge density is Q' Compute the electric field intensity vector at an
Fig. 1.12(a).
arbitrary point along the z-axis.
Solution The electric field intensity vector along the z-axis due to an elemental charge
on the semicircle [Fig. 1.12(a)] is given by the expression in Eq. (1.40). To be able to
perform the integration in Eq. (1.39), we need to decompose this vector into convenient
components. With reference to Fig. 1.12(b), we first represent dE by its horizontal and
vertical components,
<2'd/
dE = dEh + d£ z z,
d Ez
=
=
d£h
d£sina,
sin
dEh
the semicircle),
=
dE* x
we need
dEcosa,
=
cos a
+ dEy y,
dE*
(the range for the azimuthal angle
the relationship d/
= ad0
0
decompose
to
is
(segment d /
—n/2 < 0 <
d Ex
-
Figure 1.12 Evaluation of the
electric field
charge
in
due
to a line
the form of a
semicircle; for
Example
1.7.
=
E3
Q'az_
1
47T£ 0
(i.e.,
as the point P'
= - dEh sin 0
d£y
7r/2).
From
(1-47)
the above expressions and
cos 0
d0
=-
a
Q"-3
2
ItteqR 5
,
(1.48)
sin0d0
=
0.
(1.49)
J
J-jt/2
—IT/
r
n/2
’
E J- r/2
3
its
rx/2
2
/
47T£ 0
evaluated
further [Fig. 1.12(c)],
x/2
Q’a 2
O'
is
.46)
we obtain
AtTSqR} /_ x/2
dEy
-Jf
=-
(1
R
an arc of radius a defined by the angle d0, and
length thus equals the radius times the angle),
/
it
= - dEh cos0,
is
=
R'
Since the direction of the vector dEh varies as the integral
moves along
a
y
4£ 0 E 3
’
(1.50)
s
)
Section
Electric Field Intensity
.5
1
so that the final expression for the electric field vector
E=
Example
2 0 R3 V
l
is
R = Vz 2 +a 2
.
E
an arbitrary point
field at
in
space due to a straight line of
medium
uniformly charged with a total charge Q. The ambient
Solution
(1.51)
.
/
Straight Line Charge of Finite Length
1.8
Find the expression for the
length
|
2
tr
IB
Vector due to Given Charge Distributions
Let the line charge and the
is air.
point (P) be in the plane of drawing (Fig. 1.13).
field
The geometry of this problem can be defined using three parameters: angles 9\ and 62 and the
perpendicular distance from the line to the point, d (for the particular position of the point P
,
shown
in Fig. 1.13, 6\
<
>
0 and 62
=
In Eq. (1.39), Q'
0).
R—
Q/l,
s/z
2
+ d2
,
dl
=
R = cos0 x — sin# z,
where
tively,
z (zi
<
z
<
Zz)
and 6
<
( 0\
9
<
dz,
and
(1-52)
62 ) are the length
each determining the position of the source point,
and angular coordinates, respecP',
along the
line.
From
Fig. 1.13,
the following trigonometric relationship exists between these two coordinates:
z
=
tand
(1.53)
d'
and
its
differential
form (by taking the
dz
cos 2 9
d
d
(.
which, given that cos#
=
d/R,
=
const),
d6
dz
Finally, substituting the expressions
O
from Eqs.
—
(1.52)
( f 02
(
/
(1.55)
~d’
cos 9 d9 x
and
t
-
^
[(sin
(1.55) into Eq. (1.39),
sin 0
dO
z
J9l
6*2
-
sin
4neold
x
+
(cos 02
— cos0i) zl.
An
Infinite Line
1.9
infinite line
electric field
Charge
charge of uniform density
intensity vector at
(1.56)
an arbitrary point
electric field
due
to
a
line
charge of finite length
Note that the expression in Eq. (1.56) can be combined for computing the
due to an arbitrary structure assembled from straight line segments of charge.
Example
we have
®1
/
Unsold \J9l
E=
(1.54)
equivalent to
is
K2 =
E =
both sides of the equation) reads
differential of
d0
Q
resides in
Determine the
air.
electric field
in space.
Figure 1.13 Evaluation of the
electric field
charge of
Example
1
due to
a line
finite length; for
.8.
14
Chapter
1
Electrostatic Field in Free
Solution
Space
Let r be the perpendicular (radial) distance of a
(observation) point,
field
P,
from
the line charge, and r the unit vector along this distance directed from the line to the point
We use
and
d=
the expression for the field due to a finite line charge in Eq. (1.56) and
n/2 (the line extends to
and x = r. What we obtain
82 -»
z -*•
r,
is
with density
electric field
line
due
to
an
— oo and
let 9\
z -> oo, respectively), as well as
P.
-> —tc/2
Q/l
=
Q',
exactly the field expression for the infinite line charge
Q
Q'
E=
infinite
(1.57)
2neor
charge
We see that the vector E is radial with respect to the line charge axis, and its magnitude varies
in space as a function of the radial distance r only, with E being inversely proportional to r.
z.,
Disk of Charge
Example 1.10
.,dE
Consider a very thin charged disk (i.e., a circular sheet of charge), of radius a and a uniform
surface charge density p s in free space. Calculate the electric field intensity vector along the
disk axis normal to its surface.
,
Solution
Instead of directly applying Eq. (1.38),
of width dr, as
shown
along the z-axis
is
in Fig. 1.14.
The
field
due
R = yr2 + z 2
4neoR i
d<2
The surface area of the
By
=
P
=
R2 =
RdR =
so that the substitution of rdr by
PsZ
2 e0
disk
where
strip of length
2nrdr.
2eo
Taking the differential of the relationship
E=
(1.59)
equal
(1.60)
superposition,
Js
a charged
a) at a point
p s dS.
r
E =/dE=M S
to
<
(1.58)
can be computed as the area of a thin
and width dr, that is,
dS
due
r
,
ring, dS,
to the ring circumference, 2nr,
electric field
<
with d Q standing for the charge of the ring, given by
the electric field due to a
charged disk; for
Example 1 .10.
subdivide the disk into elemental rings
[see Eq. (1.44)]
dE=-^%z,
Figure 1.14 Evaluation of
we
to a ring of radius r (0
7?| r
Lf
_o
Example
=
0
dR
R2
yfz
1.11
Find the electric
2
i—
RdR
in
2
1
2e 0 \
r) r=0
+ z 2 we
,
obtain
(1.62)
r dr,
Eq. (1.61) yields
\
E
to allow a negative z as well
Infinite
(1.61)
R7
a
PsZ /
|z|,
r
Jr = o
*
J
z
(
2eo \|z|
(-oo <
z
<
yja 2
(1.63)
+ z2 /
oo).
Sheet of Charge
field intensity
vector due to an infinite sheet of charge of density p s in free
space.
Solution We note that the infinite sheet of charge can be obtained from the disk in Fig. 1.14
by extending its radius to infinity. Consequently, the field due to the infinite sheet can be
obtained from the field due to the disk by letting a —> oo in Eq. (1.63). With this, the second
term in parentheses in the final field expression in Eq. (1.63) vanishes. The first term, z/|z|,
o
Section
is
either
1
(for z
>
Electric Field Intensity
.5
15
Vector due to Given Charge Distributions
-1 (for z < 0), and we conclude that the field intensity around the
uniform (constant) in both half-spaces cut by the sheet, and given by
0) or
charge
infinite sheet of
1
is
(1.64)
due
electric field
to
an
infinite
sheet of charge
with respect to the reference directions indicated in Fig. 1.15. Namely, for a positive charge
(p s
>
0) of the sheet, the actual orientations of
E are those in Fig.
1.15 (field lines are directed
outward from the positive charge), while the situation for a negative ps
field vector points toward the negative charge).
Example 1.12
A
is air.
E
just opposite (the
Hemispherical Surface Charge
hemispherical surface of radius a
medium
is
Compute
uniformly charged with a charge density p s
is
.
The
the electric field intensity vector at the center of the hemisphere
(center of the corresponding
full
sphere).
Figure 1.15
Solution
We perform a similar procedure as in Example
The radius of a
into thin rings, as depicted in Fig. 1.16.
is
defined by an angle 9 (0
<
9
< n/2)
dQ
=
is
ax
ps dS
=
a sin
ring
9. Its
and subdivide the hemisphere
whose position on the hemisphere
1.10
charge
sheet of
is
= p 2nasm6 ad9,
(1
s
C
Infinite
charge; for Example 1.11.
.65)
d/ r
r
C
r and d/r denote the ring circumference and width, respectively. Using Eq. (1.44),
the electric field intensity dE at the point O due to the charge d Q is found as follows:
where
Zr
= — a cos 9„
Rn =
andj
—
a
Jl7 =
dE
>
—
d<2(— flCOS#)
r
n
cc\
(1.66)
„
z,
dE
with z T being the local z-coordinate of the point
O with respect to the ring center (Fig.
1.16).
Figure 1.16 Evaluation of
Therefore, the resultant field vector amounts to
the electric
/•tt/2
sin 9 cos 9
Example 1.13
is
its
Solution
From
that
is,
unit
is
(1.67)
z.
hemispherical surface
charge; for Example
s
is
distributed in free space over a surface in the
a,
as
shown
positioned along the semicylinder
in Fig. 1.17(a).
1 .1
2.
form of a half
A line charge of uniform
Find the electric force on the
axis.
line
charge
length.
Eqs. (1.23) and (1.31), the per-unit-length electric force on the line charge,
on one meter (unit length) of the structure (line charge) divided by 1 m (the
given by
the force
N/m),
is
(1.68)
per-unit-length
on a
with
E
axis.
We
standing for the electric field intensity vector due to the charged semicylinder at
subdivide the semicylinder into elemental
superposition principle.
The
field
strips,
of width d /, and find
E
its
using the
due to each
long line charge with charge density dQ'
=
strip can be approximated by that of a very
ps d /; namely, the charge of the elemental strip
of length h, p s h d /, must be equal to the charge of the equivalent line charge of the same
length, so d Q'h, which yields this expression for dQ'. Having in mind Eq. (1.57), the electric
field
the
Force on a Line Charge due to a Charged Semicylinder
A uniform charge of density p
per unit of
=
(sin20)/2
of a very long circular cylinder of radius
density Q'
d9 z
4 £Q
Je =
field at
center (point O) of a
Ps
vector due to the line charge
semicylinder axis
[Fig. 1.17(b)]
is
radial with respect to the line,
comes out
dE
=
to
and
its
magnitude
at the
be
dQ!
2neoa
_
ps d
<f>
2neo
(1.69)
’
line
Coulomb
charge
(unit:
force
N/m)
E
16
Chapter!
Electrostatic Field in Free
=
since r
Space
a and
computed
=
dl
Ex = f d
Jl
where
denotes the
/
By symmetry,
ad<p.
the resultant field has an .^-component only,
as
x
= f d£cos0 =
Ji
costf>d <p
=
27r£o J (p = — 7r /2
—
(1.70)
,
rr £q
representing the semicylinder cross section. Finally,
line (semicircle)
Eqs. (1.68) and (1.70) result in
w
F„ =
Q'Ps
-
(1.71)
x.
7T£ 0
Companion Website):
Problems'. 1.13-1.28; Conceptual Questions (on
MATLAB
DEFINITION OF THE ELECTRIC SCALAR POTENTIAL
.6
1
The
1.3-1. 6;
Companion Website).
Exercises (on
electric scalar potential
tric field intensity
the potential
is
a scalar quantity that can be used instead of the elec-
Dealing with
Three difgeneral for the evaluation of E, one integration
vector for the description of the electrostatic
mathematically simpler than dealing with the
is
field.
field vector.
needed in
component, while a single integration is required for the potential,
can easily be found from the potential by differentiation. In addition,
ferent integrations are
for each vector
d/
and E,
in turn,
using the electric potential
(b)
The
Figure 1.17 Evaluation of
the electric force on a line
charge positioned along
the axis of a charged
semicylinder: (a) three-
we
are able to connect the electric field with the voltage,
fundamental bridge between the
as a
electric scalar potential
field, that
is,
by the
electric force,
dW
done by F e while moving
8
the dot product of F e and dl,
Fe
Qp
e
in
,
moving
a test point charge,
Qp
.
The work
along an elementary vector distance dl equals
dW =
dimensional view showing
theory and the circuit theory.
field
defined through the work done by the electric
is
Fe
e
dl
= Fe d/ coso?,
(1.72)
the structure geometry and
(b) cross-sectional
field
for
view for
computations;
Example
1 .1
where a
We
3.
d
|
is
the angle between the two vectors in the product, as
dW
note that
W
e
is
|
on Eq.
shown
in Fig. 1.18.
can be negative (for n/ 2 < a < n), meaning that the work
e
being performed against the electric
(1.72), the electric scalar potential,
defined as the work done by the field
in
V
field
at a
,
moving
by an external agent. Based
P in an electric field is
point
a test charge
from P to a reference
point 7Z (Fig. 1.18),
W
=
e
divided by
Qp
.
Having
mind Eq.
V=
definition of the electric
potential (unit:
in
W—
—
c
=
Qp
V)
Thus, the potential
to not
8
|
depend on
V
Qp
(1.23), this
r
-
becomes
— -dl=
r
/
n
E
•
dl.
ip
Qp
P with respect
and to be equal to the
(1.73)
dl,
n Fe
/
ip
at the point
Fe
to the reference point 7Z turns out
line integral of vector
The dot product (also known as the scalar product) of
a being the angle between a and b.
vectors a and b
is
E from P
to 1Z?
b
=
to a point B,
is
a scalar given by a
•
a b cos a,
y
I
j
|
The
line integral of a vector function (field) a
defined as fpa
dl
=
a
dl.
where
dl
is
along a line (curve)
/.
from a point
A
the differential length vector tangential to the curve (as in
Section
Obviously, the potential at the reference point
is
Definition of the Electric Scalar Potential
.6
1
zero (integral from
P
=
17
TZ to TV).
and hence the unit V/m for the
electric field intensity. Note that <f> is also used to denote the electric potential.
By the principle of conservation of energy, the net work done by the electrostatic field in moving Q from a point A to some point B and then moving it back to
p
A along a different path is zero (because after the round trip, the system is the same
The
unit for the potential
is
volt (abbreviated V),
means
as at the beginning). This
that the line integral of the electric field intensity
vector along an arbitrary closed path (contour)
is
zero,
Figure 1.18 Displacement
of a test charge in an
electrostatic field.
£
which constitutes Maxwell’s
first
E
•
dl
=
(1.75)
0,
electrostatic field
equation for the electrostatic
circulation (closed line integral) of
E
conservative nature of the
in electrostatics is
field.
We see that the
always zero, and hence the
electrostatic field belongs to a class of conservative vector fields.
10
Eq. (1.75) can alternatively be derived from Coulomb’s law,
i.e.,
from the
expression for the electric field intensity vector due to a single point charge, Q,
given by Eq. (1.24) and Fig. 1.7. To do this, we break the contour C up into elemental segments ( dl -* 0) parallel and normal to E, as indicated in Fig. 1.19. We realize
that contributions to the overall line integral in Eq. (1.75) occur only for the seg90°
ments parallel to E, while no contribution for the segments normal to E (a
=
and
E
•
dl
= 0).
In specific, along the segments parallel to
Q E
respect to the charge
),
•
dl equals either
Edl
resulting in a zero net line integral (sum) along C.
(a
By
E
(segments radial with
— 0)
or
—Edl
(a
=
180°),
superposition, this result
is
90°
Figure 1.19 Derivation of
Maxwell's
first
equation for the
electrostatic field starting with
Coulomb's
Fig. 1.18)
oriented from
A
contour (and usually mark
of a along C.
10
By
The reference
toward B.
it
If
the line
is
closed (for example, a circle or a square),
and the corresponding
line integral,
§c a
•
dl, is
we
call
it
termed the circulation
direction of dl coincides with the orientation of the contour.
definition, a vector field
arbitrary shape.
C),
is
said to be conservative
when
its
circulation
is
zero for a closed path of
law.
18
Chapter
Electrostatic Field in Free
1
Space
also valid for any charge distribution (which, as we know, can be represented as a
system of equivalent point charges, and for which the resultant field can be obtained
as a vector
sum
of individual fields due to point charges).
The line integral of E between two points in an electrostatic
depend on the path of integration. To prove this, let us refer to Fig.
the circulation of
E
<£
E
C
•
dl
electrostatic field vector
between two points is the
same for any path of
path independence
line integral of
E
•
dl
E
-/
J AnB
the
is
same
does not
and write
•
dl
+
E
f
dl
•
vBnA
0-76)
dl,
A to B along the path containing the point
so that, from Eq. (1.75), the integral from
m
integration.
dl= f
E
J AmB
= f
JA
AmBnA
-L
1.20
C as
along a contour
Jc
Figure 1.20 For the proof
that the line integral of an
field
as the integral along the path with the point n,
for the
MmB
E
w
II
E
f
•
dl,
(1.77)
7 AnB
means that, for an adopted reference point, the
any point in the field is a uniquely determined quantity, having
the same value for any path of moving the test charge.
Finally, let us see what happens with the potential at an arbitrary point P in the
or along any other path. This also
electric potential at
field after a
new
reference point,
11', is
pTZ
plZ'
V=
E
/
new
adopted. The
•
dl
=
is
rlZ'
E
/
•
dl
Jp
Jp
potential
+
E
/
dl,
(1.78)
Jiz
'
'
v
V
and thus the change
change
new
in
in potential
amounts
AV =
potential due to a
to
V-V =
(1.79)
reference point
very important to note that
It is
in the field; if the
the
same constant
AV does not depend on the position of the point P
is changed, the potential
by an additive constant.
reference point
value,
i.e.,
at all points
changes by
Problems 1.29 and 1.30; Conceptual Questions (on Companion Website):
and 1.8; MATLAB Exercises (on Companion Website).
:
1 .7
1.7
ELECTRIC POTENTIAL DUE TO GIVEN CHARGE
DISTRIBUTIONS
Let us determine the electric scalar potential due to a point charge
general,
it is
customary to adopt the reference point,
when
1Z, at infinity
As we
in free space. In
whenever
possi-
an example, this
choice is impossible, or at least inadequate, for infinite charge distributions, such as
an infinite straight line charge. From Fig. 1.21 and Eqs. (1.74) and (1.24), the potential at a distance R from the charge Q, and with respect to the reference point at
ble,
namely,
the charge distribution
is finite.
shall see in
infinity, is
a point charge
due
to
v=
E
JP
poo
rOO
plZ
electric potential
•
dl
=
/
Jx=R
Edx=
I
JR
Q
r
47r£ 0 * z
dr
V=
(1.80)
4n£()R
.
Section
R
Q
E
P
1
Electric Potential
.7
x
as
charge
potential function (of
(finite)
Shown
same
in free
due to
a point
space.
obtained for a reference point at an arbitrary
distance from the charge differs from this result by an additive constant,
determined by Eq.
vector
R)
19
Distributions
Figure 1.21 Evaluation of the
electric potential
The
due to Given Charge
E
is
(1.79).
in Fig. 1.22
is
tangential at
potential,
V=
a sketch of electric-field lines (in general, lines to which
all
points)
const, at
all
and equipotential surfaces (surfaces having the
Of
points) around the point charge in Fig. 1.21.
“beams” starting at the
charge and equipotentials are spherical surfaces (graphed as circles - equipotential
course, based
lines
-
on Eqs.
in the figure)
(1.24)
and
(1.80), field lines are radial
centered at the charge, respectively. Field lines are perpen-
and this conclusion holds always, for an arbitrary
Namely, since a movement of a probe charge Q p (in Fig. 1.18)
over an equipotential surface results in no work by electric forces (no change in V),
E does not have a component along that movement, meaning that it is perpendicdicular to equipotential surfaces,
electrostatic field.
ular to the surface.
The density
the equipotential surface)
it
is
of field lines (the
number
of lines per unit area of
representative of the magnitude of the field vector,
decreases as 1/R 2 (the area of equipotential surfaces increases as
i.e.,
R 2 ) away from
the charge in Fig. 1.22.
Starting with Eq. (1.80)
and applying the superposition principle, the expression
for the resultant electric potential, similar to the field expression in Eq. (1.25),
obtained for the system of
is
N point charges, which reads
N
(1.81)
as well as for the three characteristic continuous charge distributions, corresponding
to Eqs. (1.37)-(1.39) for the field vector,
v=
J_
47T£()
f
Jv
(1
R
.82)
electric potential
volume charge
V - const
Figure 1.22 Field and
equipotential lines for a point
charge
in Fig.
1
.21
due
to
20
Chapter
electric potential
1
due
Electrostatic Field in Free
Space
f Ps
1
1/-
to
electric potential
due
v—
to line
rQ'di
1
(1.84)
R
h
47re 0
charge
(1.83)
’
R
47T £ 0 Is
surface charge
Obviously, these integrals are substantially simpler than the respective
field inte-
and the same holds true for the resulting solutions for the potential due to
charge distributions on various characteristic geometries, for which we have already
grals,
evaluated the electric
Example 1.14
field
vector in Section
due to
Electric Potential
a
1.5.
Charged Ring
Find the electric scalar potential along the axis of a uniformly charged ring in free space
normal to the ring plane. The line charge density of the ring is Q' and its radius is a.
Solution
(for
— oo <
From Eq. (1.84) and
z < oo) is given by
Fig. 1.11, the potential at
J_I &
4nE 0 Jc
Example 1.15
Compute
Potential
d/
R
=
Q'
an arbitrary point P on the z-axis
due to an
Q' a
1 dl=
4ne0 RJc
(1.85)
2e 0 s/z 2
Infinite Line
+ a2
Charge
the electric potential at an arbitrary point in space due to an infinite uniform line
charge of density Q'
Solution
in air.
The expressions
that the reference point
is
for potential computations given in Eqs.
(1
.81 )— (
1
.84) all
evaluation of the potential due to
to an infinite charge distribution with respect to the reference point taken at infinity
nite). In
such cases,
we invoke Eq.
straight line charge, the field of
(1.74) instead
is r,
we apply Eq.
(1.74)
-
to
compute the
potential,
is
given by Eq. (1.57),
is
this charge, at a point
a typical,
whose
radial distance
R
Figure 1.23 Evaluation of the
due to an
charge - cross
electric potential
infinite line
section of the structure; for
Example 1.15.
and an
is infi-
infinite
and a very important,
from the
to a convenient integration path shown in Fig. 1.23, which
which
example. To find the potential due to
charge line
imply
These expressions cannot therefore be used for the
infinite charge distributions (because the potential due
at infinity.
Section 1.8
an arc between points P and
consists of
21
Voltage
M and a straight line segment between points M and
and obtain
1Z,
/>
.
E
=
dl
JP
The
rH
rM
72
V=
/
E dl+
/
•
=
dl
JM
JP
line integral
rrn
E
between points P and
/
dx.
A=r
M
is
qi
rr-jz
Edx=
zero because
E
is
(1.86)
2 ;t£o*
Jr
perpendicular to dl along the
from the line charge with respect to the reference
comes out to be
path. Thus, the potential at a distance r
point that
is
a distance rn
away from
it
rR
Q!
V=
Note
r
V—
oo.
>•
is
in
space
is
and
(1.79)]. In
an
infi-
given by Eq. (1.87)
r)
plus an infinite constant, as a consequence of the change of the reference point from
Fig. 1.23 to infinity [see Eqs. (1.78)
to
also “correct” in a sense
can be understood as the potential distribution (function of
it
due
charge
which implies that
Note, furthermore, that this result, so an
nite potential regardless of the location of the observation point,
that
electric potential
infinite line
that the adoption of the reference point at infinity in this case,
oo in Eq. (1.87), would result in
rji
(1.87)
In
2ti eq
other words, the potential at
all
7Z in
points
changed by the same constant value after the new reference point (at infinity) is
infinite (AC -> oo), masks the correct potential distribution. However,
adopted, which, being
this result,
although “correct,”
is
useless for the analysis, since the actual function C(r) can-
not be extracted from an infinite additive constant, and that
is
why we
say that in potential
evaluations due to infinite charge distributions reference point cannot be taken at
Problems: 1.31-1.36;
MATLAB Exercises (on Companion Website).
VOLTAGE
1.8
By
infinity.
between two points is the potential difference between
The voltage between points A and B is denoted by Tab, so
definition, the voltage
them. The unit
is
V.
we have
^AB =
where
V\ and Cb
are the potentials
Va~ Cb
at point A
(1.88)
,
same reference point.
Combining Eqs. (1.74) and (1.88), we
and point
definition of voltage (unit:
V)
B, respectively, with
respect to the
Vab —
/
E
•
dl
—
•
get
dl
=
JA
E
•
dl
+
rE
•
dl
(1.89)
or
Eab=
Edl.
(1
.90)
JA
Therefore, the voltage between two points in an electrostatic field equals, in turn,
the line integral of the electric field intensity vector along any path between these
points. Obviously,
new
Cba = -Cab-
adopted for the electric potential in a system, voltTo prove this statement, we recall that the
potential at all points in the system changes by the same constant value (AC) after
the new reference point is adopted [see Eq. (1.79)]. Since a voltage is the difference
of the potential values at the respective points in the system, Eq. (1.88), the new
voltage between points A and B is given by
If
a
reference point
is
ages in the system remain unchanged.
^ab =
i.e., it
- ^b - (^A + AC) - (CB + AC) = CA - Cb = Cab,
equals the old voltage between these points.
(1.91)
voltage via a line integral of
E
1
22
Chapter
Electrostatic Field in Free
1
Note
Kirchhoff's voltage
Space
that, in
terms of voltages, Eq. (1.75) can be written as
E
law
along
which,
if
(1.92)
applied to a closed path in a dc circuit, represents Kirchhoff’s circuital
law for voltages. This law
closed path in a circuit
is
v'=°C
is
tells
zero.
11
us that the algebraic
sum
of voltages around any
Kirchhoff’s voltage law for dc circuits, Eq. (1.92),
therefore just a special form of Maxwell’s
first
equation for the electrostatic
field,
Eq. (1.75). We shall see in a later chapter that with some restrictions and approximations, Kirchhoff’s voltage law in the form in Eq. (1.92) can also be used for the
analysis of ac circuits.
Problems:
1.37;
Conceptual Questions (on Companion Website):
1.9.
HISTORICAL ASIDE
The
Gustav Robert Kirchhoff
SI unit for the vol-
tage,
electric
and electromotive force
(to
be studied in a
who
is
Berlin, Breslau,
famous
for his invention of the
the electricity and voltage
made
—
the
first
voltmeter.
a series of electrochemical cells
as a pile of alternating silver
and zinc plates
separated by cardboard disks soaked in salty water
(electrolyte).
(Portrait:
Edgar Fahs Smith
Collection,
and Hei-
Kirchhoff
rec-
eived his doctoral degree from the University
of Konigsberg
under
experimental
of
ments, he used his tongue to sense and measure
was
chemist,
taught at Universities of
delberg.
electric battery (voltaic pile) in 1800. In experi-
Volta’s battery
and
was named in honor of
Alessandro Volta (17451827), an Italian physicist
and inventor, a prophysics at the University
of Pavia,
physicist
later chapter), volt (V),
fessor
German
(1824-1887), a
potential,
in
Neumann
1847,
(1798-
Extending
the
Ohm (17891854), he formulated, in 1850, the fundamental
relations between currents and voltages in an
electric circuit with multiple loops, which we call
Kirchhoff’s circuital laws for currents and voltages.
1895).
work
Kirchhoff also
of
made seminal contributions
troscopy and thermal emission.
(Portrait:
to spec-
Edgar Fahs
Smith Collection, University of Pennsylvania Libraries)
University of Pennsylvania Libraries)
1
.9
DIFFERENTIAL RELATIONSHIP BETWEEN THE FIELD
AND POTENTIAL IN ELECTROSTATICS
Eq. (1.74) represents an integral relationship between the electric field intensity vecand the potential in electrostatics, which enables us to determine V if we know
tor
E. In this section,
we
shall derive
these two quantities, and then use
1
Algebraic
the voltage
sum means
is
an equivalent, differential, relationship between
it
for evaluating
that the voltages in the
sum can be
E from
V.
of arbitrary sign, where the sign with which
taken depends on the agreement or disagreement of the orientation (polarity) of the
voltage with the orientation of the contour.
Section 1.10
A
Gradient
23
which the potential is EA Let
— ck along the x-axis, to a
point B, as shown in Fig. 1.24. The resulting change in the potential amounts to the
potential at B (new potential) minus the potential at A (starting potential), that is,
Consider a point
us
move from
in
an electrostatic
field at
.
that point for an elementary distance dl
dE=EB -EA
(1.93)
.
d/
= dx
Er
1
1
On the other hand, Eq. (1.90) tells us that the potential difference (voltage) between
points A and B is related to the electric field intensity vector as
La — Lb = E
•
dl
= E dl cos a = E cos a
dx
— Ex dr,
(1
.94)
between E and
V
in
electrostatics.
~~e7~
with
Figure 1.24 Derivation of
the differential relationship
Ex standing for the x-component of E, which equals the projection of E on the
E cos a. Combining Eqs. (1.93) and (1.94), we have
x-axis,
(1.95)
7
-D
differential relationship
between
Similarly, the projections of the vector
E on
the other two axes of the Cartesian
coordinate system are obtained as
Ey =
M
E =
7
dy
is
„
V 9x
dV
(1.97)
dy
an important general means
In an electrostatic system, the potential
Cartesian x-axis, while
it
ones because the potential
is
a
V=
L(x, y, z). This diffor computing the field E from
V2
Since
,
1.38;
constant in any individual plane normal to the
= V0 +
V^ + L
2
and d are constants. Find the
dE/dy
only, according to
Exercises (on
is
varies along that axis as
V(x)
1.10
— xH
1-D Problem
Example 1.16
Problems:
I
E in electrostatics.
the potential
Solution
given by
three coordinates (multivariable function),
ferential relationship
E
is
partial derivatives instead of ordinary
all
where Eq, Vi,
E
/ 3L
E = Ex x + Ey y + Ez z =
function of
(1.96)
'
dz
Hence, the complete vector expression for
where we use
— dE
=
0 and
dE/d z
=
0,
(1.98)
^,
electric field intensity vector in the system.
we
are left with an
x-component of the vector
Eq. (1.97), which comes out to be
dE _
Ei
2 E2 x
dx
d
cP
(1.99)
'
Conceptual Questions (on Companion Website):
1.10;
MATLAB
Companion Website).
GRADIENT
The expression
function (E).
in the parentheses in
It is
Eq. (1.97)
is
called the gradient of the scalar
sometimes written as grad E, but much more frequently we write
E and V
/
24
Chapter
Space
Electrostatic Field in Free
1
it
using the so-called del operator or nabla operator, defined as
V =
del operator
So,
E
from
V
—
8x
a
a
„
a
„
yH
J
x-l
dy
„
z.
(1
dz
.
100 )
we have
E = — grad V = — VF,
in electrostatics
where,
in the
Cartesian coordinate system (Fig.
grad
gradient in Cartesian
V = VV =
dv
—
A
+
x
dx
coordinates
(1.101)
1.2),
dv
—
„
y
dy
+
(1.102)
The other two best-known and most commonly used coordinate systems are the
and the spherical. An arbitrary point (M) in the cylindrical coordinate
system is represented as (r, 0, z), where r is the radial distance from the z-axis to
the point M, 0 the azimuthal angle measured from the x-axis in the xy-plane, and
z the same as in the Cartesian coordinate system, as shown in Fig. 1.25. The ranges
cylindrical
of the coordinates are
0
<
r
<
0
oo,
< 0 < 2n
(or
— 7r < 0 <
— oo <
it),
z
<
oo.
(1
.103)
r
z
The coordinate
unit vectors,
increasing
and
r,
0,
r,
0,
and
z, respectively.
are,
by
definition, directed in directions of
They are
all
mutually perpendicular (the cylin-
z,
system belongs to the class of orthogonal coordinate systems), and
cylindrical coordinates can be expressed as
drical coordinate
the vector
E
in
E = Er r +
X
Figure 1.25 Point M(r.
and coordinate
in
<p,
z)
unit vectors
the cylindrical coordinate
-f-
£ z z.
(1.104)
Since 0 is not a length coordinate but an angular one, an incremental distance d
corresponding to an elementary increment in the coordinate, d0, equals dl = rdrf)
and this exactly is the
computing the change in potential dV in Eqs. (1.93)—
(1.95) now in the 0 direction. Therefore, the 0-component of the electric field
vector at the point
in Fig. 1.25 equals E ^ = — dV/dl = — dV/(rdrp) and not just
— dF/d0. The other two cylindrical coordinates, r and z are length coordinates, so
no modification is needed, Er — — dV /dr and E z — — dV /d z. Consequently, having
in mind Eqs. (1.104) and (1.101), the gradient of V — F(r, 0,z) in the cylindrical
coordinate system is, in place of Eq. (1.102), given by
(the length of an arc of radius r defined by the angle d0),
displacement dl
system.
E<p§
in Fig. 1.24 in
M
,
gradient
grad
in cylindrical
F = VF =
dV
—
„
r
+
dr
coordinates
In the spherical
l
r
dv
90
M
coordinate system, a point
dv
4>
T
H
„
z.
(1.105)
dz
is
defined by
(r,
0,0), with r
being the radial distance from the coordinate origin (O) to M, 9 the zenith angle
between the
z-axis
and the position vector of M, and 0 the same as
coordinate system, as illustrated
in Fig. 1.26.
0</-<oo,
O<0<7r,
and the component representation of the vector
O<0< 2n,
E
r,
0,
in directions
and
4>
(1
are
.106)
reads
E = Er i + E0 Q + E^.
where
in the cylindrical
The ranges of spherical coordinates
(1.107)
are mutually perpendicular coordinate unit vectors (directed
of increasing
r,
9,
and 0, respectively). As depicted
in Fig. 1.10, the
Section 1.10
25
Gradient
corresponding incremental distances dl along these vectors equal to dr, rdd, and
rsin0 d0, and hence the following expression for the gradient of V = V(r, 6, <p) in
the spherical coordinate system:
grad
3F„
—
V = VF =
important to note that there
It is
1
3F-
r
36
is
—
1
-
0 H
r H
dr
:
r sin
no equivalent
dV
—
tions
So,
V
all
(1.108)
gradient
in spherical
coordinates
in the cylindrical or spheri-
cal coordinates to the expression for the del operator, V, in
because
7
4>
9 30
Eq. (1.100), essentially
the unit vectors in Figs. 1.25 and 1.26 except z (of course, their direc-
and not magnitudes) depend on the location (on coordinates) of the point M.
can be formally treated as a vector, given by Eq. (1.100), only in Cartesian
coordinates.
The expression
in
Eq. (1.102) and the corresponding expressions associated
VF for any multivariable scalar
with other coordinate systems enable us to calculate
function
F
We
(not necessarily the electric potential).
shall
now
derive the rela-
tionship between the gradient and so-called directional derivative of a scalar field,
which
will
new
provide us with a
physical interpretation of the gradient operation
in general.
x
Combining Eqs.
dF
(1.94), (1.95),
Ex = E
•
x
and
= — VF
(1.101),
we have
dF
•
x
can
now
= VF-x.
(1.109)
dr
due
We
Figure 1.26 Point M(r,
6,
0)
and coordinate unit vectors
in
the spherical coordinate
system.
formally proclaim the Cartesian x-axis to be an arbitrarily positioned
linear axis in space,
and
call
it
an
/-axis,
with respect to which (Fig. 1.27)
— = VF-1= V
|
F| cos
(1
/3,
.
110 )
directional derivative
dl
where 1 is the associated unit vector (|1| = 1) and f5 the angle between the vector VF
and the /-axis. The derivative dF/d / is referred to as the directional derivative of F
in the / direction. It represents the rate of change of F in this direction, and equals
the projection (component) of the vector VF on (along) the /-axis. Eq. (1.110) is
an equivalent mathematical definition of the gradient of a scalar, and, although
obtained specifically for the electrostatic potential, it is valid for any scalar field
F (for which the necessary derivatives exist).
In addition, Eq. (1.110) has a very important physical meaning and practical
implications.
is
We
a fixed vector.
note that for a given scalar
By
field
steering the /-axis around
= 0,
VF
at a point
dF
~dd
=
|VF|
P
in Fig. 1.27
by varying the angle /}, we
For = ±90°, dF/d/ = 0 [from
dF/d / at the location P.
however, we reach the maximum
obtain different derivatives
Eq. (1.110)]; for
which becomes
F,
P, i.e.,
08
=0
in
dF/d /
(cos
ft
(1
).
=
the directional derivative of
1),
a scalar field.
.
111
)
max
VF
VF
maximum
gradient of a scalar field
which
F
space rate of change in
F
is
F
(fi
=
changes most rapidly and the magnitude of the
all
0).
We
conclude that the
a vector that provides us with both the direction in
change. This property of the gradient operation
applications, in
physical
gradient
This means that (1) the magnitude of
equals the maximum space rate of change
in the function F per unit distance [|VF| = (dF/d/) max ] and (2)
points in the
direction of the
Figure 1.27 Relationship
between the gradient and
is
areas of science and engineering.
maximum
space rate of
extensively used in
numerous
meaning of the
26
Chapter
1
Electrostatic Field in Free
Space
Field
From
from
Potential,
Charged Ring
the electric potential due to a charged ring given by Eq. (1.85), find the electric field
intensity vector along the ring axis
By symmetry,
Solution
normal
the vector
E
to
plane.
its
along the z-axis
that a combination of Eqs. (1.101), (1.105),
and
(Fig. 1.11)
has a z-component only, so
(1.85) gives
(1
which, of course,
the
is
same
result as in
Eq.
.
112 )
(1.44).
Conversely, according to Eq. (1.74), the expression for the potential due to the ring can
be found from the expression for the
the point
P
field, in
Eq. (1.44), by integration along the z-axis from
—
(with a coordinate z) to the reference point at infinity (z
oo), as
(1.113)
and
it is
a simple matter to verify that the solution of this integral
is
exactly the potential in
Eq. (1.85).
Problems: 1.39-1.45; Conceptual Questions (on Companion Website):
MATLAB Exercises (on Companion Website).
An
D AND 2-D ELECTRIC DIPOLES
3
1.11
1.11;
is
a very important, fundamental electrostatic system consisting
of two point charges
Q of opposite polarities separated by a distance d. We alterna-
electric dipole
tively refer to this
system as a three-dimensional (3-D) dipole, to distinguish it from
line charges, which will be analyzed later in this
an analogous 2-D system combining
derive the expressions for the electrostatic scalar potential
section. So, let us
first
and
vector due to a 3-D dipole at large distances compared to
field intensity
Introducing a spherical coordinate system whose origin
as
shown
in Fig. 1.28
and using Eq.
is
d.
at the dipole center
(1.80) for the electric potential
due to a single
point charge and superposition, the potential due to the dipole at a point
P
is
(1.114)
Since
r
d,
/*2
— r\ ^ dcosO
electric dipole potential
and
/q/q
Q
d cos 0
4;i£o
r2
d d
Figure 1.28
Electric dipole.
r
2
,
p
and Eq. (1.114) becomes
cos 6
p
r
'
Ansor
2
An e^r2
(1.115)
Section
3-D and 2-D
1 .1 1
Electric Dipoles
Figure 1.29 Cross section of a
line dipole.
(1.116)
is
the dipole
moment
=
(jp
Qd). The unit for p
Applying the formula for the gradient
the expression for
V in Eq.
1
3
1/r
,
E
and
•
m.
in spherical coordinates, Eq. (1.108), to
dV
*
2 cos 6
e
r
note that
C
moment
(1.115) yields
E = — VE =
We
is
electric dipole
E due
4k sqT'
dQ
+ sin 0 0
r
(1.117)
electric dipole field
1
to an electric dipole vary with distance as 1 /r
1
and
respectively.
We
fundamental for the
concept is used in electro-
shall see in the next chapter that electric dipoles are
discussion of dielectric polarization. In addition, a dipole
magnetic interference (EMI) considerations, since the quasistatic (low-frequency)
electric field produced by an electrical device can often be approximated by the field
due to a single electric dipole, given in Eq. (1.117).
The combination of two equal parallel infinite line charges Q' of opposite polarities separated by a distance d constitutes a line or two-dimensional (2-D) electric
dipole. Let us find the potential due to this system.
Fig. 1.29 shows the cross section of a line dipole with the following per-unitlength dipole
moment:
=
P
The
system
cylindrical coordinate
(1.118)
<2 d.
adopted such that the z-axis
is
is
normal
to the
plane of drawing and coincides with the central line of the dipole. Taking the point
O (r = 0) for the reference point for potential and applying Eq.
E=
where
/fti
=
d/2 and
charges 1 and
In
2k so
^2
Q!
.
In
h
’
—=
*R2
,
In
,
r\
=
r
2;r£o
n+d cos 0
O'
n
2k Sq
<<C 1,
Q'd.
,
In
/
1
+
dcosd)
-
(1
.
120 )
the final expression for the electric potential due to a
out to be
P' cos
0
(1
27r£or
=
(1.119)
r\
d/2 are the distances of the reference point from
large distances from the dipole (r^>d),
v^
with p'
in's
27T£q
ri
we have
At
2k sq
+ x) ^ x for x
line dipole turns
-Q!
rnx
i
2, respectively.
V
Since ln(l
Q!
(1.87),
.
121
)
line dipole potential
27
28
Chapter
Space
Electrostatic Field in Free
1
Note
that for quasistatic electric field evaluations, transmission lines (telecom-
munication
power
lines,
lines,
computer buses,
can sometimes be approx-
etc.)
imately modeled by a single line electric dipole, with the potential computed
by Eq. (1.121). Such evaluations are important
between transmission
Problems 1.46-1.50;
'
:
lines, as well as
MATLAB
Exercises (on
in
studying undesired couplings
EMI
associated
field levels.
Companion
Website).
FORMULATION AND PROOF OF GAUSS' LAW
1.12
Gauss’ law represents one of the fundamental laws of electromagnetism. It is an
Coulomb’s law and a direct consequence of the mathe-
alternative statement of
matical form of the electric field intensity vector due to a point charge. Gauss’ law
involves the flux (surface integral) of the vector
12
and can equivalently be formulated
E
through a closed mathemati-
form, which is
based on a differential operation called divergence on E. This important equation,
in either form, provides an easy means of calculating the electric field due to highly
symmetrical charge distributions, including problems with spherical, cylindrical, and
planar symmetry, respectively.
Gauss’ law states that the outward flux of the electric field intensity vector
through any closed surface in free space is equal to the total charge enclosed by
that surface, divided by sq. To prove it, let us consider first a single point charge
Q in free space and evaluate the flux of vector E through an arbitrarily positioned
cal surface,
in a differential
surface element (differentially small patch) of area dS,
dS, as
shown
This elementary flux
in Fig. 1.30.
dV E = EdS,
where n
is
is
dS
whose surface area vector
is
given by
=
dSn,
(1.122)
the unit vector normal to d S directed from the charge outward. Using
Eq. (1.24), we have
Figure 1.30 Evaluation of
the flux of the vector
to a point charge
through
surface element
d.V,
area vector, dS,
is
d^/.'
E due
from the charge outward.
-
—
(1.123)
^r>
7
47r£ 0
R2
a
whose
oriented
= EdScosa = EdS n =
dS n standing
with
(Fig. 1.30).
at the
charge.
The
solid angle of the
cone
dft
elementary solid angle
[a solid
dS on the plane perpendicular
for the projection of
to
R
This projection can be considered as the base of a cone with the apex
angle
is
measured
in
—
by definition.
is,
d
(1.124)
R2
steradians (sr or srad)], yielding
d4> £
= -^-
dQ.
(1.125)
4jT£()
Assume now
depicted
l2
The
(to
in Fig.
that the charge
1.31(a).
flux of a vector function a
be discussed
in this
Q
From Eq.
section)
is
is
enclosed by an arbitrary closed surface 5, as
outward flux of the
(1.125), by integration, the
through an open or closed surface S
is
defined as
fs a
the vector element of the surface perpendicular to
accordance to the orientation of the surface.
it,
dS, where dS
and directed
in
,
Section
1
Formulation and Proof of Gauss' Law
.12
29
HISTORICAL ASIDE
Johann Karl Friedrich
Gauss (1777-1855) was
a
German mathematician
and a director of the
Gottingen Observatory.
Gauss was born in Braunschweig (Brunswick)
as a son of a gardener
and a servant girl, and he
showed his mathematical genius very early. At
the age of seven, to the
astonishment of his elementary school teacher,
he summed the integers from 1 to 100 within
seconds by realizing that the sum equaled 50
pairs of numbers from opposite ends of the list,
each pair adding up to 101 (1 + 100 = 101, 2 +
= 5,050. He attended
99 = 101,
.), so 50 x 101
Universities of Braunschweig, Gottingen, and
Helmstedt, and received his doctoral degree in
.
vector
E
.
through the entire surface S (n
'!'£
=
<£ dvI/ £
JS
is
1799.
Gauss made numerous seminal contribu-
number
tions to the
greatest mathematicians of
<k d£2
all
time, others being
and Newton
developed in 1813 the divergence
theorem, providing the mathematical form for the
famous law of electricity that relates the flux of the
electric field intensity vector through a closed surface to the enclosed net charge and now carries his
name - the law which first came out from experiments with charged concentric metallic spheres
by Faraday (1791-1867) and was later translated
into the Gaussian mathematical form by Maxwell
(1831-1879). Upon Weber’s (1804-1891) arrival
at Gottingen in 1831, they worked together on a
variety of topics in electricity and magnetism (see
Weber’s biography in Section 4.8).
Archimedes
(287 B.C.-212 B.C.)
He
(1642-1727).
directed from the surface outward)
= -0—
theory, algebra, geometry,
and calculus, proved several fundamental theorems in different branches of mathematics, and
is considered by many to be one of the three
=
is
(1.126)
4tt£q
47T£ 0 Js
where £2 is the full solid angle, which, in turn, can be interpreted as the angular
measure for a spherical surface of arbitrary radius r and area So = 4 nr2 resulting in
(for spherical surfaces, dS n = dS)
,
(1.127)
As
the
full solid
angle turns out to equal 4;r
-i
EdS = —
£o
(sr),
the total flux
for a point charge
full
solid
angle
becomes
Q enclosed by S.
(1.128)
Figure 1.31 Evaluation of the
outward
flux of
E due
to a
point charge through a closed
surface S, the charge being
enclosed by S
outside S (b).
(a)
or located
30
Chapter
1
Space
Electrostatic Field in Free
Q
For a point charge
fluxes through the
outside the surface S,
two surface elements shown
we
elementary
which have opposite
realize that the
in Fig. 1.31(b),
orientations, are
d vl
(
,
=
£;)through dSj
Q
^
and
^£2
(
Q
=
d'Ff) through
£
/\j[£
respectively, so that their contributions to the flux integral cancel each other,
the total flux of
E
through S
=
Figure 1.32 Arbitrary closed
surface containing a volume
charge distribution
By means
E dS =
•
<j)
and
is
for a point charge outside S.
0,
(1.130)
of the superposition principle, Eq. (1.128) can then readily be
generalized to the case of
N point charges enclosed by a (closed) surface S:
in free
® E dS = ®
space.
•
Js
=
<£
El
•
dS
+
Js
where
E
<£
E 2 dS +
•
•
•
•
+
Js
-
(
is
(Ei
+ E2 +
(f
E n dS =
+
Eyy)
•
dS
Js
—+—+
Js
£()
the field due to a charge Qi,i
=
Qn
+
£0
£()
(1.131)
’
1,2,..., N. With the notation
N
Qs = J^Qi,
(1.132)
i=l
we have
Gauss' law
(1.133)
There may also be charges located outside the surface S; recall from Eq. (1.130),
however, that the contribution to the total flux resulting from such charges is zero.
As we know, any (continuous or discontinuous) charge distribution can be represented as a system of equivalent point charges. This means that Eq. (1.133) holds
true for any charge distribution (in free space), where
Qs =
Eq. (1.133)
total
is
charge enclosed by S
(arbitrary charge distribution).
E-dS =
volume charge
distributions
is
the
volume charge
which Gauss’ law can be written as
distribution, Fig. 1.32, in terms of
for
34)
the mathematical formulation of Gauss’ law.
The most general case of continuous charge
Gauss' law
(1 .1
1
(1.135)
£(J
with v denoting the volume enclosed by the surface S and p the volume charge
density. This particular form of Gauss’ law is usually referred to as Maxwell’s third
equation for the electrostatic
The
field in free space.
principle of conservation of energy in the electrostatic field, Eq. (1.75),
and Gauss’
law, Eq. (1.135), are the
two fundamental
equations) describing the electrostatic
13
Gauss’ law
Nevertheless,
tion in the
in later
is
13
As shown here and
the second equation out of a total of two Maxwell’s equations used
we term
complete
chapters.
integral equations (Maxwell’s
field in free space.
it
Maxwell’s third equation because
set of four general
it is
in electrostatics.
customarily positioned as the third equa-
Maxwell’s equations for the electromagnetic
field, as
we
shall see
,
.
Section
in Fig. 1.19,
both equations can be derived from Coulomb’s law,
i.e.,
1 .1
3
Applications of Gauss'
Law
31
from the
expression for the electrostatic field due to a point charge in free space.
Problems'. 1.51
and
1.52;
Conceptual Questions (on Companion Website):
1.12-1.16.
LAW
APPLICATIONS OF GAUSS'
1.13
now
Let us
discuss procedures for using Gauss’ law to determine the electric field
is known. Gauss’ law is always true, and we can
any charge distribution and any problem. However, it enables us to analytically solve only for the field due to highly symmetrical charge distributions. To
understand this, note that Gauss’ law is mathematically formulated by an integral
equation [Eq. (1.133)] in which the unknown quantity to be determined (E) appears
inside the integral. We can use the law to obtain a solution, therefore, only in cases
in which we are able to bring the field intensity, E, outside the integral sign, and
solve for it. These cases involve highly symmetrical charge distributions, for which
we are able to choose a closed surface S, known as a Gaussian surface, that satisfies two conditions: (1) E is everywhere either normal or tangential to S and (2)
E = const on the portion of S on which E is normal. When E is tangential to the
surface, E dS in Eq. (1.133) becomes zero. When E is normal to the surface, E dS
becomes E dS, and since we have also that E is constant, it can be brought outside
the integral sign in Eq. (1.133). We shall now apply these basic ideas to problems
with spherical, cylindrical, and planar symmetry, respectively, in four characteristic
intensity
apply
it
if
the charge distribution
to
•
•
examples.
Example of
Example 1.18
The
class of
first
Problem with Spherical Symmetry
a
problems for which Gauss’ law can be used to solve for the
field involve
charge distributions that depend only on the radial coordinate in the spherical coordinate
system.
vector
Due
to spherical symmetry, only the radial
present,
is
and
this
component
these facts, the solution procedure
radius a with a uniform
distribution both inside
is
is
component of the
electric field intensity
a function of the radial coordinate only. Based
simple to perform.
volume charge density p, in
and outside the sphere and
As an example,
and determine
free space,
on
consider a sphere of
(a) the field
(b) the electric scalar potential at the
sphere center.
with spherical symmetry -
Solution
(a)
The
Figure 1.33 Application of
Gauss' law to a problem
Example
electric field vector at
an arbitrary point in space
E=
is
1
E{r) r
(1.136)
E
in
a problem with spherical
symmetry
where
r is
vector, as
the radial coordinate in the spherical coordinate system and r
shown
in Fig. 1.33.
centered at the origin.
dS=
The outward
•
S
is
is
the radial unit
a spherical surface of radius r
E dS = Er dSi = EdSr
dSr
dS
surface
On S,
•
flux of the vector
®E
Js
The Gaussian
=®
JS
E(r)
r
= EdS.
(1.137)
E through S is hence
dS
=
E{r)
(p
dS
=
E{r)
S
.18.
of the form
=
1
E{r) 4 nr
(0
<
r
<
oo)
Js
(1.138)
Since p = const (uniform charge distribution), the charge enclosed by S is computed,
according to Eq. (1.30), as p times the corresponding volume, which is that of either the
i
32
Chapter
1
Electrostatic Field in Free
entire
Space
domain
<
inside S, for r
outside this sphere), for r
>
or the charged sphere of radius a (there
a,
Qs amounts
so
a,
is
no charge
to
pAnr3 /3
for r
p\nc? j'b
for r
<
>
a
(1.139)
a
I
According to Gauss’
law, 4>£
=
Qs/so yielding
,
E(r)
Note
that the field for r
>
a
pr/Oso )
pa 3 /(3e0 r2 )
=
for r
for r
<
>
a
(1.140)
a
'
identical to that of a point charge
is
p47ra 3 /3 (the total
Q=
charge of the sphere) placed at the sphere center. This means that the point charge and
the charged sphere are equivalent sources with respect to the region outside the sphere.
Concept of equivalent sources
(b)
To
is
often used in electromagnetics.
find the electric potential at the sphere center (with respect to the reference point at
infinity
14
),
we
first
conclude that the potential outside the charged sphere
is
identical to
the potential of the equivalent point charge, because the fields are identical, so that the
potential at the sphere surface
(r
=
a)
is
obtained directly from Eq. (1.80):
V(a)
Q
=
(1.141)
Aneoa
Invoking Eq. (1.90), the potential difference (voltage) between the sphere center and
E from the center to the surface, which results in the
surface equals the line integral of
following for the potential at the center:
E(0)
=
Example of
Example 1.19
E(r) dr
a
+
=
V(a)
3
Q _
8nsoa
pa 2
(1.142)
2 eq
Problem with Cylindrical Symmetry
problems we deal with using Gauss’ law involve infinitely long charge
depend only on the radial coordinate in the cylindrical coordinate system.
As an example, consider an infinitely long charged cylinder of radius a and volume charge
The second
class of
distributions that
density
r
2
pW = Po -y
a
(1-143)
1
in free space, where po is a constant and r the radial coordinate
cylinder axis), and find the electric field everywhere.
Solution
Shown
of the form given in
associated unit vector.
Due
the distance from the
symmetry
and
Eq. (1.136), r being here the radial cylindrical coordinate and r the
The Gaussian surface is a cylinder of radius r and height (length) h,
in Fig. 1.34
is
a cross section of the cylinder.
of the charge distribution in this case, the field
is
(i.e.,
is
to cylindrical
radial with respect to the cylinder axis
positioned coaxially with the charged cylinder.
Figure 1.34 Application of
Gauss' law to a problem
with cylindrical symmetry -
Example
1
To determine E inside the charge distribution, we apply Gauss’
Gaussian cylinder of radius r < a.
law, Eq. (1.135), to a
2
E(r) 2nrh
.19.
-if —E
po
£()
Jr’={
a*
2jiE d Eh,
(1.144)
dv
Sc
p
14
Whenever
infinity,
[e.g.,
the reference point for the potential to be determined
is
not specified,
we assume
except for infinite charge distributions, where such assumption would give us an
Eq. (1.87) for
n
r
->•
oo].
that
it
is
at
infinite potential
Section
where Sc
is
the lateral surface area of the cylinder, and dv
1
shell of radius r (0
<
explained in Section
<
1
r
1.4).
the
is
volume of
1 .1
3
Applications of Gauss'
Law
33
a thin cylindrical
thickness d F, and height h (we adopt as large as possible dv, as
r),
This volume
is
computed
as that of a thin flat rectangular slab with
1
edges equal to 2 nr (circumference of the cylindrical
shell), h and d/ (for the purpose of the
volume computation, the shell is flattened into a rectangular slab). The flux of the vector E
through the top and bottom surfaces of the Gaussian cylinder is zero since E is tangential to
those surfaces (E dS = 0). By integration in Eq. (1.144),
E(r)
Qr
= - ^z
,
,
forr<n.
(1.145)
4eo«
For a
field
point outside the charged cylinder ( r
Eq. (1.144) becomes
a,
which
>
a),
the upper limit in the integral in
a.
(1.146)
results in
2
E(r)
=
for r
>
4e0 r
Note
that this field
is
identical to the field
due
to an infinite line charge positioned
along the cylinder axis with the same charge per unit length O' as the cylinder [see
Eqs. (1.31) and (1.57)].
Example of
Example 1.20
a
Problem with Planar Symmetry
and the following example illustrate how problems involving charge distributions
depend on one Cartesian coordinate only - problems with planar symmetry - can be
solved using Gauss’ law. Consider first a layer of volume charges in free space with the
volume charge density being the following even function of the Cartesian coordinate x\
Finally, this
that
P(x)
where po and a
(a
>
0) are constants
intensity vector everywhere,
Solution
=
i.e.,
M
Po
<
(1.147)
a.
and p(x) = 0 for |x| > a. Determine the
and outside the charge layer.
electric field
inside
shows the charge distribution. Due to planar symmetry, vector E at an
x-component only, and Ex depends on the coordinate x only
any plane x = const), so we can write
Fig. 1.35
arbitrary point of space has an
(it is
constant in
E = Ex {x) x.
In addition, since the charge distribution
tor
E in the
plane defined by
x'
= x (x >
is
(1.148)
symmetrical with respect to the plane x
0) in Fig. 1.35
is
= 0, vec-
equal in magnitude and opposite in
Figure 1.35 Application of
Gauss' law to a problem
with planar symmetry -
Example
1
.20.
E
in
a problem with planar
symmetry
:
34
Chapter
1
'
Electrostatic Field in Free
Space
direction to that in the plane x'
(rectangular box) that
is
=
—x. The Gaussian surface
— x and x' = —x,
field intensity Ex (x)
areas So, positioned in the planes x'
a right-angled parallelepiped
is
=
cut symmetrically by the plane x
0 and has two of
faces, with
its
respectively.
Let us first determine the
inside the charge layer (— a < x < a).
Noting that E and dS are oriented in the same direction at both faces So, as well as that
E and dS are mutually perpendicular on the remaining portions of the Gaussian surface,
we have
Ex (x) S 0 =
2
—
p(x')S0 dx\
f
(1.149)
v—
-
£o Jx’=-x
dv
where dv
volume of a
the
is
thin slice of the parallelepiped with thickness dx'.
The
integration
in x' yields
E*(x)
now
Let us
If
we then
nothing
start
will
=
^
^1
-
~
for
<
|x|
(1.150)
a.
j,
bring the parallelepiped faces So onto the boundaries of the charge layer.
moving them toward
change
infinity,
in the application of
symmetrically with respect to the plane x
Gauss’ law, since there
is
=
0,
no charge beyond the layer
boundaries. This reasoning gives us directly the field intensity outside the charge layer:
Ex (x) = Ex (a)
—
- ——
=—
2poa
—
-
tor
,
x >
(1.151)
a,
3e 0
Ex (x) = Ex (—a)
2po«
f
for x
,
< —a.
(1.152)
3e 0
Example
Planar Symmetry, Antisymmetrical Charge Distribution
1.21
Repeat the previous example but for the following odd function of x as the charge density of
the 2n-thick layer:
p(x)
=
p0 ~,
\x\
<a\
(1.153)
a
there
is
no charge outside the
Solution
As
this
is
layer.
=
0,
zero, which implies that the electric field outside the layer
is
an antisymmetrical charge distribution with respect to the plane x
the total charge of the layer
is
zero; namely, this field
due
is
as
if
to an equivalent infinite sheet of charge (Fig. 1.15) with zero
surface charge density (p s = 0). In other words, if we subdivide the charge layer described by
Eq. (1.153) into differentially thin layers of thicknesses dx' and add (integrate) the fields,
given by Eq. (1.64), due to
all
these thin layers (observed outside
and negatives exactly cancel out, and the result
place of Eqs. (1.151) and (1.152) we thus have
p(x') dx'
£
=0
(total field for
Ex (x) =
Positioning the Gaussian rectangular surface in Fig.
1
0
.35
all
(|x|
of them), the positives
—a
x <
>
>
or x
a)
is
zero. In
(1.154)
a).
such that one of its parallel faces
with area So is in the region x' < —a (on the left of the charge layer) and the other is in the
plane x' — x where — a < x < a (inside the layer), Gauss’ law gives [instead of Eqs. (1.149)
and (1.150)]
£*(x)So
=
!/'
Po
e 0 J x '=-a
since the flux of
E through
volume integration
— So dx'
(*)
2
=
Je~a
a
the face outside the charge layer
effectively starts at x'
—
(*
is
- fl2
0 155 )
-
)
zero,
from Eq.
(1.154),
Problems 1.53-1.64; Conceptual Questions (on Companion Website):
1.18;
MATLAB
Exercises (on
and the
-a, where the charge distribution begins.
Companion Website).
1.17
and
Section
1.14
1
.14
Differential
Form
35
Law
FORM OF GAUSS' LAW
DIFFERENTIAL
r
an integral equation in the spatial domain (the integrations involved
are carried out with respect to spatial coordinates), and, in the general case, it repGauss’ law
of Gauss'
is
resents an integral relationship
between the
electric field intensity vector, E,
E*(x)
_____
Ex (x + dx)
and
the volume charge density, p [Eq. (1.135)]. In this section, we shall derive an equivalent, differential, relationship between E and p, that is, the differential form of
x + dx
x
x
dx
Gauss’ law.
Assume first that p
D
a function of the Cartesian coordinate x only, p
charge distribution), so that the only present component of E is x =
is
E
— p(x) (1Ex (x). Let
S
us apply Gauss’ law, Eq. (1.135), to a rectangular closed surface S, the dimension of
dx and the sides normal to that direction are So in area,
field being constant on both surfaces So, no integration
is needed on the left-hand side of Eq. (1.135). In addition, since dx is differentially
small, we can take p(x) as constant in the volume enclosed by S, so that no integration is needed on the right-hand side of Eq. (1.135) either. Noting that on the
left-hand side of S, E and dS are directed in opposite directions, we have
which
in the
x direction
as indicated in Fig. 1.36.
is
The
—Ex (x) So + Ex (x +
dx) So
=
—1 p(x) So dx.
(1 .1
Figure 1.36 For derivation
of the one-dimensional
Gauss' law
in differential
form.
56)
£o
The
differential of
Ex corresponding to the displacement
dEx = Ex (x+
divided by dx,
Eq. (1.156),
is,
by
dx)
- Ex (x),
definition, the derivative of
this derivative
comes out
dx,
Ex
(1.157)
with respect to
x.
From
to be
d Ex
_
dx
p_
(1.158)
7
-D
differential
Gauss' law
£q
is a differential equation in the spatial domain, and it represents the onedimensional Gauss’ law in differential form. It states that, in cases when the charge
distribution changes with a single linear spatial coordinate in free space, the rate of
This
change of the
electric field intensity with that coordinate equals the local density of
volume charge, divided by £oThe generalization of Eq. (1.158) to the three-dimensional case is straightforward. The charge density is now a function of all three coordinates, p = p(x, y, z),
and the Gaussian surface S has to be differentially small in all three dimensions, as
shown in Fig. 1.37. We break the flux integral over S in Eq. (1.135) up into three
pairs of integrals, each pair being carried out over two parallel sides of S. All sides
being differentially small, the flux over each of them can be approximated by taking
a constant value of the field component normal to the side and multiplying it by
the side area. This gives us the result for each individual pair of integrals that has
same form as in the one-dimensional case above [Eqs. (1.156)— (1.158)].
For the first pair of integrals, we have that the outward flux of E through sides
normal to the x direction (front side and back side in Fig. 1.37) is
exactly the
Figure 1.37 For derivation
of Gauss' law in differential
form
/
J front
E dS +
/
E
•
dS
=
(change of
Ex
across dx) x
(
dy d z)
7 back
BEx
dx (dy dz),
dx
for
an arbitrary charge
distribution.
(1.159)
-
36
Chapter
1
Electrostatic Field in Free
,
Space
and adding the
results for other
two pairs of
dE
—
E dS =
dE v
- cLt:(dydz)
3*
5
we
integrals,
obtain the total flux
dE
dy (dxdz)
H
dz (dvdy).
H
(1.160)
dz
3y
— dx dy dz, with which the above equation becomes
dE y
dE z
/ 3 Ex
Av (Av — 0).
=
- +
^ +
(1.161)
\dx
dy
dz
The volume enclosed by S is A v
_
E dS
•
(
5
Av
Since
is
— — —
very small, the total charge in
it
can be found with no integration
in
Eq. (1.135), simply as
Qs — p Av = p (dvdy dz).
Finally, interconnecting the flux
Gauss' law
in differential
form
(1.162)
and charge by means of Eq.
dEx
dEy
dx
dy
3
Ez
(1.135),
we have
p
(1.163)
'
This
dz
Gauss’ law in differential form,
is
for the electrostatic field in free space.
i.e.,
It is
£q
Maxwell’s third differential equation
a partial differential equation or
PDE
(partial derivatives with respect to individual coordinates enter into the equation) in
unknowns
three
(three
components of the vector E). This equation provides us with
way E varies in space. It relates the rate of change
valuable information about the
components with
of field
spatial coordinates to the local charge density.
We
see
components along the direction of that component only
and not along y and z, etc.) contribute in this relationship.
that changes of individual
(change of
Ex
Problems
1.65
:
along
and
x,
1.66;
Conceptual Questions (on Companion Website):
1.19;
MATLAB Exercises (on Companion Website).
DIVERGENCE
1.15
The expression on the
left-hand side of Eq. (1.163)
vector function (E), and
is
is
called the divergence of a
written as div E. Applying formally the formula for the
dot product of two vectors in the Cartesian (rectangular) coordinate system,
a
•
b
=
(a x
x
+ ay y + a z z)
•
(b x x
to the del operator, Eq. (1.100),
+ by y + b z z) =
ax bx
+ ay by + a z b z
,
(1 .1
64)
and vector E, we get
V E=
•
(1.165)
and this is exactly div E, in Eq. (1.163). Note that the divergence is an operation that
performed on a vector, but the result is a scalar. The differential Gauss’ law now
can be written in a short form
is
divE
Gauss' law using divergence
=
or
V E=
•
where,
in
coordinates
Cartesian
in the
—
(1.166)
eo
£o
notation
divergence
—
Cartesian coordinate system,
div
E= V E=
dEx
dEy
dx
dy
•
(1.167)
)
Section
1
.15
Divergence
In nonrectangular coordinate systems, the corresponding formulas are more
complex, due to curvature (or nonrectangularity) of differential volume cells in
place of the one in Fig. 1.37. For instance, if, in analogy to the situation in Fig. 1.36,
we consider a one-dimensional spherical volume charge distribution with p — p{r),
r
being the radial spherical coordinate in Fig. 1.26, the only component of E is
as in Eq. (1.136), and applying Gauss’ law, Eq. (1.135), to the inner
Er = Er (r),
and outer (radius
(radius r
r
+
dr) surfaces of a thin spherical shell in Fig. 1.9,
Eq. (1.156) becomes
— Er (r
)
A nr2
+ Er (r +
dr) An(r
+
dr)
— p(r) Anr
=
2
2
dr,
(1 .1
68)
£0
where the use is made of the expression for dv in Eq. (1.33). After eliminating An
on both sides of the equation, the expression on the left-hand side is the differential
Er corresponding to the coordinate increment dr,
which divided by dr represents the derivative of r 2 E r with respect to r, and hence
[see Eq. (1.157)] of the function r
2
d(r
is
Er _
)
p_
(1
’
r2
analogously to Eq. (1.158). This
2
dr
.
169 )
£q
a one-dimensional differential Gauss’ law in the
3-D case of p = p(r,6,(p)
mimics the derivation in Eqs. (1.159)— (1.163) accommodated to the elemental
spherical cuboid in Fig. 1.10, and similarly for p — p(r, 0, z) (Fig. 1.25). The resulting
and
spherical coordinate system,
its
generalization to the
formula for the cylindrical coordinate system
divE
=V E=
•
1
-
—9
(
is
rEr )
r or
and that
1
dE,
r
o<p
-
</>
+
dEz
(1
dz
.
170 )
divergence
in cylindrical
coordinates
for spherical coordinates
divE
As
+
=V E=
•
\
rz
(^Er)
or \
+
(sin
rsin#
/
6Ee +
)
ot)
the vector expression for the operator
V
Cartesian coordinate system, the dot product in
dE,
-4
<\>
rsu
sin0
in Eq. (1.100)
V E
•
actually
is
90
(1
.
171
)
divergence
in spherical
coordinates
valid only in the
makes sense only
in
we shall still use extensively the notation V E in
cylindrical and spherical coordinate systems as well - to simply mean the divergence
rectangular coordinates. However,
•
of E (div E) and a symbolic representation of the respective formulas in Eqs. (1.170)
and (1.171). In general, we shall often draw conclusions about gradient, divergence,
and other operations involving vector field quantities by treating V as a vector,
Eq. (1.100), and performing formally vector operations with it, like in Eq. (1.165).
Most importantly, these conclusions, properties, and relations, although derived in
Cartesian coordinates, hold true (as identities) in all possible (cylindrical, spherical,
and other) coordinate systems, because the properties of a physical quantity and
relations between two or more quantities are the same regardless of the choice of a
coordinate system.
Combining Eqs.
(1.161)
and
div
(1.167),
we
E=V E=
•
obtain the equation
lim
Av—>0
which tells us that the divergence of E
at a
LE
—
dS
(1
Av
given point
is
.
172 )
physical
meaning of the
divergence
the net outflow of the flux of
E per unit volume as the volume shrinks about the point. Eq.
(1.172)
is
an equivalent
37
:
38
Chapter
1
Electrostatic Field in Free
Space
mathematical definition of the divergence of a vector. From
it,
we may regard
div
E
measure of how much the field diverges from the point. In general,
a positive divergence of any vector field indicates a local source of the field at that
point producing radial field components with respect to the point - the outward flow
of flux is positive, which means that the vector diverges (spreads out) at the point.
Possible field components that are tangential to S do not contribute to the flux, and
physically as a
are not related to the divergence. Similarly, a negative divergence indicates a sink at
the point (local negative source) - the outward flux flow
is
negative,
and the vector
converges at the point. Finally, the divergence is zero where there is neither
source nor sink of locally radial field components. Gauss’ law simply tells us that
field
these positive and negative sources in the case of the electric
field,
E, are positive
and negative charges, p Av. Quantitatively, the divergence represents the volume
density of sources, and in our case this density is p.
Let us now replace p in the integral form of Gauss’ law, Eq. (1.135), by its equal,
£qV E - from the differential form of the law, Eq. (1.166):
•
E iS =
divergence theorem
V Edv.
-
fs
(1.173)
•
l
is called the divergence theorem (or Gauss-Ostrogradsky theorem).
Although we have obtained it specifically for the electrostatic field in free space, the
theorem is true for any vector field (for which the appropriate partial derivatives
exist) and is one of the basic theorems of vector calculus. It applies to an arbitrary
closed surface S and, in words, states that the net outward flux of a vector field
through 5 equals the volume integral of the divergence of that field throughout the
volume v bounded by S.
This equation
Problem with Spherical Symmetry Using
Example 1.22
Redo Example
1.18, part (a),
but
now employing
Gauss’ law
Differential Gauss'
in differential
Law
form.
Solution As we already know, from the spherical symmetry of the problem, that the electric
vector both inside the charged sphere and outside it has the form in Eq. (1.136), the
formula for the divergence in spherical coordinates, given by Eq. (1.171), retains only the
first term. For a point inside the charge distribution in Fig. 1.33, Gauss' law in differential
field
form, Eq. (1.166), thus reduces to the following differential equation in a single coordinate, r
V E=
•
— f^Efr)] = —
rL dr
L
J
which can be directly solved by integration. Since p
p-E(f)
=
—
e J[
2 dr
r
~+
+ C\ =
—
<
r
const,
—
C\
3e 0
0
(0
<
a)
(1.1 74)
,
£o
E(r)
we have
=
+
3£ 0
(1.175)
"Tm
and the integration constant, Ci, is found from the “initial” condition E(0) = 0 at the center
= 0). Namely, given that there is not a point charge ( Qo ) at the point O
in Fig. 1.33, the field at this point is zero, and hence C\ = 0 [otherwise, the constant would
amount to C\ = 0o/(4tt£o), from Eq. (1.24)], which yields the result for E in Eq. (1.140).
Outside the charge distribution in Fig. 1.33, p = 0, which substituted in Eq. (1.174)
of the sphere (for r
results in
E(r)j
=
0
= C2
2
r E(r)
—
»
E(r)
=
(a
<
r
<
oo).
(1.176)
integration, C 2 is determined by means of the “boundary” condition
boundary of the charged sphere, r = a; the field intensity on the inner side of the
The new constant of
at the
—
,
Conductors
Section 1.16
in
boundary, computed from Eq. (1.175), must be the same (since there exist no surface charges
on the boundary) as that on the outer side of it, from Eq. (1.176),
=
E(a-)
E(a + )
=
3
—
E{a)
~.
C2 =
(1.177)
3eo
With
this,
the result for
(1.176) agrees with the solution in Eq. (1.140).
Problem with Planar Symmetry by the
Example 1.23
Redo Example
E in Eq.
Differential Gauss'
Law
1.20 but with the use of the differential Gauss’ law.
Solution As this is a problem with planar symmetry, the electric field intensity vector everywhere is of the form in Eq.(1.148). Hence, Gauss’ law given by the differential equation in
Eq. (1.158) applies, which we solve by integration with respect to x (also see the previous
example). For the observation point inside the charge layer {—a <x<a) in Fig. 1.35,
Ex (x) =
p(x)dx + Ci =
—
eq Jf
Because of symmetry
—
e0
[l
-
+ Q,
-^r-J
3a 1
for
|x|
<
(1.178)
f
\
(Fig. 1.35),
Ex (a) = -Ex (-a)
—
—
_
+ Cl =
3eo
3eo
Cl
— Q=
(1.179)
0,
which, substituted in Eq. (1.178), gives the same result as in Eq. (1.150).
For the point outside the layer (|x| > a), p = 0, so that Eq. (1.158) yields
Matching
this field
dition at the layer
Ex {x) = C2
.
value to the one from Eqs. (1.178) and (1.179) in the “boundary” con-
boundary defined by x
= a, Ex (a + = Ex (a ~ = Ex (a), we
)
)
obtain
C2 =
Ex {x) =
2po«/(3eo) f° r x > a, the same as in Eq. (1.151). On the other
side of the layer, the vector E has this same magnitude but opposite direction, resulting in
2poa/(3Eo), and thus
the field expression in Eq. (1.152).
Note that the application of Gauss’ law in differential form to a problem with planar
symmetry and an antisymmetrical charge distribution is presented in Example 2.6.
Problems'. 1.67-1.74;
1.16
So
CONDUCTORS
IN
THE ELECTROSTATIC FIELD
we have considered electrostatic fields in free space (a vacuum or air). We
now extend our theory to electrostatic fields in the presence of materials. As
far,
shall
we
MATLAB Exercises (on Companion Website).
shall see,
most of the formulas derived and solution techniques developed and
used for the electrostatic
field in free
space are directly applicable to the analysis of
electrostatic fields in material space, although
some require
modification. Materials
can broadly be classified in terms of their electrical properties as conductors (which
conduct electric current) and dielectrics (insulators). In the rest of this chapter,
we
shall study the interaction of the electrostatic field with conductors, in
case essentially
no
theoretical modification
whereas the behavior of
is
needed to the
which
electrostatic equations,
dielectrics in the electrostatic field will
be discussed
in the
next chapter.
Conductors have a large proportion of freely movable
electrons and ions) that
make
electric charges (free
the electric conductivity (ability to conduct electric
current) of the material. Best conductors (with highest conductivity) are metals
(such as
silver,
copper, gold, aluminum,
etc.).
In
many
applications,
we
consider
the Electrostatic Field
39
p
40
Chapter
1
Electrostatic Field in Free
Space
metallic conductors as perfect electric conductors (with infinite conductivity). There
many
are also
other, less conductive conductors such as water, earth (ground),
so-called semiconductors
(glass,
and germanium),
(e.g., silicon
some
paper, rubber, etc.) have
etc. Practically all insulators
(usually extremely low) conductivity,
and thus
theoretically are (very poor) conductors, although almost always (in electrostatics)
may be considered
they
as perfect dielectrics (with zero conductivity),
conductors. In our studies of electrostatic
i.e., nonby conductor we normally mean a
fields,
metallic conductor.
Consider an isolated conductor, shown in
uncharged or electrically neutral (its net charge
Eext
Eext
trostatic field
E exl
electric force [see
+
- E=0 +
+
“Ps
zero).
Assume
When
that
it
is
an external elec-
applied, the free charges in the conductor are influenced by the
Eq. (1.23)]
= GE ext
Fe
(1.180)
.
Ps
+
—
is
1.38(a).
Fig.
is
-f
+
This force pushes the positive charges along the direction of
to right, while the negative charges
move
E ex
t,
that
is
from
left
opposite direction. Consequently,
in the
becomes progressively more positive, and the
more negative. In fact, for a metallic conductor, what
actually happens is movement of free electrons (negative charges) toward the left
side leaving an equal number of positive charges, namely, a deficiency of electrons,
on the other side of the conductor. Accumulated free charges form two layers of
surface charge, of densities — s (ps > 0) and ps on the conductor sides. Creation
the right-hand side of the conductor
left-hand side progressively
(b)
Figure 1.38 (a) A conductor
in an external electrostatic
field, (b)
After a transitional
process, there
is
no
,
of surplus charges in the body caused by an external electrostatic field
electrostatic field inside the
electrostatic induction.
conductor.
tric field in
The induced charges,
the conductor, Ej nt which
,
is
in turn, set
is
called the
up an internal induced
elec-
directed from the positive to the negative
oppositely to E ex t- As p s increases, Ej n becomes progressively stronger,
opposes the migration of charges from left to right. In the equilibrium, Ej nt
completely cancels out E ex in the conductor, so that the total field E in the conductor is zero, and the motion of charges stops, as illustrated in Fig. 1.38(b). Note
that the conductor remains uncharged as a whole. The entire transitional process is
extremely fast, and the electrostatic steady state is established practically instantaneously. In fact, based on the length of the time needed for this process of movement
layer,
and
i.e.,
t
it
t
of charges to the surface of a material body,
that the total electric field inside the
material
is
i.e.,
way
we determine whether a
their redistribution in such a
body becomes
zero,
we shall see in a later chapter,
commonly
most
used metallic conductor -
a conductor or dielectric. For example, as
the time to reach the equilibrium for the
copper - is as brief as ~ 10~ 19 s, whereas it takes as long as ~ 50 days for the charge
rearrangement across a piece of fused quartz (very good insulator).
In the case of a conductor that had been charged (with a positive or negative
excess charge) prior to being situated
in
the external
field,
a similar process takes
place. All free charges (for a metallic conductor, free electrons of the conductor,
which abundantly exist in the material also when it is electrically neutral as a whole,
15
plus excess charge ) are exposed to the force Fe and produce the internal field that
cancels out the externally applied field
15
Nole that excess charge on
a metallic
ative excess charge) or by taking
number
some
the electrostatic equilibrium.
body may be produced by bringing electrons
of
of these extra or missing electrons
of the body.
in
its
is
free electrons
always
much
away
to the
(positive excess charge),
body (negwhere the
smaller than the total count of free electrons
Section
We
.16
Conductors
conclude that under electrostatic conditions, there cannot be electric
in
41
the Electrostatic Field
field
conductor,
in a
E=
This
it,
1
the
is
we
first
derive
all
(1.181)
0.
fundamental property of conductors in electrostatics. Starting from
other fundamental conclusions about the behavior of conductors in
the electrostatic
no electrostatic
a conductor
field inside
field.
According to Eqs.
and
(1.181), (1.90),
(1.88), the voltage
points in the conductor, including points on
conductor is an equipotential body,
conductor and on its surface,
i.e.,
its
surface,
the potential
V=
is
is
the
between any two
means that a
same everywhere in the
zero. This
const.
(1.182)
From Eq. (1.181), V E = 0 in a conductor, implying that [Eq. (1.166)] there
cannot be surplus volume charges inside it,
(1.183)
interior
and
surface of a
conductor are equipotential
no volume charge
inside
a
conductor
So, any locally surplus charge of a conductor (whether
it
neutral as a whole or
is
not) must be located at the surface of the conductor.
isfy
in a
Let us now derive so-called boundary conditions that the electric field must sata conductor surface. The electric field intensity vector E near the conductor
on
vacuum can be decomposed
into the
respect to the boundary surface, as
En
respectively,
where a
is
shown
= £cosa
the angle that
normal and tangential components with
The two components are
in Fig. 1.39(a).
=
and
E makes
Esina,
(1.184)
with the normal to the surface.
We
apply Eq. (1.75) to the narrow rectangular elementary contour C in Fig. 1.39(a). The
field is zero along the lower side of C (E = 0 in conductors), and we let the contour
side
Ah
shrink to zero pressing the sides
A/
tightly
onto the boundary surface, so
is E
AI along the upper
that the only contribution to the line integral in Eq. (1.75)
C
side of
(no integration
E
Hence, there
is
is
•
needed, because
dl
=E
AI
A / is
•
small).
= EAl sin a = £
t
A/
=
(1.185)
0.
no tangential component of E over the surface of a conducting body
in electrostatics,
zero tangential electric
on a conductor surface
(a)
(b)
Figure 1.39 Deriving boundary conditions for the electrostatic
field
(E) near a
conductor surface: (a) narrow rectangular elementary contour (used for the
boundary condition for the tangential component of E) and (b) pillbox
elementary closed surface (for the boundary condition for the normal
component
of E).
field
42
Chapter
1
Electrostatic Field in Free
Space
In other words, the electric field intensity vector
on the surface of
a conductor
is
always normal to the surface,
E = £ n n,
(1.187)
where n is the normal unit vector on the surface, directed from the surface outward.
To obtain the boundary condition for the normal (the only existing) component of E,
we apply Eq.
AS
(1.133) to the pillbox Gaussian surface, with bases
and height Ah (shrinking to zero), shown
For similar reasons as in
obtaining Eq. (1.185), the flux in Eq. (1.133) reduces to E AS over the upper side
of S. Because the charge enclosed by S is p s AS,
in Fig. 1.39(b).
•
(t
E dS = E AS =
(E n n)
•
•
•
= En AS =
(ASn)
—p
JS
s
AS,
(1 .1
88)
£0
providing the relationship between the normal component of the electric
field
and the surface charge density on the
intensity vector near a conductor surface
surface:
normal
electric field
(1.189)
component on a conductor
surface
The
lines of the electric field are
normal to the surface of a conductor.
should always remember that the normal component
En
in
Eq. (1.189)
is
We
defined
n. When p s > 0, the field lines start from the
whereas they end on it (£„ < 0) when ps < 0.
In analyzing complex conducting structures, we usually do not know in advance
with respect to the outward normal
conductor (E n >
0),
the orientation of the electric field intensity vector at specific portions of conducting
surfaces. In such cases, the following expression for
obtained noting that
En =
n
•
E
from Eq. (1.187),
ps
Example 1.24
An
=
£0
is
ps
in
terms of the
field vector,
useful:
n E.
(1.190)
•
Metallic Sphere in a Uniform Electrostatic Field
uncharged metallic sphere is brought into a uniform electrostatic field,
around the sphere after electrostatic equilibrium is reached.
in air.
Sketch the
field lines
Figure 1.40 Uncharged
Solution The field lines in the new electrostatic state are sketched in Fig. 1.40. Because the
due to induced charges on the sphere surface (this field exists both inside and outside
the sphere) is superimposed to the external field, the field inside the sphere becomes zero,
and that outside it is not uniform any more. Negative induced charges are sinks of the field
lines on the left-hand side of the sphere, whereas the positive induced charges are sources of
the field lines on the right-hand side. The field lines on both sides are normal to the sphere
surface, and they therefore bend near the sphere. At points in air close to the left- and
metallic sphere in a uniform
right-hand side of the sphere, the electric field
external electrostatic
the remaining space. This
field
for
Example
1
.24.
field;
sphere, in
tive
air,
is
is
stronger (the field lines are denser) than in
obvious as well from noting that near the left-hand side of the
the field due to negative induced charges dominates over the field due to posi-
charges on the opposite side of the sphere,
adds to the external
field intensity.
The
field
it is
due
directed toward the negative charges, and
to positive induced charges
the right-hand side of the sphere, which results in the
at these points in air.
is
The
field at
dominates near
same strengthening of the external
field
distances from the sphere a few times the sphere diameter
practically equal to the external field (the field
due to induced charges
is
negligible).
.
Evaluation of the Electric Field and Potential due to Charged Conductors
Section 1.17
43
EVALUATION OF THE ELECTRIC FIELD AND
POTENTIAL DUE TO CHARGED CONDUCTORS
1.17
Assume that we know the charge distribution ps over the surface of a conductor situated in free space. The electric field intensity at points close to the conductor surface
can be evaluated from Eq. (1.189). How do we obtain the electric field and potential
at an arbitrary point in space? The answer is straightforward. Because E = 0 inside
the conductor, nothing will change, as far as the field outside the conductor
cerned,
if
we remove
the conductor and
fill
the space previously occupied by
is
con-
it
with
a vacuum, keeping the charge distribution ps on the surface unchanged. With this
useful equivalence, we are left with the problem of evaluating the field and potential
due
known surface charge distribution
to a
(1.83), (1.101), (1.133),
Example 1.25
A
and (1.165)
in free space,
and we can use Eqs.
(1.38),
to solve the problem.
Charged Metallic Sphere
metallic sphere of radius a
is
situated in air and charged with a charge Q. Find (a) the
charge distribution of the sphere, (b) the electric
field intensity
vector in
air,
and
(c) the
potential of the sphere.
Solution
Due
(a)
to symmetry, the charge distribution over the sphere surface
is
uniform, and hence
the associated surface charge density turns out to be
/9s
_Q
So
_ Q2
(1.191)
'
Ana
where So stands for the surface area of the sphere.
(b)
The
around the sphere is radial, and has the form given by Eq. (1.136).
Eq. (1.133), to a spherical surface of radius r (a < r < oo), positioned concentrically with the metallic sphere [see Eqs. (1.137) and (1.138)], we obtain
electric field
Applying Gauss’
law,
E(r)
Note
Q
=
which
We
that
is
in
agreement with Eq.
due
to a point charge
sphere, the
same
Example 1.26
(1.189).
at
Q
r
(1.192)
oo).
=
a
is
Q
Ps
Ansoa 2
(1.193)
£o’
16
thus given by Eq. (1.141). This
any point of its interior and surface
Charged
Cylindrical
+ 0)
[E(a
+ 8),
8
is
is
is
identical to
the potential.
The
the potential of the
[see Eq. (1.182)].
Conductor
infinitely
-+ 0] designates the
electric field
due
metallic sphere
the charge per unit length of the conductor
£(a + ) or E(a
= a.
<
placed at the sphere center, and so
Repeat the previous example but for an
16
r
realize that the field outside the charged metallic sphere, Eq. (1.192),
potential at the surface r
air, if
<
that
E{a + )
(c)
(a
AnsQr2
is
long cylindrical conductor of radius a in
Q
field in air
very close to the conductor surface at
to a
charged
44
Chapter
1
Space
Electrostatic Field in Free
Solution
(a)
Using Eq.
(1.31), the
charge per length h of the conductor
Qh
=
is
Q'h,
(1
.
(1
.
194 )
so the surface charge density amounts to
Q'
= Qh = Qh ^
2 nah
So
where So
(b)
The
is
2na'
the surface area of that part of the conductor.
electric field
is
radial (with respect to the conductor axis).
[Eq. (1.133)] applied to the cylindrical surface of radius r(a
h, coaxial with the
electric field
cylindrical
due
to
195 )
<
r
<
From Gauss’ law
oo) and height (length)
conductor [see the left-hand side of Eq. (1.144)],
a charged
E(r)
1
Qh
Q'
2nrh
£q
2tre 0 r
=
(a
conductor
<
r
<
oo).
(1
.
196 )
is an infinite charge distribution, and the reference point for the potential cannot be
adopted at infinity. The field in Eq. (1.196) being identical to the field due to an infinite
line charge of density Q\ Eq. (1.57), the potential due to the charged cylindrical conductor is given by the expression in Eq. (1.87). In particular, this expression for r = a
This
(c)
represents the potential of the conductor:
cylindrical
Q'
V=
.
T/
potential due to a charged
In
2tt £q
conductor
where
r-n ( rji
>
m
—
,
(r
^
>
,
a)
^conductor
—
r
a)
is
In——
~
2ne o
A metallic sphere, of radius a and charge Q (Q
Solution
is air.
As
Find the potential
>
0), is
enclosed by an uncharged concentric
at the
As we know,
and outer surface of the
Qh and
shell, respectively.
Qc
tive
The medium
is
induced surface charge on the
denote the
charge on the inner surface of the
total
induced charges on the inner
On the surface of the sphere, Qa =
line of the electric field originating at a positive
surfaces
c).
there cannot be volume charges in the metal under electro-
Eq. (1.183). Let
static conditions,
< b <
center of the sphere.
a result of the electrostatic induction, there
surfaces of the shell.
(1.197)
a),
Shell
metallic spherical shell, of inner radius b and outer radius c (a
everywhere
—
the distance of the reference point from the conductor axis.
Charged Sphere Enclosed by an Uncharged
Example 1.27
{r
a
Q. Since every
charge on the sphere terminates at a nega-
shell, the relationship
between
total
charges on two
is
Qb = -Qa-
(1-198)
This can be obtained also from Gauss’ law, Eq. (1.133), applied to a closed surface that is
= 0 [because of Eq. (1.181)], which implies
entirely inside the metal of the shell, so that
that
Qs =
0, i.e.,
Q a + Qb — 0- O n the other hand, since the shell
Qh +
<2c
=
0,
is
uncharged,
(1.199)
Q c = —Qh = Qa
Because of symmetry, charges on individual surfaces are disis in the form given by Eq. (1.136).
Distributions of the charge and field in the system are sketched in Fig. 1.41.
By means of Gauss’ law, the electric field for a < r < b and c < r < oo is
and hence
tributed uniformly, and the electric field everywhere in air
E(r)
=
(in air),
-
(1.200)
47T£o''
whereas for 0 <
r
<
a and b
<
r
<
c,
E(r)
=
0
(in metal).
(1.201)
)
Section
Figure 1.41 Charge and
1 .1
7
Evaluation of the Electric Field and Potential
due
to
Charged Conductors
field
distributions in the system of
Example
The
.27.
1
potential at the point
V=
O in Fig.
E(r) d r
1.41
is
thus [Eq. (1.74)]
-O-fS-l + l)
=
\a
47reo
Q(bc — ac
b
c
+ ab)
(1.202)
4neoabc
Example 1.28
Consider
Five Parallel Large Flat Electrodes
five parallel large metallic electrodes situated in air, as
shown
in Fig. 1.42(a).
The
thickness of each electrode, as well as the distance between each two adjacent electrodes,
The surface area of sides of electrodes facing each other is S = 1 m 2 The first,
and fifth electrodes are grounded, the potential of the second electrode with respect
to the ground is V = 2 kV, and the charge of the third electrode is Q = 2 iiC. Find the
electric field intensity between the electrodes.
is
d
= 2 cm.
.
fourth,
Solution Since the dimensions of the electrodes are much larger than the separation
between them, we can neglect the fringing effects, i.e., we can ignore the nonuniformity of the
electric field
near the electrode edges, as well as the existence (“leakage”) of the
outside the spaces between electrodes.
in spaces
field in
field lines
We assume, thus, that the electric field is localized only
between the electrode sides S, that vector E is normal to those sides, and that the
is uniform. Charge distributions over electrode sides are also uniform. In
each space
ddddddddd
So
Figure 1.42 Electrostatic
analysis of a system of five large
geometry
and (b) charges
on electrodes and fields
between them; for
Example 1 .28.
electrodes
in air: (a)
of the system
(b)
45
46
Chapter
1
Electrostatic Field in Free
Space
each space between electrodes, neighboring sides of electrodes must be charged with charges
of equal amounts, but of opposite polarities, as indicated in Fig. 1.42(b).
The
potential
can be expressed
electrode, which
is
V
in
is
(which we know) of the second electrode with respect to the ground
terms of the line integral of
grounded
potential
(its
E from
The
zero).
is
that electrode, to the
field is
to the
left,
first
uniform, and the line integral
simply
first
electrode
v=
Edl =
-Eid.
(1.203)
./second electrode
Hence, the
field
— 100 kV/m. On
E to the right, to
between the
intensity
the other side, the
same
and second electrodes
first
the fourth electrode, which
a similar token, the field
same
— -V/d =
Ei,d.
between the fourth and
(1
fifth
electrodes
is
the associated charges on the sides of electrodes facing each other
electrodes are at the
E\
also grounded, yielding
is
V = E2d +
By
is
potential can be expressed as the line integral of
zero,
(Q 4
£4
— 0),
=
0,
.204)
and so are
because these
(zero) potential.
£ 3 we need one more equation with them as
unknowns. Applying Gauss’ law, Eq. (1.133), to the surface So that encloses the third
electrode (the charge of which is known), we obtain
To
solve for field intensities £2 and
,
-£2 + £3 = —z-
(1.205)
e0 S
The
solution of the system of equations
-62.94
kV/m
and £3
=
162.94
composed from Eqs.
Problems'. 1.75-1.82; Conceptual Questions (on
MATLAB
1.18
Exercises (on
it,
Because there
bounded by
shell cavity
is
is
no
field
field
Faraday cage.
in
Companion Website):
£2
=
1.20-1.22;
in the external electrostatic field
throughout the sphere
outside the sphere.
We
a metallic shell (Fig. 1.43). This
perfectly protected (isolated)
thickness of the shell can be arbitrary, and
Figure 1.43 Metallic shell
an electrostatic field -
is
ELECTROSTATIC SHIELDING
without affecting the
field,
and (1.205)
Companion Website).
Let us consider again the metallic sphere
Fig. 1.40.
(1.204)
kV/m.
interior,
thus obtain a
means
in
domain with no
that the space inside the
from the external
its
shown
we can remove
electrostatic field.
The
shape does not need to be spherical.
Hence, an arbitrary closed conducting shell represents a perfect electrostatic shield
or screen for its interior domain. We call such a shield a Faraday cage. If the field
outside the cage is changed, the charge on the cage walls will redistribute itself so
that the field inside will remain zero.
We see that a Faraday cage provides an absolute protection to its interior from
an external electrostatic field. Let us now reverse the problem. Can an electrostatic
field be encapsulated by a metallic shell so that the domain outside the shell is protected from the sources inside it? The answer to this important question is twofold.
Let us get to it by analyzing two simple examples.
Consider first a single positive point charge Q positioned arbitrarily inside an
uncharged spherical conducting shell. The distribution of induced charges on the
shell surfaces and the field lines are sketched in Fig. 1.44(a). The total induced
charges on the inner and outer shell surfaces are —Q and Q, respectively (see
Section 1.18
Electrostatic Shielding
Figure 1.44 Single point
charge
(a)
and two point
charges amounting to a zero
total
(b)
(a)
Example
1.27).
The concentration of induced negative charges
of the inner surface closer to the point charge. Because there
on the outer
wall, the positive charge
shell surface
is
is
is
cage.
higher on the side
no
field in
the shell
distributed irrespective of the
means uniformly in this case
smooth and symmetrical).
Let us now add another point charge in the cavity, and let it be exactly — Q as
shown in Fig. 1.44(b). The total negative and total positive induced charges on the
inner surface of the shell are both less than Q in magnitude, because some field lines
originating at the positive point charge end at the negative one inside the cavity,
but mutually they are equal in magnitude and opposite in polarity. The net induced
charge on the inner surface is therefore zero, implying that there is no charges whatsoever on the outer shell surface. This means, in turn, that there is no field outside
position of the point charge inside the cavity, which
(the surface
is
,
the shell, which
is
also in
agreement with Gauss’
law, applied to a spherical surface
enclosing the shell.
We
conclude that a Faraday cage can completely encapsulate an interior elecif the total charge inside the cage is
trostatic field, with a zero field outside, only
zero. This
is
true for any interior charge distribution, provided that the object or
the system of objects (devices) inside the cage
as a whole. In cases
when
is
electrically neutral
the total interior charge
is
(uncharged)
not zero, the exterior domain
induced charges on the outer surface of
independent of the distribution of
interior sources. Its relative distribution in space depends only on the shape of the
outer cage surface, whereas its absolute values at individual points in space are also
(and neighboring objects)
the cage.
The
is
in the field of
exterior field, however,
is
totally
proportional to the total amount of interior charge.
It is
interesting to note that
trostatic shields, either in the
one
illustrated in Fig. 1.44(b).
even very thin metallic
mode
shells represent ideal elec-
of operation illustrated in Fig. 1.43 or in the
However,
as
we
shall see in a later chapter, this
is
not
where the effectiveness of a shield of
a given thickness depends on the metal conductivity and on the rate with which the
necessarily the case with time-varying fields,
field varies in
time (that
is,
frequency in the case of time-harmonic
Conceptual Questions (on Companion Website): 1.23 and
1.24.
fields).
charge (b)
in a
Faraday
7
48
Chapter
1
Space
Electrostatic Field in Free
CHARGE DISTRIBUTION ON METALLIC BODIES OF
1.19
ARBITRARY SHAPES
body of an arbitrary shape, the charge
body surface is not uniform. The determination of this distribution for a given body with nonsymmetrical and/or nonsmooth surface is a rather
complex problem. In this section, we shall get some qualitative insight about how
In the general case of a charged metallic
distribution over the
the charge
is
distributed over the surface of an arbitrarily shaped isolated conduct-
we shall present a general numerical method for
determining approximately the charge density function over conducting objects.
ing body, and in the next section
Consider a system composed of two charged metallic spheres of different radii,
whose centers are a distance d apart, in free space. Let the spheres be connected by a very thin conductor, as shown in Fig. 1.45. Assume, for simplicity, that
d
a,b, so that the electric potential of each sphere can be evaluated as if the other
one were not present. In addition, we assume that the charge along the connecting
a and b,
»
conductor
is
zero, because the conductor
is
very thin, and ignore
its
1.25
and Eq.
(1.141)]
Qa
y‘ =
and
Anega
respectively,
on
Example
influence
the field between the spheres. Therefore, the potentials of spheres are [see
Qa
where
and
Qb
(1.206)
4
The
are the associated total charges of spheres.
spheres being galvanically connected together, and thus representing a single conducting body, which must be equipotential, Eq. (1.182), these potentials are the
same,
Va = Vb
So, by equating the expressions in Eq. (1.206),
Qa _ a
~ b
Qb
Using Eq.
(1.191), the sphere charges
0 20
.
-
we
obtain
(1.208)
’
can be expressed
ing surface charge densities, with which Eq. (1.208)
charge density
oc
surface
Psa
b
Psb
a
)
in
terms of the correspond-
becomes
(1.209)
curvature
and, from Eq. (1.193), the corresponding relationship between the electric field
intensities near the surfaces of spheres turns out to be
local field intensity
a
Ea
surface
Eb
curvature
We
=
b
(1
see from Eqs. (1.209) and (1.210) that the charge
two spheres
in Fig.
Figure 1.45
Two
1
.45 in
same
potential.
210 )
is
distributed
between the
such a way that the surface charge density on and electric
metallic
spheres of different
.
a'
radii at
the
2
<
,
Jfhb
Qb
,
,
Section
field intensity
sphere radius.
1
.20
Method
of
Moments
for
Numerical Analysis of Charged Metallic Bodies
49
near the surface of individual spheres is inversely proportional to the
The surface charge is denser and the field stronger on the smaller
sphere.
The importance of Eqs. (1.209) and (1.210) is much beyond the particular system in Fig. 1.45. They imply a general conclusion that the surface charge density and
the nearby field intensity at different parts of the surface of an arbitrarily shaped
conducting body are approximately proportional to the local curvature of the sur17
This means, generally, that the largest concentration of
face, as long as it is convex.
charge and the strongest electric
Note
we
that this
phenomenon
is
field are
shall see in the next chapter.
Problems'. 1.83; Conceptual Questions (on
1
around sharp parts of conducting bodies.
essential for the operation of lightning arresters, as
.20
Companion Website):
1.25-1.27.
METHOD OF MOMENTS FOR NUMERICAL ANALYSIS
OF CHARGED METALLIC BODIES
Consider a charged metallic body of an arbitrary shape situated in free space. Let
the electric potential of the body with respect to the reference point at infinity be
Vo- Our goal is to determine the charge distribution of the body. The potential at
an arbitrary point on the body surface, S, can be expressed in terms of the surface
charge density, ps over the entire S [Eq. (1.83)]. On the other hand, this potential
equals Vo (the body is equipotential), and hence
,
Ps
4nso
This
is
L
dS
=
R
Vq
(at
an arbitrary point on S ).
surface integral equation for
(1 .21
1 )
charge distribution
an integral equation with the function p s over S as unknown quantity, to be
determined.
Eq. (1.211) cannot be solved analytically - in a closed form, but only numerwith the aid of a computer. The method of moments (MoM) is a common
numerical technique used in solving integral equations such as Eq. (1.211) in electromagnetics and in other disciplines of science and engineering.
can be
implemented in numerous ways, but the simplest
solution in this case implies
subdivision of the surface S into small patches A 5), i = 1, 2,
N, with a constant
approximation of the unknown function ps on each patch. That is, we assume that
each patch is uniformly charged,
ically,
MoM
MoM
.
Ps
~ psi
(on
A Si),
t
=
.
.
l,2,
(1
.
212 )
piece-wise constant
approximation for charge
With
this
approximation,
we reduce Eq.
N
E=
(1.211) to
dS
=
its
approximate form:
Vo,
f
in which the unknown quantities are N charge-distribution coefficients, p
i
PsN-
17
If
Shown
in Fig. 1.46
is
we
JA
A Si 4ne 0 R
an example of the application of
the surface of a conducting body
incurvature,
1
is
this
concave (curved inward), the effect
method
is
s i,
pS 2
,
•
•
•
to a metallic
deep
and a decrease of the
just opposite; for a
actually have a partial effect of a Faraday cage (cavity), Fig.1.43,
local field intensity.
(1.213)
8
50
Chapter
1
]
Space
Electrostatic Field in Free
yyyy
Psl
Ps2
AS,
as2
•
•
•
Figure 1.46 Method of
moments (MoM) for analysis
charged metallic bodies:
of
approximation of the surface
charge distribution on a cube
AS,-
by means
of
N small
square
patches with constant charge
densities.
cube, where square patches are used [for the particular subdivision
figure,
N=6
By
x
(5
x
5)
=
150,
which
is
shown
in the
a rather coarse model].
stipulating that Eq. (1.213) be satisfied at centers of every small patch,
individually,
we
obtain 18
Anpsi +-4i2Ps2
H
A 21 Psi + ^22Ps2
H
+ A iN psN =
h MnPsN =
Vo
(at the center of
AS,),
Vo
(at the center of
AS 2 ),
(1
+A N2 p 2 H
AmPst
This
is
\-A NN psN
S
a system of
N
= V0
(at the center of
linear algebraic equations in
N
.
214 )
A S N ).
unknowns.
Psl
5
Ps2>
•
•
•
>
PsN-
In matrix form.
MoM matrix equation
[A][ps ]
where
[p s ]
a
is
=
[B],
Elements of matrix
r
A id =
JASi
patches
.
215 )
column matrix whose elements are the unknown charge-distribution
elements of the column matrix B are known and all equal Vo-
coefficients, while
point-matching at centers of
(1
[
[A],
which
is
a square matrix, are given by
ds
4n£ 0 R
(at the center of
AS*),
k,i
=
1,2,
(1
.
216 )
and they can be computed irrespective of the particular charge distribution.
Physically, Aid is the potential at the center of patch AS* due to patch AS, that is uni-
C/m 2
) surface charge density. In the case of commonly
used square or triangular flat patches, this potential can be evaluated analytically
(exactly), while it is evaluated numerically (approximately) if some other surface
elements are used (for example, curvilinear quadrilateral or triangular patches).
Once matrix [A] is filled, i.e., all its elements computed, we can use matrix inversion,
formly charged with unit
(1
[Ps]
=
[A]- 1 [B],
(1
.
217 )
method for solving systems of linear algebraic equations
Gaussian elimination method), to obtain the numerical results for the chargedistribution coefficients, which constitute an approximate surface charge distribution of the body, and a numerical solution to the integral equation, Eq. (1.211).
or any other standard
(e.g.,
method of moments in which the left-hand side and the right-hand side of an integral
“matched" to be equal at specific points of the definition domain
of the equation (surface S in our case) is called the point-matching method. The idea of point-matching is
similar to the concept of taking moments in mechanics, and hence the generic name method of moments.
1
The
variant of the
equation
[in
our
case, Eq. (1.21 l)j are
l
Section 1.21
The
number of subdivisions, N, the more accurate (but more computademanding in terms of computer resources) solution.
The crudest approximation in computing the elements of matrix [A] is given by
larger the
tionally
_ ASi/(4ne 0 R ki
~ ( ^ASi/a^eo)
\
)
Aki
for
for
k^i
k=
U^
'
i
all nondiagonal elements of [A] ( k ^ i ) are evaluated by approximating the
2
charged patch AS, by an equivalent point charge, A <2, = 1 (C/m ) x AS/, placed
at the patch center, and using the expression for the electric potential due to
a point charge in free space, Eq. (1.80), with Rki being the distance between
centers of patches A S^ and AS/ (Fig. 1.46). In filling diagonal elements (self
terms) of [/l] (k = i), the potential due to a (square or triangular) patch AS/
at the center of that same patch is evaluated by approximating the patch by
the equivalent circular patch of the same surface area and radius y/ A Sj/jr, and
employing the potential expression given in Problem 1.34 with z = 0 and ps =
Here,
1
C/m 2
.
from the result for [ps ], we can now obtain any other quantity of interand field at any point in space, etc.). For instance, the total charge
of the body can be found using the approximate version of the surface integral
Starting
est (potential
expression in Eqs. (1.29):
N
Q=
J^Psi AS/.
(1.219)
;=
Problems 1.84-1.86;
:
1.21
MATLAB Exercises (on Companion Website).
IMAGE THEORY
Often, electrostatic systems include charge configurations in the presence of
grounded conducting planes. Examples are charged conductors near grounded
metallic plates or large flat bodies, transmission lines in which one of the conductors is a ground plane (such as microstrip transmission lines), various charged
objects (power lines, charged clouds, charged airplanes, lightning rods, etc.) above
the earth’s surface, and so on. There is a very useful theory (theorem) by means
of which we can remove the conducting plane from the system, and replace it by
the equivalent charge distribution in free space. In this section,
this
theorem, and apply
it
we
shall derive
to problems which otherwise could not be analytically
solved.
Consider two point charges of equal magnitudes and opposite polarities, Q and
in free space. By symmetry, the total electric field intensity vector due to the
charges is normal to the plane of symmetry of the charges (the tangential components due to individual charges are of equal magnitudes and opposite directions,
so they cancel each other out), as shown in Fig. 1.47(a). The plane of symmetry is
—Q,
and at the potential V — 0
same magnitudes and opposite
due to individual charges
and they cancel out). Hence, nothing will change in the entire space if we insert an infinite grounded metallic foil
(conducting plane) in the plane of symmetry, as is done in Fig. 1.47(b), because
the boundary condition in Eq. (1.186) is automatically satisfied and V = 0 over the
plane of symmetry (the foil is grounded). In the new system, surface charges will be
induced on both sides of the foil, according to Eq. (1.190). We note that the foil actually separates the entire space onto two completely independent half-spaces, i.e., it
equipotential,
(the potentials
are of the
signs,
Image Theory
51
r
52
Chapter
1
Electrostatic Field in Free
,
Space
Figure 1.47 Deriving image theory: systems
(a), (b),
and
(c) are
equivalent with respect to the electric
field in
the upper half-space.
between the two half-spaces.
change in the upper half-space if we, furthermore,
remove the point charge — Q and the induced charge on the lower side of the foil
from the system. (Note that we can put whatever we want below the foil, and the
field above it will remain the same, in the electrostatic state.) We are thus left with
the point charge Q above the foil and the induced charge on its upper side (and
nothing in the lower half-space), as depicted in Fig. 1.47(c).
We conclude that, as far as the electrostatic field in the upper half-space is concerned, systems in Figs. 1.47(a) and (c) are equivalent. This is so-called image theory,
which, generalized to more than one point charge, i.e., to a (discrete or continuous)
charge distribution, states that an arbitrary charge configuration above an infinite
grounded conducting plane can be replaced by a new charge configuration in free
space consisting of the original charge configuration itself and its negative image in
the (former) conducting plane. The equivalence is with respect to the electric field
above the conducting plane, whose component due to the induced charge on the
acts as a perfect electrostatic screen (see Section 1.18)
Because of
(a)
plane
is
that,
equal to the
Example 1.29
A
nothing
point charge
Q
will
field of the
image.
Induced Charge Distribution on
is
a
Conducting Plane
placed in air at a height h above a grounded conducting plane, (a)
Determine the density of induced surface charges
the total induced charge on the plane.
at
an arbitrary point on the plane, (b) Find
Solution
(a)
(b)
Figure 1.48 Computation
of the induced surface
charge density, pS( on a
conducting plane
underneath a point charge
Q
:
(a) original
The surface charge density ps
at a point
M
on the conducting plane,
in Fig. 1.48(a). is
given by Eq. (1.190), with the unit vector n being vertical and directed from the plane
upward, and E representing the electric field intensity vector in air, at a point that is
M
from its upper side. This field is produced by the point charge Q
and the induced surface charges on the conducting plane. According to the image theory,
however, the field due to the induced charges equals that due to the negative image of
Q in free space, as shown in Fig. 1.48(b). Let the position of the point be defined by a
radial distance r from the projection of Q on the plane (point O). Vector E is given by
“glued” to the point
M
system and
(b) equivalent system using
image theory; for
Example 1 .29.
E—
Eq =
Eoriginai
Q
4jr eoR 2
+
Ej ma
ge
R=
— 2Eq cosa
yj
2
+ h2
,
(
n)
cos a
= —
A
(1
.
220 )
53
Problems
and the charge density comes out to be
ps (r)
= e 0 n-E =
2tt (r2
(b)
The
induced charge on the conducting plane
total
2ind
= J ps (r)2nrdr =
as
Qh
rdr
f
~&~~ Qh
Lo
(1.221)
2
+ h2 )
is
f
00
dR
_
~ Qk
1=0
00
1
R
= -G,
(1.222)
r=0
where dS is the surface area of an elemental ring of width dr and radius r (0 < r < oc)
around the point O (see Fig. 1.14), and the use is made of Eq. (1.62) to change variables
in integration. The result in Eq. (1.222) is expected, because all field lines terminating on
the image, — Q, in the equivalent system [(Fig. 1.48(b)] terminate on the surface charges
of the conducting plane in the original system [Fig. 1.48(a)].
Example 1.30
An
Infinite Line
Charge above
charge of uniform density Q'
infinite line
conducting plane at a distance h from
unit of
its
it.
is
a
Conducting Plane
situated in air
Compute
and
is
parallel to a
grounded
the electric force on the line charge per
length.
Solution Under the influence of the electric field of the line charge, surface charges are
induced on the conducting plane. The electric force on each meter of the line charge is
therefore [Eq. (1.68)]
K = Q'E
where
E2
(1.223)
2,
represents the electric field vector at points along the line charge due to the induced
charges on the plane.
By image theory, this field is equal to the field due
to a line charge in free
space obtained as a negative image of Q', Fig. 1.49. and hence the following for
and the per-unit-length
[see Eq. (1.57)]
E2 =
Q
2nso(2h)
its
intensity
force:
2
= Q
’
K = Q’E
2
(1
.224)
4jT£()h
Figure 1.49 Force on a
charge above
(the distance
between the
original
and the image
is
2 h).
The force
is
plane; for
attractive.
a
line
conducting
Example
1
.30.
Problems: 1.87-1.89; Conceptual Questions (on Companion Website): 1.28-1.30;
MATLAB Exercises (on Companion Website).
Problems
1.1.
Three unequal charges in a triangle. Repeat
Example 1.1 but assuming that one of the three
charges in Fig. 1.3(a) amounts to (a) 3 Q and (b)
—3 Q,
1.2.
respectively.
Three charges in equilibrium. The distance
between point charges Q\ = 36 pC and Qi =
9
pC
system are in the electrostatic equilibrium,
that the resultant Coulomb force on each
charge is zero.
this
i.e.,
is
placed
D = 3 cm.
at the line
If
O 3 is
Q 2 at a
the third charge,
connecting Q\ and
Qi
D
,
,
shown in Fig. 1.50, find
03 and d which ensure that all the charges in
distance d from Qi, as
Q3
Qi
Figure 1.50 Three point charges along a
line;
for
Problem
1
.2.
/
54
1.3.
Chapter
Electrostatic Field in Free
1
Space
Four charges at rectangle vertices. Four small
charged bodies of equal charges Q = — 1 nC
are placed at four vertices of a rectangle with
sides <7 = 4 cm and b = 2 cm. Determine the
direction and magnitude of the electric force on
1.11.
face charge
Five charges in equilibrium. Four small
charged balls of equal charges Q\ — 5 pC are
positioned at four vertices of a square, whereas
the fifth ball of unknown charge Qi is at the
square center. Find Q2 such that all the balls
1.6.
sum
of
the three forces,
all
1.12.
Q
—Q
Compute
1.13. Field
1.14. Point
circle.
A fifth charge
the electric force
is
E z (z), —00 <
q
and x
is
the length coor-
at the axis of a ring, (a)
Q
>
0,
For
and
electric field intensity along
maximum,
z < 00
(b) Plot the function
.
charge equivalent to a charged semi-
Show
that far
away along the
z-axis,
the semicircular line charge in Fig. 1.12(a)
is
equivalent to a point charge with the same
amount of charge located
on the top
at the
coordinate
origin.
charged bodies of equal charges Q exist at the
vertices of a cube with sides of length a, in free
space. Find the magnitude and direction of the
Charged contour of complex shape. Fig. 1.51
shows a contour consisting of two semicircular
parts, of radii a and b (a < b ), and two linear parts, each of length b
situated in air
on one of the charges.
due
=
a constant
which the
the z-axis
1.15.
electric force
is
charge along a rod. A rod of
charged with a line charge of
Q' [ 1 — sin(7rx//)] (0 < x < /),
maximum
find z for
Eight charges at cube vertices. Eight small
1.8. Electric field
line
in air is
the charged ring in Fig. 1.11, assume
i
charge.
1.7.
The charge density
dinate along the rod. Calculate the total charge
mid. All sides of the pyramid have the same
length, a.
/
density Q'(x)
positioned at the top vertex of the pyra-
is
Nonuniform
where Q'0
are positioned in air at the corners
of the square base of a pyramid.
sur-
—
ps {r)
length
Fe + F e 2 + Fe 3 ?
Five charges at pyramid vertices. Four point
charges
A
of the rod.
Three point charges in space, (a) For the three
charges from Example 1.3, find the resultant
electric force on the charge Q2 (F e 2 ). (b)
Determine the force F e 3 (on Q3). (c) What is
the
disk.
distributed in free space over a
2
Pso^/a (0 < r < a), where r is the
radial distance from the disk center, and p s 0 is
a constant. Obtain the total charge of the disk.
is
are in the electrostatic equilibrium.
1.5.
is
circular disk of radius a.
each of the bodies.
1.4.
Nonuniform surface charge on a
and
distributed along
tric field intensity
to three point charges in
— a. The
carries a charge
its
length.
contour
is
Q uniformly
Compute
the elec-
vector at the contour center
(point O).
from Example 1.3,
determine the magnitude and direction of the
space. For the three charges
electric field intensity vector at (a) the coor-
dinate origin and (b) the point at the z-axis
defined by z
1.9.
=
100 m.
Nonuniform volume charge
infinitely
is
in a cylinder.
An
long cylinder of radius a in free space
charged with a volume charge density p{r)
—
po(a
and
r
r)
a (0
<
the radial
<
—
where po is a constant
distance from the cylinder axis.
r
a),
Figure 1.51 Uniformly charged
contour with two semicircular and
two linear parts; for Problem 1.15.
Find the charge per unit length of the cylinder.
1.10.
Nonuniform volume charge
in a cube.
of edge
space
length
a
in
free
is
A cube
charged
volume with a charge density p(x) =
sin(7TJc/a),
0 < x < a, where po is a constant
Po
and x is the normal distance from one of
over
its
the cube sides.
the cube.
Compute
the total charge of
1.16.
Nonuniform
line charge along a semicircle.
Consider the geometry in Fig. 1.12(a), and
assume that the charge along the semicircle is nonuniform, given by Q\4>) = Q' s\n4>
< tt/2), where Q' is a constant,
(— n/2 <
()
<f)
(a)
0
Find the total charge of the semicircle.
55
Problems
(b)
Prove that the
charged over
electric field intensity vector
along the z-axis equals
E=
2
—Q'qCi y/[8eo(z
2
+
a 2 ) 3 / 2 ].
1.17.
is
3/4 of a circle of radius a
The
total
charge of the arc
is
situated in
total charge of the sphere
field intensity
charged sheet with a circular hole. An
charge with a constant density p s has a hole of radius a in it. The sheet
is in the xy-plane of the Cartesian coordinate system and the center of the hole is at
the coordinate origin. The ambient medium
is air. Under these circumstances, determine
the electric field intensity vector at an arbitrary point along the z-axis - in the following
A
of a circle of radius
Determine the
a.
elec-
vector at an arbitrary point
along the axis that contains the arc center and
is
normal
to the arc plane.
1.19. Semi-infinite
A
charge.
line
uniform charge density Q'
line
is
two ways, respectively: (a) integrating the fields
due to elementary rings as in Fig. 1.14 and (b)
combining the results of Examples 1.11 (infinite sheet of charge, with no hole) and 1.10
charge of
distributed in
space along the negative part of the
free
x-axis
in
0).
system
coordinate
Cartesian
the
(— oo < x <
(charged disk).
Find the expression for the
electric field intensity vector at
vector at the sphere center.
infinite sheet of
Line charge along a quarter of a circle.
charge of density Q' in free space is distributed
uniformly along an arc representing a quarter
tric field intensity
(a) the
(b) the electric
1.25. Infinite
electric field intensity vector at the arc center.
1.18.
Compute
and
air.
Q. Calculate the
is
surface such that the charge
is
defined as in Fig. 1.10 or 1.16.
A
Line charge along three-quarters of a circle.
uniform line charge in the form of an arc that
is
its
given by p s (0) = ps o sin 20, where p s o
a constant and the angle 0 (0 < 0 < n) is
density
an arbitrary
1.26.
point in the xy-plane.
Force on a charged semicylinder due to a line
charge. For the structure composed from a line
charge and a charged semicylinder shown in
1.20. Half-positive, half-negative infinite line charge.
A line charge in free space
is
Fig. 1.17(a)
distributed along
and described
in
Example
1.13, find
the force per unit length on the semicylinder.
the x-axis in the Cartesian coordinate system.
The
line
— oo <
charge density
>
Q' (Q'
is
x < 0 and —Q' for 0 < x <
oo.
0)
for
1.27.
Derive
the expression for the electric field intensity
vector at a point
where d >
1.21.
M with coordinates
problem
as in Fig. 4.11 in
0.
Find the electric
is
Consider an infinitely long unistrip of width a and surface
charge density p s in air. Using the geometristrip.
cal representation of the cross section of the
(0, d, 0),
field intensity
at a distance a
given by Eqs. (4.43) and (4.44), obtain the
expression for the E field at an arbitrary point
in space
vector at a point
from each of the square
Chapter 4 (also see
and change of integration variables
Fig. 1.13)
Charged square contour. A line charge of uniform charge density Q' is distributed along a
square contour a on a side. The medium is air.
that
Charged
formly charged
1.28.
Two
due to
this charge.
parallel oppositely charged strips.
parallel,
vertices.
with charge densities ps and
1.22. Point
charge equivalent to a charged disk.
Consider the charged disk in
show
is
that for
|z|
a,
2
ps 7ta placed
1.14,
and
Eq. (1.63)
of a point charge Q =
the
equivalent to the field
E
Fig.
field in
(ps
0).
shown
the same
The
—ps
,
respectively
cross section of the structure
in Fig. 1.52.
The width of the
as the distance
and the medium
is air.
between them ( a
Find the
is
strips is
—
d),
electric field
at the disk center.
a
due to a nonuniformly charged disk.
Consider the disk with a nonuniform charge
1.23. Field
distribution
>
Two
very long strips are uniformly charged
from Problem
1.11,
and
find the
Ps
electric field intensity vector along the disk axis
normal
to
its
•
plane.
A
d
Figure 1.52 Cross section
two parallel, very long
of
1.24.
Nonuniformly charged spherical surface. A
sphere of radius a in free space is nonuniformly
_
f
I
charged strips; for
Problem 1.28.
a
56
^
Chapter
Space
Electrostatic Field in Free
1
intensity vector at the center of the cross sec-
1.29.
in an electrostatic field. What is the work
done by electric forces in moving a charge Q =
1 nC from the coordinate origin to the point
(1 m, 1 m, 1 m) in the electrostatic field given
m)
=
2
—
in the
(x x
the straight line joining the
1.30.
V/m
z)
surface
center of the nonuniformly charged spherical
1.37.
M
marked
2
in
1.38.
Sketch
field
potential
M,
V
from
in a
in Fig. 1.54.
The
potential.
region
point
electrostatic
a function of a sin-
is
gle rectangular coordinate x,
P~
Two
Q\—l /rC
and Q 2 = —3 q,C, are
located at the two nonadjacent vertices of a
square contour a = 15 cm on a side. Find the
voltage between any of the remaining two vertices of the square and the square center.
Find the work done by
charge Qi = — 1 nC
the figure.
Voltage due to two point charges.
charges,
electric forces in carrying a
from the point Mi to the point
due to a nonuniform spherical surface
surface from Problem 1.24.
A
in Fig. 1.53.
0).
charge. Determine the electric potential at the
two points?
in the field of a point charge.
point
charge Q\ = 10 nC is positioned at the center of a square contour a = 10 cm on a side,
shown
=
1.36. Potential
(x, y, z in
Work
as
(z
+y y
Cartesian coordinate system along
y, z)
hemispherical
a
to
Consider the hemispherical surface
charge from Example 1.12, and find the electric scalar potential at the hemisphere center
charge.
Work
by E(x,
due
1.35. Potential
tion (point A).
V (x)
and
Sketch the electric
is
shown
field intensity
1
Ex {x)
&
i
Figure 1.53
\
a
Q2
of a charge
q
field of a
in this region.
Movement
in
the
charge Q\
positioned at the center
of a square contour; for
M2
a
1.31. Electric potential
due
Problem 1.30.
to three point charges in
space. For the three charges
from Example
1.3,
calculate the electric potential at points defined
by
(0,0,2 m)
(a)
and
(b)
(1
m,
1
m,
1
m),
respectively.
Figure 1.54 1-D potential distribution; for
1.32.
Point charge and an arbitrary reference point.
Derive the expression for the potential at a distance r from a point charge Q in free space
with respect to the reference point which is
an arbitrary (finite) distance rji away from the
Problem
1.39. Field
charge.
1.33.
1.34.
Q'
/ (4eoV z
2
+ a2
1.40. Field
following expression
for
potential along the e-axis
+ z2 -
|z|)/(2e 0 )-
the
electric
(—00 <
z
scalar
< 00 ): V
=
from
V
in
potential, charged semicircle. For
(a) obtain the expression for
Ez
Example
in
1.7,
Eq. (1.50)
from the expression for V given in Problem
1.33 and (b) explain why it is impossible to
obtain the expression for Ex in Eq. (1.48) from
this same expression for V.
).
due to a charged disk. For the
charged disk from Example 1.10, derive the
2
obtain the expression for
the semicircular line charge from
Potential
Ps(%/«
in free space,
Eq. (1.24) from the expression for
Eq. (1.80).
Potential due to a semicircular line charge.
Prove that the electric scalar potential at
an arbitrary point along the e-axis in the
field of the semicircular line charge shown in
Fig. 1.12(a) and described in Example 1.7 is
V=
in
.38.
from potential, point charge. For a point
charge
E
1
1.41.
from potential, charged
charged disk from Example
Field
the expression for
expression for
E
V given
disk.
For the
1.10,
obtain
in
Eq. (1.63) from the
in
Problem
1.34.
57
Problems
from potential, charged hemisphere.
For the hemispherical surface charge from
Example 1.12, explain why we cannot obtain
the expression for E at the hemisphere center
(z = 0), given in Eq. (1.67), from the expres-
1.42. Field
sion for
1.43.
V computed in Problem
1.49.
in
1.50.
1.35.
due to a
line
the expression for
V
Eq. (1.121).
Near and
far potential
Two
dipole.
Q
and
field
due to a
infinite line charges,
line
with densi-
100 pC/m and Q'2 = —100 pC/m, are
positioned along lines defined by (1 m, 0, z) and
are perpendicular to equipotential surfaces (as
coordinate
in Fig. 1.22).
Calculate
ties
h(x,y )
lOOxlny [m] (x, y in km), where x and
y are coordinates in the horizontal plane and
1 km < x, y < 10 km. (a) What is the direction
km,
of the steepest ascent at (3
steep, in degrees,
Maximum
The
3
km)?
(b)
(—1 m,
a
maximum
Q\
1
1.52.
0), respectively.
Flux for a different placement of the point
charge. If the point charge
ous problem
is
Q
from the previ-
placed at the center of a side
of the electric field vector due to the charge
m,
1
electric dipole.
Two
nC and Q 2 — — 1 nC,
at points
through the surface composed of the remaining
point
are
five sides of the cube.
sit-
along the z-axis
of a point charge from Gauss’ law. Using
Gauss’ law, derive the expression for the elec-
1.53. Field
tric field intensity
vector at the point defined by Cartesian
(0, 0, 0),
(100 m, 100 m, 100 m),
and
by
(b)
The cube edges are a
outward flux of the electric field
intensity vector due to this charge through each
of the cube sides.
m
1.47. Potential
and
long. Find the
defined by z = 1
and z = — 1 m, respectively.
Compute the electric potential and field inten-
coordinates (a)
0, 0)
of the cube, determine the total outward flux
uated in free space
sity
at the point defined
m, —1 m).
Large and small
=
E
increase in the electric scalar
potential at a point (1
charges,
and
,
A
by E(x, y, z) = (4 x — z 2 y +
2yzz)V/m (x,y,z in m). Find the direction
of the
V
< 00 in the Cartesian
The medium is air.
of a cube in free space.
How
intensity vector in
z
system.
(100 m, 100 m,
given
is
—00 <
1.51. Flux of the electric field vector through a cube
side.
point charge Q is located at the center
increase in electrostatic potential.
electrostatic field
region
0, z),
Cartesian coordinates (a) (2 m,
the ascent in (a)?
is
=
x
given by a function
is
=
field
(b) (0,
1
m,
0),
and
(c)
vector of a point charge in
free space [Eq. (1.24)].
1.54.
respectively.
Uniformly charged thin spherical
shell.
An
infinitely thin spherical shell of radius a in free
due to a small
space
electric
An electric dipole with a moment
= 1 pCm z is located at the origin of a spheri-
uniformly charged over
is
its
surface with
dipole.
a total charge Q. Determine: (a) the electric
p
field
cal
is
coordinate system. The length of the dipole
d
=
1
cm. Find
V
and
E
m, n/2, n/2), (c) (1 m,
(d)
m,
jt/
(e)
(1
(10 m, zr/4, 0),
4,0),
(f) (100 m, jr/4, 0).
n, 0),
nonuniform
line
(1
m,
0, 0), (b) (1
Dipole
equivalent
to
a
1.55.
far
and
charge density given by Eq. (1.32).
an infinite line charge from Gauss’ law.
Using Gauss’ law, derive the expression for the
electric field intensity vector of an infinite line
1.56. Field of
from Problem
away along the z-
(|z|»a) this charge distribution can
be replaced by an equivalent electric dipole
moment,
at
the
coordinate origin.
p, of the equivalent dipole.
and outside the sphere with the volume
inside
charge in free space [Eq.
axis
located
Sphere with a nonuniform volume charge. Find
the distribution of the electric scalar potential
distribution along the semicircle
and show that
and outside the
the potential of the shell, and (c) the
potential at the shell center.
charge. Consider the nonuniform line charge
1.16,
intensity vector inside
shell, (b)
at the following
points defined by spherical coordinates: (a)
1.48.
field
Angle between field lines and equipotential
surfaces. Using the concept of gradient, prove
that in an arbitrary electrostatic field, field lines
elevation in a region
1.46.
E from
the expression for
1.44. Direction of the steepest ascent. The terrain
1.45.
Expression for the electric
dipole. For the line dipole in Fig. 1.29, obtain
Find the
1.57.
(1.57)].
Uniformly charged thin cylindrical shell. An
infinitely long and infinitely thin cylindrical
shell of radius a
shell
is
is
situated in free space.
charged over
its
The
surface with a uniform
58
Chapter
Electrostatic Field in Free
1
Space
charge density ps Find the electric
.
sity
1.58.
field inten-
1.66.
vector inside and outside the shell.
with
Cylinder
Compute
volume
uniform
charge.
the voltage between the surface and
is
the axis of a uniformly charged infinite cylinder
of radius a in free space,
density in the cylinder
is
if
the
volume charge
1.67.
of an infinite sheet of charge from Gauss’
Using Gauss’ law, derive the expression for
the electric field intensity vector of an infinite
the
1.68.
Charge from
field, cylindrical
symmetry. From
component
Two
only given by Eqs. (1.145) and (1.146), obtain
the corresponding charge distribution in free
charged sheets.
parallel oppositely
—p
s
Two
space [Eq. (1.143)].
are situated in free space, (a) Find
between the sheets,
the voltage between the sheets?
side the space
(b)
What
1.69.
Charge from
field,
spherical symmetry. Using
Gauss’ law in differential form, show that the
field with a radial spherical component only
is
layer of charge in free space has a uniform
given by Eq. (1.140) is produced by a uniformly
charged sphere of radius a and charge density
volume charge density p and thickness
p in free space.
Equivalent sheet of charge.
An
infinitely large
d. (a)
Compute
the electric field vector inside the
layer, (b)
Show
side the layer
is
1.70.
that, as far as the field out-
concerned, the layer can be
charge, and find the surface charge density, p s
of this sheet.
,
Layer with a cosine volume charge distribution.
density of a volume charge in free space
depends on the Cartesian coordinate x only
and is given by p{x) = pocosinx/a) (|x| < a)
and p(x) — 0 (|jc| > a), where po and a (a > 0)
are constants, (a) Determine the electric field
intensity vector in the entire space, (b) Find the
voltage between planes x — —a and x = a.
The
the differential Gauss’ law.
1.71.
layer).
space
A
is
volume charge
nate system as p(x)
0,
distribution in free
described in the rectangular coordi-
and p(x)
—
poQ
x ^a
= — po e - *? 0 for x
for
>
0,
x <
0,
p(0)
=
with po and a
being positive constants. Calculate the electric
field intensity
Uniform
is
vector for
— oo < x <
oo.
volume charge density
in that
What
region?
symmetry by differen-
1.72.
Problem with planar symmetry using differenGauss’ law. Redo Problem 1.61 employing
tial
differential Gauss’ law.
Antisymmetrical charge, differential Gauss’
law. Redo Example 1.21 applying differential
Gauss’ law.
1.74.
Gauss’ law in differential and integral form. In
is given by
a certain region, the electric field
(4xyx
-(-
2x 2 y
+
z)
V/m
(x,
y
in
m). The
medium is air. (a) Calculate the charge density,
(b) From the result in (a), find the total charge
enclosed in a cube situated
dinate octant (x,y,z
>
0),
in the first coor-
with one vertex at
the coordinate origin, and the edges, of length
1
m, parallel to coordinate
axes, (c)
Confirm
the validity of Gauss’ law in integral form and
electric field. In a certain region, there
a uniform electric field, Eo-
cylindrical
Gauss’ law. Redo Example 1.19 but with the
use of Gauss’ law in differential form.
E=
Exponential charge distribution in the entire
space.
Problem with
tial
1.73.
Layer with a sine charge distribution. Repeat
the previous problem but for the following
charge density function: p(x) = po sin(7rx/o)
for |jc| < a (there is no charge outside the
Nonuniformly charged sphere using differenGauss’ law. For the nonuniform volume
charge distribution in a sphere defined by
Eq. (1.32) and analyzed in Problem 1.55 (based
on Gauss’ law in integral form), compute the
electric field intensity vector everywhere using
tial
replaced by an equivalent infinite sheet of
1.65.
From
the field with a radial cylindrical
the electric field intensity vector inside and out-
1.64.
planar symmetry.
field,
expressions in Eqs. (1.150)— (1.152), obtain
sheet of charge in free space [Eq. (1.64)].
p s and
1.63.
Charge from
the corresponding charge distribution in free
parallel infinite sheets of charge with densities
1.62.
.
space [Eq. (1.147)].
law.
1.61.
e0
field
p.
1.59. Field
1.60.
Charge distribution from 1-D field distribuFind the volume charge density p(x) in
the electrostatic system from Example 1.16,
assuming that the permittivity of the medium
tion.
is
the
the divergence theorem by evaluating the net
outward flux of E through the surface of the
cube defined in (b).
59
Problems
1.75.
Excentric charged sphere inside an unchar-
ged
Consider
shell.
Example
the
and assume that the sphere
1.27,
moved toward
unit length of the
first
and the third conductor,
Q\ and
from
structure
is
the shell wall so that the centers
of the sphere and the shell are separated by a
distance d. Find the potential of the shell in the
new
(b)
electrostatic state
=
d
b
—
is
=
d
(a)
if
a (the sphere
(b
—
a)/2 and
pressed against the
shell wall).
charge inside a charged
1.76. Point
2Q
charge
shell.
A
point
placed at the center of an
is
air-
spherical metallic shell, charged with
filled
and situated
in air.
The inner and outer
<
of the shell are a and b (a
b). (a)
Q
radii
What
is
the total charge on the inner and on the outer
surface of the shell, respectively? (b) Find the
Figure 1.55 Detail of the cross section of a system
potential of the shell.
of four cylindrical conductors; for
1.77.
Three concentric
shells,
concentric spherical metallic shells are situ-
ated in
is
a
=
air.
30
The outer radius of the inner
mm, and
its
Q—
charge
1.80.
1
.79.
10 nC.
Three concentric conductors, one grounded.
Shown
shell
in
Fig.
1.56
is
a
system consisting
of three concentric spherical conductors (the
The
inner and outer radii of the middle shell are
inner conductor
= 50 mm and c = 60 mm, and its potential
V = 1 kV with respect to the reference point at
remaining two are spherical shells). The radius
of the inner conductor is a — 2 mm. The inner
radius of the middle conductor is b — 5 mm,
b
infinity.
The inner and outer
shell are
d
=
90
radii of the outer
mm and e =
100
mm, and
and (b) the voltage between the inner
and the outer shell.
dle shell
tial.
etry
shells,
two
at the
that the charges of the inner
are
Q\
= 2 nC
and
and outer
Q 3 = —2 nC,
= V3
).
Under
c
— 6 mm.
ground are V\ — 15 V and V2 = 10 V,
respectively. Determine total charges of the
inner and middle conductors, Q\ and 02to the
shells
respectively,
as well as that their potentials are the
(V\
a solid sphere, while the
the inner and middle conductors with respect
same potenConsider a structure with the same geomas in the previous problem, and assume
Three concentric
is
The inner radius of
the outer conductor is d = 8 mm. The space
between the conductors is air-filled. The outer
conductor is grounded, and the potentials of
and outer
it is
uncharged. Calculate (a) the charge of the mid-
1.78.
Problem
one uncharged. Three
same
these circumstances, com( Q 2 ) and
and the middle
pute (a) the charge of the middle shell
(b) the potentials of the inner
shells (V\
point at
1.79.
and
V2 ) with respect to the reference
infinity.
Figure 1.56 System
Four coaxial cylindrical conductors. Four very
long conductors, each in the form of a cylindrical shell with thickness d= 1 cm, are posi-
of three concentric
spherical
conductors; for
tioned in air coaxially with respect to each
other, as indicated in Fig. 1.55,
detail of the cross section of the system.
first
The
and the fourth conductor are grounded,
and the potential of the third conductor with
respect to the ground is V 3 = 1 kV. The second
conductor
is
Problem
1
.80.
which shows a
uncharged. Find the charges per
1.81.
Charged metallic
metallic foil
is
foil.
An
infinitely large flat
situated in air
and charged
uni-
formly with the surface charge density ps =
2
1 nC/m
Find the electric field intensity vector
everywhere.
.
N
60
Chapter
1 . 82 .
Two
Electrostatic Field in Free
1
metallic slabs.
An
slab of thickness d =
1
Space
infinitely large metallic
cm
is
situated in air
1.86.
and
charged such that the surface charge density
2
at each of the slab surfaces is p s = 1 p C/m
field intensity
.
Fig. 1.46),
3 cm. In the
new
all
four surfaces of the slabs, (b)
electric field intensity vector
(c) the
1 . 83
.
dicular to
everywhere, and
and
metallic spheres at the same potential.
Consider the system in Fig. 1.45, and assume
that a — 5 cm, b = 1 cm, and d = 1 m, as well
as that the total charge of the two spheres
600 pC. Find
(a) the potential of the
spheres and (b) the electric
field intensities
1.87.
Ea
1.88.
plate into
a
=
1
m
N
Subdivide the
square patches, and assume that
and Vq
=
1
V. (a) Tabulate and plot
85
.
100). (c)
image. Find
its
Q
in
Imaging a
For the structure defined
determine the distribution
of induced surface charges on the conducting
line charge.
Example
1.30,
plane.
1.89.
Charged wire
shows
Fig. 1.57
parallel
to
corner screen.
a
a cross section of the structure
consisting of a metallic wire of radius a
90° corner metallic screen in
air.
and a
The distance
of the wire from both the horizontal and ver-
each dimension), (b) Compute the total charge
where h
(i) N = 9,
N — 25,
N = 49, and (iv) N = 100, respectively.
MoM computation for a charged cube. Write a
of the plate, taking
.
N—
the results for the surface charge density (p s )
of the patches, taking
100 (ten partitions in
N=
1
plane at points that are a/2, 2a,
Force on a point charge due to
in
determine the charge distribution on a
very thin charged square plate of edge length a
at a potential Vq, in free space.
its
the electric force on the point charge
sented in Section 1.20, write a computer proto
approx-
Fig. 1.48(a).
MoM-based computer program for a charged
plate. Using the method of moments as pregram
is
center.
and Eb near the surfaces of the spheres.
1 . 84 .
the cube in
100fl, respectively, distant
surface (for
Two
is
(e.g.,
distribution
from the plate
Also compute the
electric field inside the cube from the previous problem (Problem 1.85), at a quarter of
its space diagonal (body diagonal) and at its
voltage between the slabs.
Q—
whose charge
program, compute the electric field along the
axis of the plate from Problem 1.84 perpen-
elec-
trostatic state, calculate (a) the surface charge
densities at
vector at an arbitrary point in
imately described by Eq. (1.212). (b) Using the
expression in (a) and the associated computer
distance between the surfaces of the two slabs
D—
Write the approximate inte-
space due to a charged body
placed parallel to the charged slab such that the
is
integral expression for the elec-
gral expression for the evaluation of the electric
Another metallic slab of the same thickness,
which is uncharged, is then introduced and
facing each other
Approximate
tric field vector, (a)
computer program
(ii)
for the
(iii)
tical
half-planes constituting the screen
»
a. If
the wire
is
is
h,
charged with Q' per
unit length, calculate the voltage
between the
wire and the screen.
method-of-moments
analysis of a charged metallic cube, Fig. 1.46,
=
m, and compute the
= 1 V and ten,
or as many as possible (given available computational resources), subdivisions per cube
= 600 if ten subdivisions per edge are
edge (
section of a charged
adopted).
for
with edge length a
total
1
charge of the cube for Vq
Figure
1
.57 Cross
metallic wire parallel to a
metallic corner screen;
Problem
1
.89.
Dielectrics, Capacitance,
and Electric Energy
r
Introduction:
n
chapter,
this
I fields
we
shall
analyze electrostatic
vector and the electric potential due to polarized
and study
dielectrics using free-space formulas and techniques
from the previous chapter. Gauss’ law will be generalized for an electrostatic system that includes
arbitrary media (conductors and dielectrics), and
in the presence of dielectrics
several important related topics. Dielectrics or insulators are
little
nonconducting materials, having very
free charges inside
them
(theoretically, per-
have no free charges). In addition,
redistribution of any free charges (e.g., electrons)
fect dielectrics
deposited inside the material
lasts
much
longer
than in metallic conductors, a typical example, as
already indicated in Section 1.16, being the charge
rearrangement time of ~ 50 days for fused quartz
compared to ~ 10~ 19 s for copper. We shall see,
however, that another type of charges, called bound
or polarization charges, exist in a polarized dielectric.
We
shall first investigate the
mechanisms of
the polarization of dielectrics, caused by an external electric field.
By
introducing the macroscopic
quantities such as the polarization vector,
and the
volume and surface density of bound charges,
is
it
possible to evaluate the electric field intensity
characterization of dielectric materials in terms
of their linearity, homogeneity, and isotropy will
be presented. Dielectric-dielectric boundary conditions will be derived and used. The chapter will also
introduce Poisson’s and Laplace’s second-order
differential equations for the potential and their
solution.
Having both conductors and dielectrics, we
then put them together to form capacitors and
related electrostatic systems. The capacitor is a fundamental element in electrical engineering. Its basic
shall
property
itors
is its
capacitance.
We
shall analyze capac-
with electrodes of different shapes and with
different types of dielectrics,
and evaluate capac-
itance per unit length of various two-conductor
61
62
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
we
transmission lines as well. In addition,
shall
of
tric
conducting bodies,
operation of the structure prior to
an eventual breakdown.
ties for the safe
and other systems
capacitors, transmission lines,
and introduce the
maximal permissible extents of quanti-
line) define
evaluate the electric energy contained in charged
The
elec-
energy density to help us find and quantify
electrostatic
transmission
lines, to
analysis of capacitors and
determine their capacitance,
the localization and distribution of the energy in
energy, and
such systems. Dielectric breakdown, occurring for
a culmination of the theory of the electrostatic
exceedingly strong electric fields in a dielectric
field. It
material causing
it
to
become conducting,
many
will also
be discussed for various electrostatic structures. We
shall analyze structures with electric fields close to
breakdown
breakdown
represents,
characteristics, represents
on the other hand,
a
gateway to
practical applications of this theory. Finally,
a clear understanding of concepts that will be pre-
sented in this chapter
levels in order to predict critical val-
is
essential for
many
similar,
ues of voltages and other circuit quantities for the
and analogous concepts in other areas of electromagnetics, which are to be introduced later in
structure at breakdown. Such parameters
the text.
breakdown voltage of
(e.g.,
dual,
the
a capacitor or a transmission
POLARIZATION OF DIELECTRICS
2.1
Each atom or molecule
in a dielectric is electrically neutral.
For most
centers of “gravity” of the positive and negative charges in an
coincide - in the absence of the external electric
When
field.
dielectrics,
atom or molecule
a dielectric
is
placed in
E ex
t. however, the positive and negative charges shift
in opposite directions against their mutual attraction, and produce a small electric
dipole (Fig. 1.28), which is aligned with the electric field lines. The moment of this
equivalent dipole is given by p = Qd [Eq. (1.116)], where Q is the positive charge of
the atom or molecule (-Q is the negative charge), and d is the vector displacement
an external
of
field,
of intensity
Q with respect to — Q. The charges are displaced from their equilibrium positions
by forces
Eel
respectively,
and thus
=
£?E ex
and
t
Fe2 = — £?E ex t,
(2.1)
-
Q shifts in the direction of E ext
,
while
—Q moves in the oppo-
so that p and E ex t are collinear and have the same direction. The
displacement d is very small, smaller than the dimensions of atoms and molecules.
site direction,
Q and — Q
bound
by atomic and molecular forces and can
field. So, the two charges in
an equivalent small dipole cannot separate one from the other and migrate across
the material in opposite directions run by the electric field. Hence, these charges are
The charges
only
are
shift positions slightly in
called
bound charges
Some
in place
response to the external
(in contrast to free charges).
such as water, have molecules with a permanent displacement between the centers of the positive and negative charge, so that they act as
small electric dipoles even with no applied electric field. Such molecules are known
as polar molecules, and the dielectrics are called polar dielectrics (those with no
dielectrics,
built-in dipoles are
nonpolar
dipoles are oriented in a
electric dipole,
is
dielectrics). In the
random way.
If
absence of the electric
a polar molecule, which
field, all
the
we model by an
brought into an electric field, however, the forces on the two dipole
The torques (moments) of
charges, given in Eq. (2.1), act as indicated in Fig. 2.1.
forces with respect to the center of the dipole (point
Tj
=
ri
x F e]
and
T2 =
O)
r2
are
x
F e2
,
(2.2)
Section 2.2
ri
and
r2
center.
We
notice that
with
63
Polarization Vector
Q and — Q with respect to the dipole
and hence the resultant torque on the dipole
denoting the position vectors of
r\
—
t2
= d,
turns out to be
Ton
dipole
— Tj + T2 —
Q{*\
— *2)
X
E ex — Q d
t
X Eext
—
p X
E e xt»
(2.3)
torque on an electric dipole
an external
electric field
where we assume that E ex is practically uniform along the dipole. Vector T on dipole
is normal to the plane of p and E ex t (the plane of drawing in Fig. 2.1), and its
magnitude amounts to
t
Ton
We
dipole
—
|P
x
E ex — P Eext sina.
(2.4)
t
|
see that the torque given by Eq. (2.3) tends to rotate the dipole about the
through the dipole center and being normal to the dipole and the plane
axis passing
about the vector Ton dipole- The action of such torques in the dielecrandom intermolecular thermic forces, and is to align the dipoles, to
some extent, in the same direction - toward the field lines. We also see that the
stronger the field, the larger Ton dipole and the larger the component of the resultant
of drawing,
tric is
i.e.,
against
>
moment of all the molecules, J2 P> along the direction of E ex A sufficiently
strong field may even produce an additional displacement between the positive and
Figure 2.1 Polarization of
negative charges in a polar molecule, resulting in a larger p.
a polar molecule in an
dipole
t-
We
conclude that both an unpolar and polar dielectric in an electric field
can be viewed as an arrangement of (more or less) oriented microscopic electric
The process of making atoms and molecules in a dielectric behave as dipoles
and orienting the dipoles toward the direction of the external field is termed the
polarization of the dielectric, and bound charges are sometimes referred to as polarization charges. This process is extremely fast, practically instantaneous, and the
dielectric in the new electrostatic state is said to be polarized or in the polarized
state. For almost all materials, the removal of the external electric field results in
the return to their normal, unpolarized, state. A very few dielectrics, called electrets, remain permanently polarized in the absence of an applied electric field (an
example is a strained piezoelectric crystal).
dipoles.
Conceptual Questions (on Companion Website):
2.2
POLARIZATION VECTOR
When
polarized (by an external electric
and the
field),
2.1
and
2.2.
a dielectric
is
a source of
its
own
an arbitrary point in space (inside or outside the
dielectric) is a sum of the external (primary) field and the field due to the polarized
dielectric (secondary field). To determine the secondary field, we replace the dielectric by a collection of equivalent small dipoles, which can be considered to be in a
vacuum, as the rest of the material does not produce any field.
Theoretically, we could use the expression for the electric field due to an electric
dipole, Eq. (1.117), and obtain the field due to a polarized dielectric by superposition. However, as many atoms or molecules in a dielectric body that many
equivalent small dipoles in it, and, with the “microscopic” approach to the evaluation of the field due to the polarized dielectric, we would need to consider every
single dipole, which is practically impossible [there is on the order of as many as
10 30 atoms per unit volume (1 m 3 ) in solid and liquid dielectrics].
We rather adopt a “macroscopic” approach, and introduce a macroscopic quantity called the polarization vector to describe the polarized state of a dielectric and
electric field,
total field at
polar dielectrics:
model
external electric field.
of
in
64
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
the resulting
average dipole
moment
We first average dipole moments in
field.
P)in dv
in
(2.5)
^in dv
an elementary volume of a
polarized dielectric
and then multiply this average by the concentration of dipoles
of atoms or molecules in the dielectric), which equals
/v,
What we
get
is,
by
=
(i.e.,
concentration
(2.6)
dv
definition, the polarization vector:
P—
polarization vector (unit:
C/m
an elementary volume dv,
P)in dv
Nv Pav
(2.7)
dv
2
)
Note that P would represent the resultant dipole moment in a unit volume (1
it were polarized uniformly (equally) throughout the volume. Note also that
Pdv= (]C p
) indl
is
the dipole
moment
ized dielectric,
i.e.,
to
m3
)
if
(2- 8 >
,
of an electric dipole equivalent to an element dv of the polarall
the dipoles within
The
1
it.
unit for
P
C/m 2
is
In any dielectric material, the polarization vector at a point
is
.
a function of the
(total) electric field intensity vector at that point,
P =
For linear
Xe
-
electric susceptibility of
P(E).
(2.9)
(in the electrical sense) materials, this relationship
P=
a
is
linear,
i.e.,
Xe^oE,
(
2 10 )
.
linear dielectric
where Xe
is
the electric susceptibility of the dielectric.
It is
a pure number,
i.e.,
dimensionless quantity, obtained by measurements on individual materials, and
always nonnegative (x e
2.3
We
>
0).
For a vacuum, Xe
= 0, whereas Xe ^ 0 for air.
BOUND VOLUME AND SURFACE CHARGE
shall
of excess
now
DENSITIES
derive the expressions for calculating the macroscopic distribution
bound charges
body from a given distribution of
obtained by averaging the microscopic
the dielectric material. These expressions will be used in the next section
in a polarized dielectric
the polarization vector, P, which, in turn,
dipoles in
a
is
is
for free-space evaluations of the electric field
due
to polarized dielectrics.
Let us first find the total bound (polarization) charge Q p s enclosed by an
arbitrary imaginary closed surface S situated (totally or partly) inside a polarized
dielectric body, as
shown
in Fig. 2.2.
Knowing
of a vast collection of small electric dipoles,
1
An
elementary volume dv, as we use
sense,
and cannot be
vector, for instance, that
“on average,” but yet
it
infinitely small in a
means
that
dv
in macroscopic electromagnetic theory, is small in a physical
mathematical sense. Within the definition of the polarization
enough to contain many small dipoles to be treated
P can be considered constant in dv from the macroscopic
number (millions) of atoms or molecules.
is
large
sufficiently small so that
point of view. Such dv
still
contains a vast
bound charge actually consists
each dipole being composed from a
that
Bound Volume and Surface Charge
Section 2.3
Figure 2.2 Closed surface S
a polarized dielectric
and a negative —Q, we realize that all the dipoles that appear inside S
Q and — Q, as well as dipoles that are totally outside S, contribute with zero net charge to Q s Only dipoles whose one end is inside S (and the
v
other end outside S ) contribute actually to the total bound charge in S. (We notice
right away that Q $ = 0 when S encloses the entire dielectric body.) To evaluate
p
positive
Q
with both their ends,
QpS
we therefore count the dipoles that cross the
we count the contribution of such dipoles as either Q or — Q
(in the general case),
In doing that,
Q, generally, differ
from dipole
to dipole),
surface S.
(note that
by inspecting which end of the dipole
is
inside S.
Consider an element dS of S and the case when the angle a between the vector
p av ) and vector dS, which is oriented from S outward, is less than 90°,
as depicted in Fig. 2.3(a). Note that negative ends of dipoles that extend across dS
with one (negative) end inside S are in a cylinder with bases dS and height
P
(or vector
h
so that the
number
= dcosa,
(2.11)
of these dipoles equals the concentration of dipoles,
Nw
,
times
dv = dSh. The dipole ends on the inner side of d S
being all negative, and with an assumption that all dipoles in dv are with the same
moments and charge, the corresponding bound charge is given by
the
volume of the
cylinder,
dQ p =
In the case
when a >
Nw dSdcosa(-Q)
(0
<a <
90°).
90°, portrayed in Fig. 2.3(b),
h
=d
cos (7T
—
a)
=
d (— cosa),
(2.1 3)
and, because the ends of dipoles on the inner side of dS are
dQ p =
(2.12)
jVv
dSd(— cos a) Q
which turns out to be the same
result as in
Eq.
(90°
<a <
(2.12).
all
positive,
180°),
(2.14)
body.
Densities
in
65
66
Chapter 2
Dielectrics, Capacitance,
and
Electric
Energy
Figure 2.3 Element of surface 5
in Fig. 2.2, in two cases with
regards to the angle a between
P and
(b) 90°
dS: (a) 0
< a <
<a <
90°
and
180°.
For unpolar
where not
all
dielectrics,
dipole
p av
moments
=
and P
p,
= Nv p = Nv Qd.
are mutually parallel,
we can
For polar
dielectrics,
consider the dipoles in
moments p av =
Hence, for an arbitrary a (0 < a < 180°), we have
small cylinders in Figs. 2.3(a) and (b) to be equivalent dipoles with
Qd, so that P
=
7Vv pav
= Nv Qd.
dQp = — NvQddS cos a = — PdScosa = — P dS
(2.15)
•
(note that the boundary case, a
we
total
bound
=
90° and d
Q p = 0, is
Qp
formula). Finally, by integrating the result for d
this
also properly included in
over the entire surface
S,
obtain
=
QpS
(polarization)
P
•
dS.
(2.16)
charge enclosed by a closed
surface
S
This
is
an integral equation similar
form to Gauss’
in
law,
Eq. (1.133).
It tells
us that
the outward flux of the polarization vector through an arbitrary closed surface in
an electrostatic system that includes dielectric materials is equal to the total bound
(polarization) charge enclosed by that surface, multiplied by —1.
Eq. (2.16) is true for any closed surface S. Let us now apply it to the surface S
enclosing an elementary volume Av inside a polarized dielectric:
-
ptin Av
(Qp)
LP
dS
(Av
Av
Av
(2.17)
0),
The expression on the
volume
bound charge,
represents the density of excess
with both sides of the equation being also divided by Av.
left-hand side of Eq. (2.17)
Pp
while the expression on
—
(Gp)in Av
(2.18)
Av
by definition [Eq.
of the divergence of the polarization vector. Hence,
bound volume charge
its
right-hand side
PP
density
If
P
=
in
a
uniformly polarized dielectric
— — div P = —V
•
(1.172)], the negative
P.
(2.19)
const inside the dielectric (uniformly polarized dielectric),
derivatives of
no volume bound charge
is,
P
all
spatial
are zero, and using Eq. (2.19),
P
=
const
PP
=
o.
(
2 20 )
.
.
Bound Volume and Surface Charge
Section 2.3
67
Densities
Figure 2.4 Elementary closed
surface used for deriving the
boundary condition for the
P on a surface
vector
dielectric-free space, Eg. (2.23).
If
P^
volume bound charge exists only if the polarizavolume of the dielectric (nonuniformly polarized
const, however, then excess
tion vector varies throughout the
way that its divergence is nonzero, otherwise p p = 0.
On the surface of a polarized dielectric, there always exists excess surface bound
dielectric) in a
charge (there are ends of dipoles pressed onto the surface that cannot be compensated by oppositely charged ends of neighboring dipoles), except on parts of the
where P and the dipoles are tangential to the surface. To determine the
bound (or polarization) surface charge density, pps we apply Eq. (2.16)
to a small pillbox surface, with bases AS and height Ah (Ah -> 0), shown in Fig. 2.4.
There is no polarization in free space (a vacuum or air),
surface
associated
,
P=
so that the flux of vector
S,
and we have
P
in
Eq. (2.16)
is
(2.21)
0,
reduced to P AS over the lower side of
Eq. (1.188)]
no polarization
in
a vacuum
or air
[also see the similar derivation in
pps AS
= -P
AS.
(
With Ad denoting the normal unit vector oriented from the
AS = — ASAd, which yields
Pps
—
Ad
•
dielectric
2 22 )
.
body outward,
bound
P.
(2.23)
fid
surface charge density;
outward normal on a
dielectric surface
This
is
the boundary condition for the vector
P on
a surface dielectric-free space,
connecting the polarization vector in the dielectric near the boundary surface
and the bound surface charge density on the
component of P contributes to p ps
surface.
Note
that only the
normal
.
Example
The
2.1
Nonuniformly Polarized Dielectric Cube
polarization vector in a dielectric cube
shown
P(x,y)
in Fig. 2.5
is
given by
= P0 ^x,
1
(2.24)
a
where Pq
is
a constant.
The surrounding medium is air. Find the distribution of bound charges
of the cube.
Solution Using Eqs. (2.19) and (1.167), the bound volume charge density inside the cube
Pp
-
dPx
T0 y
is
(2.25)
dx
whereas Eq. (2.23) tells us that bound surface charge
and its density amounts to
exists
on the front
side of the
cube
only,
Figure 2.5 Dielectric cube
with polarization P(x,
Pps
=
x-P(fl-,y)
=
(2.26)
Example
2.1
y); for
68
Chapter 2
Dielectrics, Capacitance,
with P(n
_
and
Electric
Energy
,y) denoting the polarization vector in the dielectric very close to the
the back side, p ps = 0 since P(0 + y)
0 because fid and P are mutually perpendicular.
surface at x
=
On
a.
= 0,
,
pps =
Note that, with v designating the volume of the cube and S
sides
Q P = f Ppdv+ (fpps dS = [ f — —&y-') a 2 dy + f
Jv
Jy = 0 \
JS
Jy = 0
/
as expected (the total
bound charge of a
dielectric
body
is
boundary
while on the remaining four
its
boundary
—
ady
surface,
= 0,
(2.27)
&
always zero), where, in accordance
with our general integration strategy explained in Section
the cube of thickness dy, and
dS
a strip of width dy
1.4, dv is adopted to be a
on the front cube side.
slice
of
Problems'. 2.1.
EVALUATION OF THE ELECTRIC FIELD AND POTENTIAL
DUE TO POLARIZED DIELECTRICS
2.4
In this section,
we
scalar potential
due
assume that the
state of polarization of a dielectric
shall evaluate the electric field intensity vector
and
electric
to polarized dielectric bodies in several characteristic cases.
body
is
We
described by a given dis-
From P, using Eqs. (2.19) and
volume and surface bound charge densities,
p p and p ps throughout the body volume and over its surface, respectively. Then,
we calculate the field E and potential V (and any other related quantity of interest) using the appropriate free-space formulas and equations [Eqs. (1.37), (1.38),
(1.82), (1.83), (1.133), etc.] and solution techniques suitable to specific geometries
and source distributions.
tribution of the polarization vector, P, inside the body.
(2.23),
we
first
find the distribution of
,
Example 2.2
A
Uniformly Polarized Dielectric Disk
dielectric disk of radius a
polarized throughout
its
magnitude being
its
and thickness d
is
situated in free space.
The
disk
is
uniformly
volume, the polarization vector being normal to the disk bases and
P. Find (a) the distribution of
bound charges of
the disk and (b) the
electric field intensity vector at the disk center.
Solution
(a)
Eq. (2.20) tells us that there is no bound volume charge inside the disk. According to
Eq. (2.23) and Fig. 2.6, the bound surface charge densities on the upper and lower disk
bases are
Ppsi
= fidi p = p
and
Pp S 2
-P,
(2.28)
on the side disk surface, p ps 3 = fid 3 P = 0.
due to the polarized disk equals the field due to two circular sheets
of charge with densities p ps i and p pS 2 in free space. We use the expression for the field
due to a circular sheet of charge (thin charged disk) in Eq. (1.63) and the superposition
principle to add up the fields due to two sheets, Ej and E 2 The charge densities are
p s = ±P and the distance from each sheet at the disk center (point O) is d/2, so the two
fields are the same, and the total field comes out to be
respectively, while
(b)
= fid2 p =
The
•
electric field
.
E=
E]
+ E2 =
2Ei
'
eo
(2.29)
'
.
L
2 s/a
2
+
(d/2) 2
Evaluation of the Electric Field
Section 2.4
and
Potential
due
P
P
^dl'
p
e0
Ppsl
fid3
<
»o
Pps3
£0
fi
A
charge on
d2
dielectric sphere of radius a, in free space,
Compute
P.
is
a
uniformly polarized
dielectric disk; for
Example
2.2.
Uniformly Polarized Dielectric Sphere
Example 2.3
vector
Figure 2.6 Bound surface
a
Pps2
scalar potential at the sphere center,
and
is
uniformly polarized, and the polarization
bound charge of the sphere,
(a) the distribution of
(c) the electric field intensity
(b) the electric
vector at the sphere
center.
Solution
(a)
Let us adopt a spherical coordinate system with the origin
z-axis parallel to the vector P, as
pp
=
0.
The bound
the angle 6,
We now replace
E
surface charge density at a
(b)
= nd P =
•
-Pcos Z(n d P)
,
From Eq.
(1.83)
and
is
because the
Due
i
total
a
bound charge of the
to symmetry, vector
is
0
<
6
<
n.
the function of 0 given in Eq. (2.30), and
Fig. 2.7, the potential at the point
4nSp Js
which
= Pcos<9,
(2.30)
compute
V and
sphere center (point O) using free-space concepts and equations.
F=
(c)
sphere center and the
the polarized sphere by a nonuniform spherical sheet of charge in free
whose charge density
at the
at the
The bound volume charge density is
point M on the sphere surface, defined by
in Fig. 2.7.
is
pps
space,
shown
computed
E
/
i
(2.31)
4nsoa J s
O
4ire 0 a
^
sphere, <2 p
at the point
essentially in the
O turns out to be
,
is
zero.
in Fig. 2.7 has a (negative)
same way
surface into thin rings and integrating the fields
z-component
only,
as in Fig. 1.16, subdividing the sphere
dE due to individual rings, the only two
is now a function of 9, Eq. (2.30), and
differences being that the surface charge density
1
Figure 2.7 Dielectric sphere
with a uniform polarization; for
Example
2.3.
to Polarized Dielectrics
69
t
70
Chapter 2
Dielectrics, Capacitance,
that the
and
upper
Electric
Energy
limit in the integration
E=(f>dE — —^—
is
now
=
9
n. With
p ps (9) sin# cos# d9 z
[
this,
=
Eq. (1.67) becomes
f
2eo Je = o
Js
electric field inside
'
2eo Jo
—
cos
2
——
0sin0d0z
-''
'
-t
a uniformly
(2.32)
polarized dielectric sphere
where the substitution given by u = cos 9 is used to solve the integral in 9. It can be
shown that E has this same (constant) value at any point inside the polarized sphere.
Nonuniformly Polarized
Example 2.4
Dielectric
Sphere
A nonuniformly polarized dielectric sphere, of radius a, is situated in free space. In a spherical
is
coordinate system whose origin coincides with the sphere center, the polarization vector
given by the expression
=P
V{r)
r
{r)i
= Po-
(2.33)
a
Pq is a constant). Determine (a) the bound volume and surface charge densities of the sphere
and (b) the electric scalar potential inside and outside the sphere.
(
Solution
(a)
Using the expression for the divergence in spherical coordinates, Eq.
volume charge density of the sphere amounts to
(it is
same
the
at all points inside the sphere).
= r P(0 = P0
pps
(b)
The
field
>
outside the sphere (for r
equivalent point charge
a)
is
is
= 0,
V(r)
=
V(r)
—J
is
a
Exercises (on
dielectrics.
in free
Gauss' law
for
conductors
and
a system with
dielectrics
<
r
<
,
0
(2.36)
oo.
is
<
given by
r
<
(2.37)
a.
1
E(/) dr
=
®
Companion
-
r
—
a
-
(2-38)
Website): 2.3 and 2.4;
Companion Website).
We now consider the most
and
placed at the sphere center.
thus
GENERALIZED GAUSS'
2.5
identical to the field of the
it is
eoa
Problems'. 2.2-2. 6; Conceptual Questions (on
MATLAB
is
also zero,
^ = -—
3eo
potential inside the sphere
bound
(2.35)
.
zero, because
(1.140), the electric field inside the sphere
E(r)
The
surface charge density
Q p = 0 (total charge of the sphere)
Hence, the potential outside the sphere
From Eq.
The bound
(1.171), the
LAW
general electrostatic system containing both conductors
The equivalent
field
sources are
space, and Gauss’ law, Eq. (1.133),
Qs + Qps
E dS =
•
l
now both
free
and bound charges,
becomes
£0
(2.39)
71
Characterization of Dielectric Materials
Section 2.6
where Qs and Q v s are the total free charge and the total bound charge, respectively,
enclosed by an arbitrary closed surface S. Multiplying this equation by £q, moving
Q ps to the left-hand side of the equation, then substituting it by the negative of
the flux of the polarization vector, P, from Eq. (2.16), and finally joining the two
integrals over S into a single integral, we obtain the equivalent integral equation:
E
j> (£ 0
To shorten the
we
writing,
+ P)
new
define a
= Qs
dS
•
(2.40)
.
vector quantity,
D = £qE T P,
which
is
(2.41)
called the electric flux density vector (also
ment vector or
electric flux
known
(unit:
as the electric displace-
electric induction vector). Accordingly, the flux of
(symbolized by
electric flux density vector
D
C/m 2 )
termed the
is
fl>),
41
=
f
D
•
dS,
C)
(2.42)
electric flux (unit:
(2.43)
generalized Gauss' law
JS'
where
any designated surface (open or closed). In place of Eq.
S' is
D
•
dS
= Qs
(2.40),
.
is an equivalent form of Gauss’ law for electrostatic fields in arbitrary media,
which is more convenient than the form in Eq. (2.39) because it has only free charges
on the right-hand side of the integral equation, and not the bound charges, and thus
This
is
simpler to use.
states that the
referred to as the generalized Gauss’ law, and, in words,
It is
outward
electric flux
system including conductors and dielectrics equals the total
the surface.
density,
D,
From Eq.
is
C/m2
(2.43), the unit for the electric flux
is
it
any electrostatic
free charge enclosed by
through any closed surface
in
C, so that the unit for
its
.
In the general case, free charge
is
represented by means of the volume charge
density, p, yielding
generalized Gauss' law
(2.44)
in
terms of the volume charge
density
with v denoting the volume bounded by
less of the
choice of
v,
S.
Since this integral relation
is
true regard-
the divergence theorem, Eq. (1.173), gives the differential
form of the generalized Gauss’ law:
V-D = p.
(2.45)
generalized differential
Gauss' law
Problems'. 2.7-2.11; Conceptual Questions (on
2.6
Companion Website):
2.5.
CHARACTERIZATION OF DIELECTRIC MATERIALS
The polarization properties of materials can be described by the relationship
between the polarization vector, P, and the electric field intensity vector, E,
Eq. (2.9). We now employ the electric flux density vector, D, and substituting
Eq.
(2.9) into
Eq.
(2.41), obtain the equivalent relationship
D = e 0 E + P(E) = D(E),
(2.46)
constitutive equation of
an
arbitrary (nonlinear) dielectric
72
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
which
is
more often used
for characterization of dielectric materials
and
is
termed
a constitutive equation of the material. For linear dielectrics, Eq. (2.10) applies, and
Eq. (2.46) becomes
D=
constitutive equation
of a linear dielectric
where
£ is the permittivity
(Xe
+
and
=£
1 )£qE
r
£oE
D = eE,
or
(2.47)
e T the relative permittivity of the
medium (£ r is some-
times referred to as the dielectric constant of the material). The unit for e
per meter (F/m), while
£ r is dimensionless,
£r
is
farad
obtained as
=
Xe
+
£r
>
1.
(2.48)
1,
and hence
The value of
eT
£
permittivity of a linear
dielectric (unit:
F/m)
is
(2.49)
shows how much the permittivity of a
—
(2.50)
£r£()>
higher than the permittivity of free space (vacuum), given in Eq.
space and nondielectric materials (such as metals), e r
=
1
(1.2).
For free
and
D = £qE.
constitutive equation for free
space
dielectric material,
(2.51)
Table 2.1 shows values of the relative permittivity of a number of selected materials, for electrostatic or low-frequency time-varying (time-harmonic) applied electric
2
at room temperature (20°C).
For nonlinear dielectrics, the constitutive relation between
fields,
is
nonlinear. This also
on the
means
E
electric field intensity,
independent of
E
D and E, Eq. (2.46),
that the polarization properties of the material
(for linear dielectrics, Xe
and
depend
e are constants,
).
In so-called ferroelectric materials, Eq. (2.46)
is
not only nonlinear, but also
shows hysteresis effects. The function D(E) has multiple branches, so that D is
not uniquely determined by a value of E but it depends also on the history of
polarization of the material, i.e., on its previous states. A notable example is barium titanate (BaTiC^), used in ceramic capacitors and various microwave devices
(e.g., ceramic filters and multiplexers).
Another concept in characterization of materials is homogeneity. A material
is said to be homogeneous when its properties do not change from point to point
,
region being considered. In a linear homogeneous dielectric, £ is a constant
independent of spatial coordinates. Otherwise, the material is inhomogeneous [e.g.,
in the
e
= e(x, y, z)
Finally,
in the region].
we
introduce the concept of isotropy in classifying dielectric materials.
Generally, properties of isotropic media are independent of direction. In a linear
isotropic dielectric, £
in the
same
is
a scalar quantity,
and hence
D and E are always collinear and
however, individual components of D depend differently on
of E, so that Eq. (2.47) becomes a matrix equation,
~D X
[e]
~
Dy
- permittivity tensor of an
anisotropic dielectric
LDzJ
2
medium,
components
direction, regardless of the orientation of E. In an anisotropic
~
=
different
-
£ xx £ xy £xz
~EX
£yx £ yy £ yz
Ey
_ £ zx £ zy £ zz _
lEzi
(2.52)
At higher frequencies, when viewed over very wide frequency ranges, the permittivity generally
most materials) is not a constant, but depends on the operating frequency of electromagnetic
waves propagating through the material.
(for
Section 2.6
Table 2.1
Characterization of Dielectric Materials
Relative permittivity of selected materials*
.
Material
St
Material
St
Quartz
5
5-6
Vacuum
1
Freon
1
Air
1.0005
Diamond
Wet soil
Styrofoam
1.03
Mica (ruby)
5.4
Polyurethane foam
1.1
Steatite
5.8
Paper
1.3-3
Sodium chloride (NaCl)
5.9
Wood
2-5
Porcelain
6
Dry
2-6
Neoprene
Paraffin
2.1
Silicon nitride (Si 3
Teflon
2.1
Marble
Vaseline
2.16
Polyethylene
2.25
Alumina (AI 2 O 3 )
Animal and human muscle
Oil
2.3
Silicon (Si)
Rubber
2.4-3
Gallium arsenide
13
Polystyrene
2.56
Germanium
16
PVC
2.7
Ammonia
22
Amber
2.7
Alcohol (ethyl)
25
Plexiglass
3.4
Tantalum pentoxide
25
Nylon
3. 6-4.5
Glycerin
50
soil
Fused
02 )
5-15
6.6
N4
7.2
)
8
8.8
10
11.9
(liquid)
3.8
Ice
75
Sulfur
4
Water
81
Glass
4-10
Rutile (Ti 02 )
Bakelite
4.74
Barium
*
For
silica (Si
static
or low-frequency applied electric
Thus, instead of a single scalar
e,
fields, at
titanate
89-173
(BaTiOa)
1,200
room temperature.
we have
a tensor
[e]
(permittivity tensor),
i.e.,
nine (generally different) scalars corresponding to different pairs of spatial components of
D
and E. Crystalline
dielectric materials, in general, are anisotropic; the
moments to be formed and oriented by
much more easily along the crystal axes than in
periodic nature of crystals causes dipole
means of the applied
other directions.
electric field
An example is rutile (Ti02), whose relative permittivity is e = 173
and e = 89 at right angles. For many
r
in the direction parallel to a crystal axis
r
change in permittivity with direction is small. For example, quartz has
and it is customary to adopt a rounded value e T = 5 for its average
relative permittivity and treat the material as isotropic.
The theory of dielectrics we have discussed so far assumes normal designed
regimes of operation of electrical systems - when the electric field intensity, E, in
crystals the
£r
= 4.7 — 5.1,
individual dielectric parts of a system
the intensity
E in
a dielectric cannot
is
below a certain “breakdown”
be increased
exceeded, the dielectric becomes conducting.
It
indefinitely:
if
level.
Namely,
a certain value
is
temporarily or permanently loses
down. The breaking field value, i.e., the
an individual dielectric material can withstand
without breakdown, is termed the dielectric strength of the material. We denote it
by Ecr (critical field intensity). The values of Ecx for different materials are obtained
by measurement. For air,
its
insulating property,
maximum electric field
and
is
said to break
intensity that
£C rO =
3
MV/m.
(2.53)
dielectric strength of air
73
r
74
Chapter 2
Dielectrics, Capacitance,
and
Energy
Electric
In gaseous dielectrics, like air, because of a very strong applied electric field, the
and ions are accelerated, by Coulomb forces
free electrons
ities
[see Eq. (1.23)], to veloc-
high enough that in collisions with neutral molecules, they are able to knock
electrons out of the molecule (so-called impact ionization).
The newly created
electrons and positively charged ions are also accelerated by the
field,
free
they collide
more electrons, and the result is an avalanche process of
impact ionization and very rapid generation of a vast number of free electrons that
with molecules, liberate
constitute a substantial electric current in the gas (usually sparking occurs as well).
In other words, the gas, normally a very
into an excellent conductor.
time in thunderstorms
all
Note
that
good
many
air
insulator,
is
suddenly transformed
breakdowns occur
spheric electric fields (fields due to charged clouds), reaching the
in
at
any instant of
over the earth. Basically, they are caused by large atmo-
breakdown value
Eq. (2.53), and their most obvious manifestation is, of course, lightning.
Similar avalanche processes occur at high enough electric field intensities
liquid
and
solid dielectrics.
For
of the dielectric strength (Ec
)
solids, these
in
processes are enhanced and the value
of the particular piece of a dielectric
is
lowered by
impurities and structural defects in the material, by certain ways the material
is
manufactured, and even by microscopic air-filled cracks and voids in the material.
In addition, when, under the influence of a strong electric field, the local heat due to
leakage currents flowing in lossy (low-loss) dielectrics is generated faster than it can
temperature may cause a change
and lead to a so-called thermal breakdown of the dielectric. Such breakdown processes depend on the duration of the applied strong field
and the ambient temperature. Breakdowns in solid dielectrics most often cause a
permanent damage to the material (e.g., formation of highly conductive channels of
molten material, sometimes including carbonized matter, that irreversibly damage
be dissipated
in the material, the resulting rise of
in the material (melting)
the texture of the dielectric).
The values of
Table
in
2.2. Dielectric
ECT
for
some
selected dielectric materials are presented in
strengths of dielectrics other than air are larger than the value
Eq. (2.53). Note that, by definition, the dielectric strength of a vacuum
Conceptual Questions (on Companion Website):
Table 2.2.
is infinite.
2.6.
Dielectric strength of selected materials*
£C (MV/m)
Material
r
Ect (MV/m)
Material
Air (atmospheric pressure)
3
Bakelite
25
Barium
7.5
Glass (plate)
30
Freon
~8
Paraffin
Germanium
-10
Silicon (Si)
titanate
(BaTiOi)
Porcelain
11
Gallium arsenide
~30
—30
~35
—40
Oil (mineral)
15
Polyethylene
47
Paper (impregnated)
15
Mica
200
Polystyrene
20
Fused quartz (Si02)
Teflon
20
Silicon nitride
Rubber (hard)
25
Vacuum
Wood
(douglas
fir)
At room temperature.
~
10
Alumina
(S^N,^
-1000
-1000
oo
Section 2.8
2.7
Electrostatic Field in Linear, Isotropic,
75
and Homogeneous Media
MAXWELL'S EQUATIONS FOR THE ELECTROSTATIC
FIELD
We note that Maxwell’s first equation for the electrostatic field, Eq.
depend on the material
properties,
and
is
same
the
in all
(1.75),
does not
kinds of dielectrics as
it
Eq. (2.44) is Maxwell’s third equation, and we now write down the
full set of Maxwell’s equations for the electrostatic field in an arbitrary medium,
together with the constitutive equation, Eq. (2.46) or (2.47):
is
in free space.
§c E
•
dl
=
0
Maxwell's
first
equation
in
electrostatics
&D.dS = /v pdv
D = D(E)
We
[D
=
(2.54)
.
eE]
Maxwell's third equation
constitutive equation for
shall see later in this text that these equations represent a subset of the full
set of
Maxwell’s equations for the electromagnetic
static case. In the
field,
specialized for the electro-
general case, the set contains four Maxwell’s equations and three
constitutive equations.
same form
As we
shall see, the third
equation (generalized Gauss’ law)
under nonstatic conditions. Constitutive equations are
not Maxwell’s equations, but are associated with them and are needed to supply the
information about the materials involved.
retains this
2.8
also
ELECTROSTATIC FIELD
IN LINEAR, ISOTROPIC,
AND
HOMOGENEOUS MEDIA
Most often we deal with
and homogeneous dielectrics, in which
Eq. (2.47) applies, and the permittivity e is independent of the intensity of the
applied field, is the same for all directions, and does not change from point to point.
For such media, we can bring e outside the integral sign in the integral form of the
linear, isotropic,
generalized Gauss’ law, Eq. (2.43),
E dS = Qs
Gauss' law for a
(2.55)
•
or outside the operator (div) sign in the differential generalized Gauss’ law,
Eq. (2.45),
V E= —
(2.56)
.
£
We
notice that Eqs. (2.55) and (2.56) are identical to the corresponding free-space
laws, Eqs. (1.133)
and
(1.165), except for £o being substituted
by
e.
Recall that the
expression for the electric field intensity vector due to a point charge in free space,
and with it also Coulomb’s law, can be derived from Gauss’ law (see Problem 1.53).
Based on this, we can now reconsider all charge distributions in free space we have
considered so far, and all structures with conductors in free space we have analyzed, and by merely replacing £o with £ in all the equations, obtain the solutions
for the same (free) charge distributions and the same conducting structures situated in a homogeneous dielectric of permittivity e? This is the power of the concept
3
In
what follows
we shall always assume linear and isotropic media,
medium under consideration is nonlinear and/or anisotropic.
(in this entire text),
explicitly specify that the
except
when we
homogeneous
dielectric
D
76
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
of dielectric permittivity.
sity
We
emphasize again
with using the electric flux den-
that,
we are left to deal with free charges in
of bound charges to the field is properly
vector and the dielectric permittivity,
the system only, while the contribution
added through e. Thus, for example, Eq. (1.82) implies that the potential due to a
free volume charge distribution in a homogeneous dielectric with permittivity e is
given by
v=
_L fe±.
(2.57)
R
Ane Jv
Also, the free surface charge density on the surface of a conductor surrounded by a
dielectric with permittivity e
is
[from Eq. (1.190)]
Ps
=
e
fi
•
E
(2.58)
(boundary condition for the normal component of E), and so on. Note, however,
boundary condition for the tangential component of E near a conductor
surface, Eq. (1.186), is always the same, irrespective of the properties (e) of the
surrounding dielectric.
that the
tric,
Once we
we can
and
(2.47)]
find the electric field in a structure filled with a
homogeneous
dielec-
calculate the polarization vector in the dielectric as [Eqs. (2.41)
D — £qE =
P=
polarization vector in a linear
(e
—
£q)E,
(2.59)
dielectric
and then the distribution of volume and surface bound charges of the dielectric using Eqs. (2.19) and (2.23).
Note that, from Eqs. (2.19), (2.59), (2.56), and (2.50), the bound volume charge
density, p p at a point in the dielectric can be obtained directly from the free volume
,
charge density, p,
Pp
at that point as
= —V P„ =
-(£
•
In an analogous manner,
-
£o)V
_ =
E
—
e
-—— p
eT
£()
—
—
1
-
(2.60)
p.
we derive the relationship between the bound and free
surface charge densities on the surface of a conductor surrounded by a dielectric
with relative permittivity e x
.
Shown
in Fig. 2.8
is
a detail of the surface.
Eqs. (2.23), (2.59), (2.58), and (2.50), and noting that
directed from the dielectric outward; in Eq. (2.58),
outward],
we
fi
is
=—
Combining
Eq. (2.23), Ad is
directed from the conductor
fid
fi
[in
obtain
ii
dielectric
pps
=
fi d
•
P = -(£ - £o)fi E =
•
—
-
(2.61)
ps.
£r
conductor
fid
Figure 2.8 Detail of a
conductor-linear dielectric
Although the free surface charge density, p s is actually localized on the conductor
boundary surface and the bound surface charge density, p ps is localized
on the dielectric side of the surface, they can be treated as a single sheet of charge
,
side of the
,
with the total density
surface.
Pstot
Example 2.5
A
Dielectric
— Ps + Pps —
(2.62)
Sphere with Free Volume Charge
homogeneous
dielectric sphere, of radius a and relative permittivity e r is situated in
There is a free volume charge density p(r) = por/a (0 < r < a) throughout the sphere
volume, where r is the distance from the sphere center (spherical radial coordinate) and po
is a constant. Determine (a) the electric scalar potential for 0 < r < oo and (b) the bound
air.
charge distribution of the sphere.
,
Section 2.8
and Homogeneous Media
Electrostatic Field in Linear, Isotropic,
Solution
(a)
Because of spherical symmetry of the problem, the electric flux density vector, D, is
purely radial and depends only on r. From the generalized Gauss’ law [Eq. (2.44)],
Example
applied in a similar fashion to that in
symmetry
(see also
Example
D(r)
2
Por /(4a)
=
Poa /(4r
electric field intensity vector
=
E(r)
The
potential at a distance r
V(r)
)
[also see
V(r)
=
a
for r
>
a
D(r)/(e T eo ) for
r
D(r)/eo
r>a
for
r°°
1
is
—
1
Dir
f
(2.64)
hence:
for r
’
>
(2.65)
a.
4e 0 r
)
dr*
+
Ppa
=
V{a)
1
(2.60), the
r
-
—
£r
,
1
for r
.
.
P(r)
and
<
(2.66)
a.
inside the sphere
amounts
- 1V
--P0(£r
=
to
(2.67)
sra
St
(2.23), (2.59), (2.47),
3
3e r
bound volume charge density
PpW =
,
1
+
4e 0
Jr, =r
According to Eq.
Using Eqs.
given by
is
a
Poa
=
d/
£>(/-')
/
<
magnitude
its
Eq. (1.142)]
^•r^O
(b)
<
same form, and
£0 Jr'=r
and
for r
from the sphere center
—
=
of the
is
for spherical
D is found to be
(2.63)
2
3
The
accommodated
1.19
magnitude of
1.18), the
(2.63), the
bound
surface charge density
on the
sphere surface comes out to be
Pps
D
= P(a
,
)
=
sr
-
1
D(a
=
)
po(e r
Sketched in
Model
of a
Fig. 2.9(a) is a
1 )a
.
(
4e r
£r
Example 2.6
-
2 68 )
.
pn junction
pn junction between two semiconducting
half-spaces,
doped
p-type and «-type, respectively. The volume charge distribution in the semiconductor can be
approximated by the following function:
p{x)
-
—po
for x
<
0
for x
=0
-*/ 0
for x
po e
>
0
(2.69)
,
0
where po and a are positive constants. The permittivity of the semiconductor
is s.
Find
(a)
the electric field intensity vector in the semiconductor, (b) the electric scalar potential in the
semiconductor, and
(c)
the voltage between the ends of the semiconductor, from the end on
the n-type side to the end on the p-type side of the junction.
Solution
(a)
This
is
a
problem with planar symmetry (see Examples
electric field intensity vector in the
The
semiconductor
is
differential generalized Gauss’ law, Eq. (2.56),
d Ex (x)
p(x)
dx
e
1.20, 1.21,
given by
and
1.23),
E = Ex (x)x
and the
[Eq. (1.148)].
becomes
(2.70)
77
>
78
Chapter 2
and
Dielectrics, Capacitance,
1
Energy
Electric
Figure 2.9 Model of a pn
volume charge
junction: (a)
density, (b) electric field
intensity,
and
potential; for
This
to
x
is
(c) electric scalar
Example
2.6.
a first-order differential equation in
[as in
x and we solve
,
it
by integrating with respect
Eq. (1.178)]:
X
EAx)
where C
x -> — oo.
is
=
p(x')dx’ + C,
-f
£ Jx'=-oo
the constant of integration, which represents the field
We
meaning
Ex
in the
plane
note that
p(x')
field
(2.71)
that the total charge of the
cLc'
= 0,
semiconductor
can exist far from the junction [see also Eq.
Ex (x —
=poo)
(
1
=
.
is
(2.72)
zero. This means, in turn, that
no
54)],
0,
(2.73)
e
Section 2.9
and hence
C=
0.
Dielectric-Dielectric
Substituting Eq. (2.69), the integration for the field points in the p-type
region of the semiconductor yields
Ex {x) = — —
In the «-type region,
Ex (x) =
Fig. 2.9(b)
field is
—
we have
x '/ a
dx'
x/a
—
<x <
(— oo
(2.74)
0).
&
up
to break the integration
e^dx' +
J
shows the
Q
f
J — oo
£
e~x
J
=
dx'^
Ex (x)
electric field intensity
into
e~
in the
two
parts:
x la
(0
<x <
semiconductor.
oo).
We
(2.75)
see that the
oriented from the n-type doped region to the p-type doped region. This field
is
pn junction, as it exists in the junction even when an external
voltage is not applied (e.g., when the terminals of a pn diode are not connected to an
external voltage source). In the equilibrium state established after the pn junction is
called the built-in field of a
formed, the built-in
right
(b)
field
prevents further diffusion of positive charges (holes) to the
and negative charges (electrons) to the
across the junction.
4
Let us arbitrarily adopt our reference point for potential (7Z) at the center of the junction,
x = 0, so that the potential at points in the p- type region is [Eq. (1.74)]
pH
V(x)
=
/»0
E
/
dl
=
JP
qX /ci
_
(
In the n-type region,
we
n
V{x)
^
e
)
_oo < x <
,
-
r0
y/fl
e
/
dx'
Jx
(2.76)
0.
nX
/»P
E
•
dl
=—
/
E
•
dl
Jn
Jp
The
i
n n
reverse the direction of integration for convenience,
=
_
Ex (x') dx' = —
/
Jx'=x
Po a
_
(c)
left
Po«
/"
£
Jo
e
-*7a fat
=—
Ex (x') dx'
/
Jo
_
_
e
-x/a\
o
<x <
oo.
distribution of the potential, V(x), along the semiconductor
is
shown
in Fig. 2.9(c).
The voltage between the n-type and p-type ends of the semiconductor turns out
V{x -> oo)
(2.77)
£
— V{x
-»
— oo) =
.
to be
(2.78)
e
This voltage
is
called the built-in voltage of a pn junction (diode).
Problems 2.12-2.15;
:
2.9
So
far,
MATLAB Exercises (on Companion Website).
DIELECTRIC-DIELECTRIC
we have considered boundary
BOUNDARY CONDITIONS
surfaces conductor-free space (Fig. 1.39),
and conductor-dielectric (Fig. 2.8), and analyzed the
fields close to surfaces and surface charge densities on the surfaces [Eqs. (1.186),
(1.189), (2.23), and (2.58)]. Let us now consider a dielectric-dielectric boundary
dielectric-free space (Fig. 2.4),
4
Note that initially there are excess holes to the left of the plane x = 0 (p-type semiconductor is positively
its entire volume) and excess electrons to the right (n-type semiconductor is negatively
charged by doping). In a brief transient, a diffusion of holes occurs from the p-region into the n-region
and electrons diffuse in the opposite direction, until an electric field is built up in such a direction that
charged over
the diffusion current drops to zero.
Boundary Conditions
79
2
I
80
Chapter 2
and
Dielectrics, Capacitance,
Electric
Energy
Figure 2.10 Dielectric-dielectric boundary surface: deriving boundary conditions for
(b)
(a) tangential
components
of
E and
normal components of D.
surface,
shown in Fig.
near the surface.
2.10,
We
and derive the boundary conditions
for field
components
apply the same technique as in deriving the corresponding
boundary conditions
for the conductor-free space case, Eqs. (1.186)
and
(1.189).
The main difference is that now the field exists at both sides of the boundary. Let
Ei and Dj be, respectively, the electric field intensity vector and electric flux density
vector close to the boundary in medium 1, whereas E 2 and D 2 stand for the same
medium
quantities in
2.
Applying Eq. (1.75) to a narrow rectangular elementary contour C,
we
Fig. 2.10(a),
obtain
continuity of the tangential
i
component of E
E
•
dl
= Eu A — E A —
This boundary condition
on the two
On
/
tells
0
/
2x
us that the tangential
sides of the boundary,
i.e.,
that
E
t
is
= E2t
E\t
(2.79)
.
components of E are the same
continuous across the boundary.
the other hand, an application of Eq. (2.43) to a pillbox Gaussian elemen-
tary surface, Fig. 2.10(b), gives
l
D
•
dS
= D\ n AS — D 2n A S —
(we employ vector
ponents of
to region
1,
D
ps
AS
D 2n —
D\n
(2.80)
ps
D to avoid dealing with bound charges), where the normal com-
are defined with respect to the unit normal n directed from region 2
and ps
is
the free surface charge density that
may
exist
on the
surface.
In the absence of charge,
continuity of the normal
component of D, charge-free
surface
^ln
— D 2n
(Ps
—
(2.81)
0).
D be the same
boundary with no free charge on it. In other words, D n is
continuous across the boundary free of charge.
Relationships in Eqs. (2.79) and (2.80) represent two primary boundary conditions for the electrostatic field at the interface between two arbitrary media. When
the dielectrics 1 and 2 are linear, D) = £]Ei and D 2 = e 2 E 2 so we obtain an additional pair of (secondary) boundary conditions - for the tangential components of D
and normal components of E. From Eq. (2.79), the boundary condition for D reads
This boundary condition enforces that the normal components of
on the two sides of
a
,
t
D
it
D
2l
(2.82)
£\
£
Section 2.9
Dielectric-Dielectric
81
Boundary Conditions
Figure 2.11 Refraction
of electric field lines at a
dielectric-dielectric interface.
and we see that D is discontinuous across the boundary. Similarly, Eq.
the boundary condition for E n (if p s = 0 at the interface):
(2.81) yields
t
eiE ln
= e 2 E 2n
(2.83)
,
which shows that E n is also discontinuous across the boundary.
Note that the boundary conditions for E and D, that is, for
Eqs. (2.79) and (2.80) can be written in vector form:
n x Ei — n x
n Di
E2 =
- n D2 =
E and D n
t
0,
(ii
ps
directed from region 2 to region
,
in
(2.84)
boundary condition
for
E
(2.85)
boundary condition
for
Dn
1),
,
t
which is often more convenient for use in analyzing complex structures.
Let us consider again the interface between two dielectric media and the angles
a\ and a 2 that field lines in region 1 and region 2 make with the normal to the
interface, as depicted in Fig. 2.11. The tangents of these angles can be expressed as
tanai
Eu
= ——
tan a 2
E\n
We divide the
E2
= ——
E2n
1
and
(2.86)
.
tangents and use Eqs. (2.79) and (2.83) to get
tanai
e\
tan a 2
s2
(2.87)
This
is
law of refraction of electric
field lines
the law of refraction of the electric field lines at a dielectric-dielectric bound-
is free of charge (p s = 0). Bending of field lines
unequal bound charges on the two sides of the boundary.
ary that
Finally, let us find the distribution of
From Eq.
dielectric interface (Fig. 2.10).
bound
is,
essentially, a result of
on a dielectricbound charge densities that
surface charges
(2.23), the
accumulate on the two sides of the interface are
Ppsi
=
nd i
•
Pi
and
p ps2
=
n d2 P 2
(2.88)
,
= — n and n d2 =
where n d i
n [n d in Eq. (2.23) is directed from the dielectric outward]. Hence, by adding p ps i and p ps2 together, the total bound surface charge
density at the interface is given by
p ps
This
is
= n P2 — n
•
•
(2.89)
Pi.
the boundary condition for the normal components of vector
P at a dielectric-
dielectric boundary.
Problems 2.16-2.18; Conceptual Questions (on Companion Website):
:
MATLAB Exercises (on Companion Website).
2.7
and
2.8;
boundary condition
for
Pn
82
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
AND LAPLACE
POISSON'S
2.10
EQUATIONS
S
Consider the electric field intensity vector, E, and scalar potential, V, in a homogeneous dielectric region of permittivity e. The spatial derivatives of E at a point in
the region are related to the free
point by Eq. (2.56), which
volume charge
may
density, p, that
exist at that
a first-order differential equation. E, in turn,
is
V
to the spatial derivatives of
through Eq. (1.101).
It is
related
is
obvious, then, that the
second-order spatial derivatives of
V
are related to p by
differential equation. This equation
is
Poisson’s equation, which
means of
is
a second-order
easily derived
by
substituting Eq. (1.101) into Eq. (2.56):
V.(VV0 = --.
(2.90)
£
The double-V operation
equation
in Poisson’s
performed by evaluating
is
the
first
gradient of V, and then the divergence of the result. In the Cartesian coordi-
nate system, gradient and divergence are calculated using Eqs. (1.102) and (1.167),
and we have
respectively,
,
=
V (VL)
•
9
-
fdv\
div
„
= V- |^—
grad E)
fdv\
a
dv
dv A
x+—
y+ — zj
fsv
JTr
(
fdv\
9
d
+
„
2
v
9?
+
d
2
v
W
+
2
v
a?'
d
(Z9,)
We
note that the same result would have been obtained by applying formally the
formula for the dot product in the Cartesian coordinate system, Eq. (1.164), to the
dot product of two identical vectors (V) in (V V)V, where V is expressed as in
Eq. (1.100). Hence, we can write
•
V (VV) =
The operator V V
becomes
•
is
abbreviated
V2
(V
V)V.
•
(2.92)
(“del squared”), and Poisson’s equation
V2 V =
Poisson's equation
(2.93)
£
where
vv =
,
Lapiacian in Cartesian
coordinates
In a charge-free region
(
p
—
d
2
v
^
+
d
known
as Laplace’s equation.
(2
'
94)
(2.95)
o,
The V 2 operator
see that the Lapiacian operates on a scalar
-p/e or 0).
The expressions
2
v
a?'
d
+
a7
v2 f =
is
v
0),
Laplace's equation
which
2
(e.g.,
is
called the Lapiacian.
V), and the result
is
We
another scalar
(e.g.,
for
V2 V
in the cylindrical
coordinate system
spherical coordinate system (Fig. 1.26) can be obtained in the
first
taking the gradient of
[Eqs. (1.170)
V=
Lapiacian in cylindrical
coordinates
V(r,
(p,
and
V
[Eqs. (1.105)
and
(Fig. 1.25)
and the
i.e., by
same manner,
and then the divergence
(1.108)],
(1.171)] of the result. In cylindrical coordinates, the Lapiacian of
z ) thus
comes out
v, V =
to be
19/
9VA
7TrV17)
1
+
?
d
2
V
w
2
W
d
+
V
(2.96)
0
Section 2.10
Poisson's
83
and Laplace's Equations
HISTORICAL ASIDE
and magnetism. In 1813, he published a general-
Simeon Denis Poisson (1781-1840), French mathwas a student of Laplace (17491827) and Lagrange (1736-1813), and successor
ization of Laplace’s differential equation for the
of Fourier (1768-1830) as a professor, at Ecole
so inside a solid, in mechanics or for a nonzero
He is best known for his work
charge density, so inside a charge distribution, in
ematician,
Polytechnique, Paris.
on probability (Poisson’s
distribution),
potential theory, valid for a nonzero
and his con-
mass
density,
electromagnetics.
tributions to mathematics as applied to electricity
Pierre Simon de Laplace (1749-1827), French
mathematician and astronomer, was President of
Academie des Sciences (French Academy). His
work in mathematical astronomy was summed
up
in a
monumental five-volume book on
partial
appeared.
on the
the-
under Napoleon, and was made a marquis by Louis
XVIII.
V(r,6,(f)) in spherical coordinates
——
V2 V =
him, Laplace’s equation,
also wrote a treatise
of substances. Laplace was a minister and senator
celes-
mechanics ( Mecanique Celeste ), published
between 1799 and 1825, in which the second-order
V=
He
after
ory of probability, and worked on specific heats
tial
and that of
equation for potential that
differential
we now name
r2 dr
Example 2.7
/
3
1
+
;
I
r2 sin0 3 0
.
sin 9
—
30 )
\
d
1
+
r2
2
sin 6
3
2
V
(2.97)
2
La placia n
in spherical
coordinates
Application of a 1-D Poisson's Equation
A free volume charge of a uniform density p exists in a homogeneous dielectric, of permittivbetween two flat metallic electrodes, as shown in Fig. 2.12. The electrodes are connected
and the distance between them is d. Neglecting the fringing effects, find (a)
the electric potential and (b) the electric field intensity vector in the dielectric.
ity e,
to a voltage Vo,
Solution
(a)
is equivalent to assuming that the electrodes are infinitely
which case the potential in the dielectric varies with the distance from the electrodes only. Let x be the normal distance from the left electrode (Fig. 2.12). Poisson’s
equation, given by Eqs. (2.93) and (2.94), becomes
Neglecting the fringing effects
V(x)
large, in
d
2
V (x)
dx 2
By
integrating
it
twice,
we
——
(0
<x <
(2.98)
d).
e
get
2
V(x)
=—
2s
+ C\x + C2
(2.99)
,
where C\ and
Figure 2.12 Uniform
results in
volume charge between
pd/(2s)
—
C2 are the constants of integration. The boundary condition F(0) = Vo
C2 = Vo, and the condition on the other boundary, V(d) = 0, gives C\ =
metallic plates: application
Vo/d. Hence,
of a 1-D Poisson's equation
vW
_£^ +
v0
1
(
-£).
(
2 100 )
(fringing neglected); for
.
Example
2.7.
84
Chapter 2
Capacitance, and Electric Energy
Dielectrics,
Using Eqs. (1.101) and
(b)
E(x)
(1.102),
we
obtain the electric
dV(x)
= -VF =
= \p( x
„
x
b(
dr
Problems'. 2.19-2.22; Conceptual Questions (on
2.1
FINITE-DIFFERENCE
1
In
many
vector from the potential:
d\
-2) +
Vo]
-d\
(2.101)
X.
Companion Website):
2.9.
METHOD FOR NUMERICAL
SOLUTION OF LAPLACE
ically,
field
EQUATION
S
practical cases, Poisson’s or Laplace’s equation cannot be solved analyt-
but only numerically. The most popular and perhaps the simplest numerical
method
for solving these equations (and other types of differential equations
generally)
is
the finite-difference (FD) method.
tives in the differential
It
consists of replacing the deriva-
equation by their finite-difference approximations and
solving the resulting algebraic equations.
To
illustrate this,
coaxial cable with conductors of square cross section,
and b
cross-sectional dimensions of the cable be a
(a
consider an
air-filled
shown in Fig. 2.13(a). Let the
< b). The cable is charged by
Va and Vb, are known.
V in the space between
time-invariant charges, and the potentials of the conductors,
Our
goal
is
to determine the distribution of the potential
the cable conductors.
This
is
a
two-dimensional electrostatic problem for which, according to
Fig. 2.13(b), Laplace’s equation, Eqs. (2.95)
V2 F =
d
2
v
dx
We
discretize the region
of sides
d
[Fig. 2.13(b)],
compute the
and
d
by
v
(
3y2
2
2 102 )
.
between the conductors by introducing a grid with cells
and employ the FD method in order to approximately
potentials at the grid points (nodes). Obviously, the accuracy of the
computation depends on the grid resolution,
d, the
2
(2.94), is given
more accurate
i.e.,
the smaller the spacing of the grid,
(but computationally slower) the solution.
y
d
va
2
43
i
a
y
Yb
E0
X
b
(a)
O
O
(b)
(c)
Figure 2.13 Finite-difference analysis of a coaxial cable of square cross section: (a) structure geometry, (b) nodes with
unknowns, and (c) detail of the grid for approximating Laplace's equation in terms of finite
discrete potential values as
differences.
,
Section 2.1
.
Finite-Difference
1
Method
for
85
Numerical Solution of Laplace's Equation
shows a detail of the grid in Fig. 2.13(b). At the node 1, the
backward-difference approximation for the first partial derivative of V with respect
to the x-coordinate turns out to be
Fig. 2.13(c)
dV
dx
V!-V2
„
(2.103)
d
'
1
finite-difference (FD)
’
approximation of a derivative
which, combined with the forward-difference approximation for the second partial
derivative with respect to x, yields
d
2
v
dx
/3V\
_9_
2
dx \ dx )
1
dV
1
d\dx
1
\
1
(
dx
3
1
J
V3
d \
-V
V!-V2
1
d
V2 + V3 - 2V!
(2.104)
d2
Analogously,
d
2
v
dy
v4 + V5 - 2V!
(2.105)
d2
2
1
FD
so that the
approximation for Laplace’s differential equation
at point 1 in
Fig. 2.13(c) is
V2 + V3 + V4 + F5 - 4Vi
=0
2
VZ V
Vi
d
=
i (V2
+ V3 + V4 + V5 )
FD approximation of Laplace's
equation
(2.106)
aid
The simplest technique to solve the above finite-difference equation with the
of a computer is an iterative technique expressed as
v[
When some
k+1)
=
( k)
^
[v2
+v
{k)
k=
0,1,...
(2.107)
iterative solution to the
FD
Laplace's equation
of the nodes
and 5 belong to one of the surfaces of conduc-
2, 3, 4,
the potential at such nodes
tors,
{k)
+v ]
+
is
Va
in all iteration steps equal to the respective
For the initial solution, at the zeroth
at all nodes between the conductors.
By traversing the grid in a systematic manner, node by node, the average of the four
neighboring potentials is computed at the (k 4 l)th step for each node and is used to
replace the potential at that node, Eq. (2.107), and thus improve the solution from
the kth step. This procedure is repeated until the changes of the solution (residuals)
with respect to the previous iteration at all nodes are small enough, i.e., until a final
given potential of the conductor
(k
—
0) iteration step,
(
we can adopt
or
V'
(0)
Vt> ).
=0
-
set of values for the
unknown
potentials consistent with the criterion
T/(&+l)
V
l
is
obtained, where
Once
<5^
T/(fc)
V
1
< 8y
(2.108)
stands for the specified tolerance of the potential.
is known, numerinodes can be obtained
Eqs. (1.101) and (1.102) in
the approximate solution for the potential distribution
cal results for the electric field intensity vector, E, at the grid
by approximating the gradient operator involved in
terms of finite differences. For example, E at the node
approximately as
Ei
=-
VVIj
dV
=dx
1 in Fig. 2.13(c) is
V2-V3
x
2d
+
computed
T4-F5
2d
y
(2.109)
FD approximation of the
electric field vector
)
86
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
(central-difference approximation). Additionally, the surface charge density, ps
on the conducting surfaces can be found by means of the boundary condition in
Eq. (1.190). For example, assuming that the node 4 in Fig. 2.13(c) belongs to the
surface of the inner conductor in Fig. 2.13(a), ps at that point can be approximately
,
evaluated as
Ps4
=
£0 y
'
E4 =
£q Ey4
=
dV
-£Q
so
dy
V
1
-V4
Va-Vi
£()
(
2 110 )
.
charge per unit length of each of the conductors, Eq. (1.31), can
be found by numerically integrating ps that is, summing pS( along the individual conductor contours. Thus, the per-unit-length charge of the inner conductor is
given by
Finally, the total
,
Q'
a
r
=
,
Na
psi d
Jc°
(2.1
,
where Na denotes the total number of nodes along the contour
and similarly for the outer conductor.
Problems: 2.23 and 2.24;
2.12
MATLAB
1 1
i=\
Exercises (on
Ca
of the conductor,
Companion Website).
DEFINITION OF THE CAPACITANCE OF
A CAPACITOR
shows a system consisting of two metallic bodies (electrodes) embedded in
and charged with equal charges of opposite polarities, Q and —Q. This
system is referred to as a capacitor. The principal property of a capacitor is its capability to store the charge. The potential difference between the electrode carrying
Q and the one with — Q is called the capacitor voltage. The capacitor is linear if its
dielectric is linear. The dielectric can be homogeneous or inhomogeneous.
Fig. 2.14
a dielectric,
In linear capacitors, the capacitor charge,
capacitor voltage, V,
Q
oc
Q,
is
linearly proportional to the
To prove this, assume for the moment that the
2 Q. The surface charge density, ps will remain
V.
changed to
the same way over the surfaces of electrodes, and
capacitor charge
distributed in
i.e.,
is
dielectric
Figure 2.14 Capacitor.
,
its
magnitude
will
Section 2.1 2
double everywhere. The electric
dielectric also
becomes twice
its
field intensity
vector at an arbitrary point in the
previous value; for homogeneous dielectrics this
is
obvious from Eq. (1.38) with eo replaced by e, while for inhomogeneous dielectrics
we first note that the vector D doubles [e.g., generalized Gauss’ law, Eq. (2.43)],
Ea
D
any point of the dielectric [Eq. (2.47)], we find the same
for the vector E. Finally, because the capacitor voltage equals the line integral of
E (along any path) between the electrodes [Eq. (1.90)], we conclude that the voltage doubles as well. Summarily, V doubles because Q is doubled, meaning that Q
and V are linearly proportional to each other. This proportionality is customarily
and then, since
87
Definition of the Capacitance of a Capacitor
Q
c
—r—
V
~Q
at
Figure 2.15 Circuit-theory
representation of a
capacitor.
written as
Q=
CV,
(
from which the constant C, termed the capacitance of the capacitor,
is
2 112 )
.
defined as
(
2 113 )
.
capacitance of a capacitor
(unit:
The capacitance
is
F)
a measure of the ability of a capacitor to hold the charge - per
between the electrodes. It depends on the shape, size,
and mutual position of the electrodes, and on the properties of the dielectric of the
capacitor. For nonlinear capacitors, however, the capacitance depends also on the
a volt of the applied voltage
applied voltage,
C=
C(V),
(
2 114 )
.
nonlinear capacitor
and a notable example is a varactor diode. The capacitance of a capacitor is always
(C > 0), and the unit is the farad (F). For two-conductor transmission lines
(two-body systems with very long conductors of uniform cross section), we define
the capacitance per unit length of the line, that is, the capacitance for one meter
(unit length) of the structure divided by 1 m,
positive
C—C
v-
where C,
/,
and Q' are the
Shown
(i.e.,
it is
in Fig. 2.15
is
U
i
c
— —
Q_
The
(
v’
total capacitance, length,
the structure [see Eq. (1.31)].
theory,
'-.p
unit for
C
is
2 115 )
.
capacitance
p.u.l.
of a
transmission line (unit:
F/m)
and charge per unit length of
F/m.
the circuit-theory representation of a capacitor. In circuit
assumed that the charge
is
stored only in the capacitors in a circuit
on the capacitor electrodes), while the connecting conductors are considered
as
ideal short-circuiting elements with zero capacitance (and also with zero resistance
and inductance, as we
shall see in later chapters).
Finally, let us introduce
metallic
body
another related concept, the capacitance of an isolated
situated in a linear dielectric.
It is
defined as
(
2 116 )
.
^isolated body
where
Q is the charge of the body and Violated body is its potential with respect to the
reference point at
infinity.
Note
that this definition can also be regarded as a special
case of the definition in Eq. (2.113), for the capacitance of a two-body system, with
the assumption that the second body, carrying —Q, is at infinity.
capacitance of an isolated
metallic
body
88
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
ANALYSIS OF CAPACITORS WITH HOMOGENEOUS
2.1 3
DIELECTRICS
We now consider various examples of the
dielectrics.
The examples cover
analysis of capacitors with
homogeneous
several characteristic types of capacitors and trans-
mission lines of both theoretical and practical importance. The analysis of capacitors
and transmission
presented
in the
Example 2.8
lines with different types of
inhomogeneous
dielectrics will
be
next section.
Spherical Capacitor
Consider a spherical capacitor, which consists of two concentric spherical conductors, as
in Fig. 2.16. Let the radius of the inner conductor be a and the inner radius of the
shown
outer conductor be b (b
homogeneous
Solution
+Q). Due
>
a).
Find the capacitance of the capacitor
dielectric of permittivity
Assume
that the capacitor
is
if it
is
filled
with a
e.
charged with a charge
Q (the inner electrode carries
to spherical symmetry, the electric field in the dielectric
is
radial
and has the form
given by Eq. (1.136). Applying the generalized Gauss’ law in Eq. (2.55) to the spherical surface S of radius r (a < r < b) positioned concentrically with the capacitor electrodes [see the
flux
computation
in
Eq. (1.138)],
we
E(r)
The capacitor voltage
is
obtain
=
Q
Ansr2
(a
<
<
r
b).
(2.117)
[Eq. (1.90)]
(2.118)
so that the capacitance comes out to be
^= —
Q = Aneab
C
V
b-a
capacitance of a spherical
capacitor
Example 2.9
.
(2.119)
Capacitance of an Isolated Spherical Conductor
Determine the expression for the capacitance of a metallic sphere of radius a situated
Figure 2.16 Spherical capacitor
with a
for
homogeneous
Example
2.8.
dielectric;
in air.
Analysis of Capacitors with
Section 2.1 3
We
Solution
use the definition of the capacitance of an isolated body, Eq. (2.116).
potential of the sphere
if it
carries a charge
capacitance
Dielectrics
89
The
Q is (see Example 1.25)
^sphere
K
Its
Homogeneous
=
(2.120)
~T~~
4neoa
hence
is
C=
Q
4neoa.
(2.121)
^sphere
capacitance of an isolated
metallic sphere in air
Note that an isolated sphere can be regarded as the inner electrode of a spherical capacwhose outer electrode has an infinite radius, and that the same result for the sphere
capacitance is obtained from Eq. (2.119) with b
oo.
itor
Coaxial Cable
Example 2.10
A coaxial
cable
a transmission line consisting of an inner cylindrical conductor (a wire)
is
and an outer hollow
cylindrical (tubular) conductor, with the conductors being coaxial with
Note that this structure is sometimes referred
The inner conductor has a radius a, and the outer conductor has
an inner radius b (b > a). The cable is filled with a homogeneous dielectric of permittivity e.
Find the capacitance per unit length, C, of the cable.
respect to each other, as depicted in Fig. 2.17.
to as a cylindrical capacitor.
Solution This
is
a
problem with
to the cable axis (Fig. 2.17).
surface
we
S positioned
cylindrical symmetry.
By means
where
Q
is
=
electric field
in the dielectric coaxially with the cable
obtain the field intensity, and from
E(r)
The
O'
(a
<
r
<
is
radial with respect
of the generalized Gauss’ law applied to a cylindrical
it
conductors (see Example 1.26),
the voltage between the conductors,
V=
b)
2ner
r
b
E(r) dr
Ja
=
—
b
O'
2ne
In
-
,
(2.122)
a
the charge per unit length of the cable. Hence, the capacitance per unit length of
the cable, Eq. (2.115), equals
Q_
V
_
2ne
(2.1
In (b/a)'
23)
capacitance
p.u.l.
cable
Figure 2.17 Coaxial cable
(cylindrical capacitor)
homogeneous
Example 2.10.
with a
dielectric; for
of a coaxial
1
.
90
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
We
electric field in
—
’
note that, combining together Eqs. (2.122),
V
=
E(r)
a coaxial cable
(2.124)
r\n(b/a )
“^Gauss
which
Q
is
a very useful expression for the field intensity (in terms of the voltage) in a coaxial
cable.
+ + + + + + + +i+ + +
Example
Parallel-Plate Capacitor
2.1
Consider a parallel-plate capacitor, which consists of two parallel metallic
I*
area S, charged with
neous
d,
is
Q
and —Q. Let the space between the plates be
dielectric of permittivity e, as
shown
in Fig. 2.18(a).
Assume
filled
plates,
each of
with a homoge-
that the plate separation,
very small compared to the dimensions of the plates, so that the fringing effects can be
(a)
\
\ /
Q
S
+++++++++++
neglected. Calculate the capacitance of the capacitor.
Solution Under given assumptions, the electric field in the dielectric is uniform, and there
no field outside the dielectric. Applying the generalized Gauss’ law to a rectangular
closed surface that encloses the upper plate [Fig. 2.18(a)], we get [see the left-hand side of
Eq. (1.155)]
is
-Q
ES =
—
(2.125)
,
e
(b)
Figure 2.18
and hence
E=
Q/(eS). The voltage between the plates
V=
Parallel-plate capacitor with
a
homogeneous
and
er
=
for
(2.126)
dielectric
(b) fringing field for
1;
is
(a)
Example
2.1
and the capacitance of the capacitor
1
capacitance of a parallel-plate
(2.127)
capacitor, fringing neglected
For arbitrary capacitor dimensions, there
is
a considerable fringing field extending far
outside the capacitor and the field between the capacitor plates close to plate edges
form (edge
is
Electrodes of an
from Eq.
Electric Forces
Example 2.12
air-filled
between the plates
a
=
effects), as illustrated in Fig. 2.18(b) for e r
larger than the value obtained
is
d,
With
1.
this,
is
not uni-
the actual capacitance
(2.127).
on Capacitor
Plates
capacitor are parallel square plates of edge lengths
where d <&a. The capacitor voltage
is
a.
The
distance
V. Find the electric forces
on
electrodes.
on the lower plate (Fig. 2.19), we subdivide it into
and add up (integrate) the forces on individual patches that are due to the electric field of the other charged plate (the upper plate). Each
charged patch can be considered as a point charge and the force dFe 2 on it can be calculated
To
Solution
find the net electric force
differentially small patches of surface areas dS,
using Eq. (1.23), so that
dFe 2
=
/
(2.128)
Ps2dSEi.
Js 2
—
-
d <?2
the surface charge density of the lower plate and Ej
Figure 2.19 Evaluation of
Here, p s 2
the electric force on the
vector due to the charge of the upper plate. Since
lower electrode of an
and these quantities are
is
d
is
the electric field intensity
<& a, the fringing effects can be neglected,
air-filled parallel-plate
capacitor; for
Example 2.12.
Ps2
= -^
S
and
E]
=
x
2s[)S
(
S
=
a
2
).
(2.129)
ii
Analysis of Capacitors with
Section 2.1 3
We
infinite
Ei equals a half of the total electric field intensity between the plates (the
to the charge of the lower plate) and can also be obtained as the field of an
sheet of charge, Eq. (1.64). Q is the charge of the capacitor,
is
due
Q = CV =
C is its capacitance. The force on
s 0 SV
the lower plate
definition, this
is
opposite.
The
The
5
2d2
(2.131)
(2.132)
‘
Fe 2 on
Pa
is
„
2
eoV'
unit for pressure
hence
forces are attractive.
Fe 2
the pressure of the force
is
the electric pressure.
is
2
£o a V
Fe2 = -
The force on the upper electrode
Note that, from Eq. (2.131),
(2.130)
'
d
2
By
Dielectrics
realize that
other half
and
Homogeneous
ing the total electric field intensity of the capacitor,
We term it
= N/m 2 By introduc-
the surface of the plate.
(pascal),
E=
where Pa
.
V/d, the electric pressure can be
expressed as
(2.133)
This expression
valid for
is
any conducting surface, with
intensity in the dielectric (air) near the surface.
electric
pressure (unit: Pa)
E standing for the local electric field
The pressure
acts
from the conductor toward
the dielectric.
Microstrip Transmission Line
Example 2.13
Consider the transmission
resting
on
line
shown
in Fig. 2.20. It consists of a
a dielectric substrate of permittivity s
the substrate, and
is
and thickness
h,
conducting strip of width w,
and a ground plane beneath
called a microstrip line. Neglecting the fringing effects, determine the
capacitance per unit length of this
line.
Without taking into account the fringing effects, the electric field of the line is
in the dielectric below the strip only. The capacitance of the part of the
with length / is, from Eq. (2.127),
Solution
uniform and localized
line
C = e~,
(2.134)
h
and the capacitance per unit length of the
line
comes out
to be
(2.135)
capacitance
microstrip
Note, however, that this expression
Example
is
is
accurate only for h <&w. Namely, as pointed out in
2.11, the actual electric field distribution in the structure with
quite different
from that
in Fig. 2.20. In a later
chapter (on
of a
fringing
neglected
an arbitrary ratio w/h
field analysis of
h
p.u.i.
line,
transmission
Figure 2.20 Microstrip
transmission
2.13.
line; for
Example
91
92
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
Figure 2.21 Strip transmission
line; for
Example 2.14.
lines),
we shall
present accurate empirical formulas for the electrostatic analysis of microstrip
lines for all values of
Example 2.14
w/h.
Strip Transmission Line
The transmission line consisting of a strip conductor between two conducting planes (large
plates) at the same potential, as depicted in Fig. 2.21, is called a strip line. Let the width of
the strip be w and its distance from each of the planes be h. The permittivity of the dielectric
is £. Assuming that h <5C w, find the capacitance per unit length of the line.
Solution
Neglecting the fringing fields (as h
to a rectangular closed surface of length
we have
l
<£w) and applying
the generalized Gauss’ law
(along the line) that encloses the strip (Fig. 2.21),
[see the left-hand side of Eq. (1.149)]
2
O'!
Ewl=—,
(2.136)
£
E is the electric
field intensity between the line conductors and Q' the charge per unit
The voltage between the conductors is V = Eh, from which the per-unitlength capacitance of the line amounts to
where
length of the
capacitance
line,
p.u.l.
line.
(2.137)
of a strip
fringing neglected
Accurate analysis for an arbitrary ratio w/h
Example 2.15
will
be presented
in a later chapter.
Thin Symmetrical Two-Wire Transmission Line
shows a thin symmetrical two-wire transmission line in air. The charge per unit
is Q
and the radii of the wire conductors, a, are much smaller than the
distance between the conductor axes, d. Under these circumstances, compute the capacitance
Fig. 2.22
length of the line
'
,
per unit length of the
Solution
By
line.
superposition, the total electric field intensity vector in air
E=
Ej
+ E2,
is
given by
(2.138)
I
Figure 2.22 Two-wire
transmission line with d
air;
for
Example
2.1 5.
» a in
Homogeneous
Analysis of Capacitors with
Section 2.1 3
Dielectrics
where Ei and E 2 are the fields due to the individual charged conductors. Since d^> a, these
fields can be evaluated independently from each other, so we use the expression for the field
of an isolated charged wire conductor in
air,
Eq. (1.196). At a point
the axes of conductors, Fig. 2.22, vectors Ei and
intensity turns out to
E2
M in the plane containing
are collinear, and thus the resultant field
be
E=
+ £2
E\
(7.1
3°
5
where x stands for the coordinate defining the position of the point M.
The voltage between the conductors is
d -°
V= f
Eix
Jx=a
d
±-t
Jk-\r
ya
%
27re 0
—x
a d (d
)
d—x
Ja
|
2^ V aA
=
—
Q'
of the line
is
made
d
- lnW -
d-a
0'
a
71 £q
ln
(
~ ln
zh)
(2.140)
In
,
Cl
d-a
of the approximate relation
^
d.
The capacitance per
unit length
is
C-IO
V
Let us
potential
-ir
d
In
tceq
where the use
-~“
now
due
TZEO
(2.141)
ln (d/a)
two-wire
obtain the same result in a different way, using the expression for the
to an isolated charged wire conductor in air, Eq. (1.197) for r
>
a,
and the
We adopt a point M 2 on the surface of the right conductor (Fig. 2.22)
to be the reference point for potential (Vm = 0). The potential at a point Mi on the surface
superposition principle.
2
of the other conductor (with respect to the reference point)
VMl =
Q!
,
d-a
ln
2n£o
+
-Q!
amounts
to
a
,
ln
(2.142)
,
d
2 tzbq
a
—a
with the two terms representing the potentials [Eq. (1.197)] due to the conductors with perunit-length charges Q' and — 0', respectively. The capacitance per unit length of the line is
=
C'
which yields the same expression
Example 2.16
Three
as in
O'
^
=
—
VV
O'
(2.143)
,
Eq. (2.141).
Parallel Equidistant
Wires
Three parallel thin wire conductors are situated in air, as depicted in Fig. 2.23. The wire radii
are a = 1
and the distance between the axes of conductors is d = 50 mm. Two wires
are galvanically connected to each other. Determine the capacitance per unit length of this
mm
system.
Solution
nite length.
is
This
The
is
a two-conductor transmission line, which
first
the isolated (free) wire (wire
1),
we
analyze as a capacitor of
infi-
conductor of the transmission line)
while the other electrode (the other line conductor) con-
electrode of the capacitor (the
first
of the two short-circuited wires (wires 2 and 3), which are at the same potential. Let the
electrodes be charged by Q' and —Q' per unit of their length, respectively. Because of symsists
metry, the charge
—0'
of the second electrode
as indicated in Fig. 2.23.
is
distributed equally
(Fig. 2.23),
first
between wires 2 and
3,
We assume that there is no charge on the short-circuiting conductor.
Let us proclaim the second electrode of the capacitor to be
potential of the
electrode,
i.e.,
the potential at the point
with respect to the reference point
M
2
capacitance
at a
Mi on
zero potential. The
the surface of wire
taken on the surface of wire 3
(Fm 2 =
1
0)
p.u.l.
line
of a thin
93
7
94
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
Figure 2.23 Transmission
line
consisting of an isolated (free)
wire and two galvanically
interconnected wires
for
is
the
in air;
Example 2.16.
sum
of the corresponding potentials due to wires
Eq. (1.197) for
r
>
VMl =
In
2neo
The capacitance per
-
+
a
(a)
2.1
shown
In
+
—(272
—
=
Q!
and
3, respectively.
By means
of
d
=
3 Q'
d
In
47reo
-.
a
(2.144)
is
47T£q
_
VMl
a
In
2neo
unit length of the line (capacitor)
- 9.48
pF/m.
(2.145)
3 In (d/a)
Thin Wire Conductor above a Ground Plane
Consider a transmission
ing plane, as
d
d
Q'/2
2tteq
C'
Example
1, 2,
a.
line that consists of a thin
in Fig. 2.24(a).
The medium
height of the wire axis with respect to the plane
wire conductor and a grounded conduct-
is air,
is
h,
the wire
is
parallel to the plane, the
and the wire radius
is
a (a
<5C
h).
Find
the capacitance per unit length of this line.
Solution Assume that the charge per unit length of the line is Q', namely, that the wire is
charged by Q' and the plane by — Q' per unit length. By image theory (Figs. 1.47 and 1.49),
we can replace the conducting plane by a negative image of the charged wire and obtain the
equivalent thin two-wire line in Fig. 2.24(b), with the distance between the conductor axes
d = 2 h. Of course, the two systems are equivalent only in the upper half-space. The voltage
between the conductors in the original system (wire-plane) equals a half of the voltage in
the equivalent system (wire-wire). Hence, the capacitance per unit length of the wire-plane
transmission line comes out to be (see Fig. 2.24)
&
(b)
Figure 2.24
V
(a)
Thin
wire conductor above a
grounded conducting
plane in air and (b)
equivalent two-wire
for
Example
2.1 7.
where C'c
from Eq.
is
_
Q'
Vc /2
=
2
—
=
V
c
2
C
e
=
2jr£ °
(2.146)
ln(2A /a)'
the capacitance per unit length of the equivalent two-wire line, which
is
computed
(2.141).
line;
Problems 2.25-2.41; Conceptual Questions (on Companion Website): 2.10-2.18;
:
MATLAB
Exercises (on
Companion Website).
Section 2.14
Analysis of Capacitors with
Inhomogeneous
Dielectrics
ANALYSIS OF CAPACITORS WITH INHOMOGENEOUS
2.14
DIELECTRICS
In this section,
we
deal with systems (capacitors) containing
dielectrics, and, specifically,
mogeneity. The
first
class includes
(abruptly or continuously
system
if air-filled.
5
)
inhomogeneous
with two basic classes of systems with dielectric inho-
systems in which the dielectric permittivity varies
in the direction of the electric field lines of the
same
In the second class of systems, the dielectric permittivity varies
normal to the electric field lines of the air-filled system. As we shall
which the electric flux density vector, D, varies in the first class of
dielectrics is the same as in the corresponding air-filled system, while in the second
class of dielectrics this is true for the electric field intensity vector, E. Hence, the
two classes of systems are referred to as D- and E-systems, respectively.
To illustrate these concepts, consider the two parallel-plate capacitors with
in the direction
see, the
way
piece-wise
tric
in
dielectrics shown in Fig. 2.25. The permittivities of dielecAssume that the fringing effects are negligible, and that the
homogeneous
pieces are s\ and £ 2
field in
each dielectric piece
is
uniform.
The capacitor in Fig. 2.25(a) belongs to the first class of systems with dielectric
inhomogeneity - the dielectric permittivity (abruptly) changes in the direction of
the field lines, perpendicularly to the capacitor plates, so it is a D- system. From the
generalized Gauss’ law, Eq. (2.43), applied to a rectangular closed surface enclosing
the plate charged with Q, with the right-hand side positioned in either one of the
dielectrics,
we have
D\
= £>2 = D = Q
(2.147)
s'
D = const,
Fig.
capacitor in
2.25(a)
where S is the surface area of the plates. That D\ = £>2 is also obvious from the
boundary condition for the normal components of D, Eq. (2.81), applied to the
interface between the two dielectrics. The electric field intensities in the dielectrics
are not the same,
Ei
—
D
—
Q= —
£1
£10
D
—
E2 =
and
Q
= —
The voltage of the capacitor comes out
to be
V = Eidi+E2 d2 =
fS
(2.149)
\s 1
e2
J
with d\ and d2 standing for thicknesses of the layers. The capacitance
c_ Q =
V
(2.148)
e2 S
£2
is
£1^2 S
'
e 2 d\
+ s\d2
In the capacitor in Fig. 2.25(b), the change of dielectric characteristics
normal to the
-
(2.150)
is
in the
an E-system (the second class of systems). By means of Eq. (1.90) applied to a straight path between the capacitor
plates, positioned in either one of the dielectrics,
direction
field lines
E\
5
Inhomogeneous
-
this is
E2 — E
V
1
'
composed of a number of homogeneous pieces of different permittivities
homogeneous dielectrics; continuously inhomogeneous dielectrics, on the other
dielectrics
are called piece-wise
side, are characterized
(2.151)
by continuous spatial variations of permittivity.
E=
Fig.
const, capacitor in
2.25(b)
95
96
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
Figure 2.25 Capacitors
with two-layer dielectrics
representing a D-system
(a)
and £-system
(a)
(b).
where
is
V\
+
-Q
Q
V2
+
is
the separation between plates [E\
from the boundary condition
D\
c\
d
the capacitor voltage and
also obtained
for tangential
= E2
components of E,
Eq. (2.79), applied to the dielectric-dielectric interface]. The electric flux density
is discontinuous across the interface,
-Q
Q
V
(b)
c2
=
e,V
—
—
=
e\E
e2
D 2 = e 2 E=
and
d
V
(2.152)
d
Applying the generalized Gauss’ law to a rectangular surface positioned about the
0—
Mb
plate with the charge
Q yields
V
+
Q=D
(a)
1
S1
+ D 2 S2 =
(eiSi
+ £ 2 ^2
)
—
d
(2.1
,
53)
with S\ and S2 being surface areas of the parts of the plate interfacing the individual
dielectric layers,
and the capacitance
is
Q _
V~
The systems
view.
sufficient for
Note
Figure 2.26 Equivalent
two capacitors
and parallel (b).
series (a)
some
is
usually
+ e 2 S2
(2.154)
d
can be explained also from the circuit-theory point of
much simpler than the full field-theory view, and is
evaluations. Let us
compute the capacitances of capacitors
in
Fig. 2.25 using the associated equivalent circuits.
(b)
circuits for
in Fig. 2.25
Such a view
£iSj
that the interface
between the
dielectric
layers of the capacitor in
equipotential, and nothing will change in the system
Fig. 2.25(a)
is
this surface,
i.e.,
in
insert a metallic foil
between the
layers.
We
if
we
metalize
thus obtain two capac-
itors in series, as indicated in the equivalent circuit in Fig. 2.26(a).
The charges of
both capacitors are the same, and therefore
v=
from which the
equivalent capacitance of two
capacitors in series
Vi
Q
Q
+ v2 = ^- + ^- = Q
L|
L2
total capacitance of the
1
+
(2.155)
C2
system amounts to
c=e =
V
1
C[
C\C2
C\
+ c2
(2.156)
Section 2.14
With the use of Eq.
S
On
allel,
Inhomogeneous
Dielectrics
97
(2.127), the capacitances of individual capacitors are
Ci=ei—
a\
same expression
resulting in the
Analysis of Capacitors with
£2
S
—
a
(2.157)
,
2
C as
for
C2 =
and
Eq. (2.150).
in
the other hand, the system in Fig. 2.25(b) represents two capacitors in par-
the equivalent circuit of which
is
shown
The voltages
in Fig. 2.26(b).
of both
capacitors are the same, so that
Q=
The
QV + c
+ 02 =
Qi
total capacitance of the
2
system
v=
+ c2
(Ci
v.
)
(2.158)
is
C=^ = Ci + C
2
(2.159)
,
equivalent capacitance of two
capacitors in parallel
where,
Ci=ei-j
C2 =
and
a
which gives the same
result as in
A spherical capacitor
is filled
Two
Concentric Dielectric Layers
with two concentric dielectric layers.
=
2.
The
radius of the inner electrode
a
is
=
The
relative permittivity
= 4,
and that of the other layer
10 mm, the radius of the boundary surface
of the layer near the inner electrode of the capacitor
e T2
(2.160)
-f,
a
Eq. (2.154).
Spherical Capacitor with
Example 2.18
S2
e2
is
e r\
between the layers is b = 25 mm, and the inner radius of the outer electrode is c = 35 mm.
If the voltage between the inner and outer electrode is V = 10 V, what are (a) the charge of
the capacitor and (b) the total bound charge on the interface between the layers?
Solution
(a) This
is
a D-system,
and the
electric flux density vector
D = D(r)
where
r
and
r are
is
of the form
(2.161)
r,
the radial coordinate and the radial unit vector in the spherical coordi-
nate system with the origin at the capacitor center, as shown in Fig. 2.27.
Gauss’ law applied to a spherical surface placed in either the
first
The generalized
or the second layer
tells
us that
D(r)
where
=
®
a
4nrz
<
r
<
(2.162)
c,
Q is the capacitor charge - to be determined.
The
electric field intensity vector
the second one.
The capacitor voltage
=
Ei
is
D/ei
in the first layer
and
E2 = D /e 2
in
given by
is
= [ Ei(r)dr+ f E2 (r)dr = Q_
4n
Ja
Jb
1
.
£l
l
1
,
U
(
U
bj
*YI
c) J
(2.163)
from which
Q = 47T60
/b
—
a
r
\e r i ab
+
c
— b\
1
—r
oc )
V=CVr= 53.7 pC,
(2.164)
e x2
where C = 5.37 pF is the capacitance of the capacitor.
Note that the capacitor charge can also be found by computing C as the equivalent
total capacitance of a series connection of two spherical capacitors with homogeneous
dielectrics,
Eq. (2.156), where, using Eq. (2.119),
Ci
=
4ns T i£oab
b
—a
7.4
pF
and
C2 =
—=
4jr£ r2 £ohc
19.5 pF.
(2.165)
98
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
Figure 2.27 Spherical
capacitor with two concentric
homogeneous
for
(b)
Example
From
dielectric layers;
2.1 8.
Eqs. (2.59) and (2.47), the intensities of the polarization vector in the dielectric
layers with respect to the positive (outward) radial direction (Fig. 2.27) are
Pi ( r )
e
= J±Z±
D(r)
P2 (r) =
and
e rl
By means
of Eq. (2.89), the
between the layers turns out
Pps
= n P2 -
and hence the
Qp =
total
Pi
D(r).
(2.1
66)
bound surface charge density on
the boundary surface Sb
to be
= -P2 (b + + Pi(b~) =
)
(
V
£
£
-^~
- -^l)
D(b),
e rl
e r2
(2.1
67)
(2.1
68)
/
bound charge on the surface
PpsSb
=
P P s4nb
2
=
-
(
E 2
J
-Zl) Q =
Spherical Capacitor with a Continuously
Example 2.19
A
n
^
£ r2
13.43 pC.
Inhomogeneous
Dielectric
is filled with a continuously inhomogeneous dielectric, whose perdepends on the distance r from the capacitor center and is given by the function
e(r) = 3eo b/r, a < r < b, where a and b are radii of the inner and outer capacitor electrode,
respectively. The outer electrode is grounded and the potential of the inner electrode is
V. Find (a) the capacitance of the capacitor and (b) the bound charge distribution of the
spherical capacitor
mittivity
dielectric, (c)
By
the dielectric
is
integrating the charge densities in (b), prove that the total
bound charge of
zero.
Solution
(a)
Let us subdivide the dielectric into thin spherical concentric layers of radii
nesses d r, a
<
r
<
b, as
shown
in Fig. 2.28.
Each
r
and
thick-
thin layer can be considered as being
now obvious that this capacitor represents a genwhich has only two such layers. Therefore, the electric flux
density vector in the two capacitors is the same, given by Eqs. (2.161) and (2.162). Here,
E = D/e(r), and the potential of the inner electrode with respect to the ground (outer
electrode) and the capacitance of the capacitor can be obtained as
homogeneous, of permittivity
eralization of that in Fig. 2.27
h
V= f
E(r) dr
=
Q_
4tt
f
Ja
e(r). It is
,
dr
r 2 e(r)
Specifically, for the given function e(r),
C
Q = 47T
v
C = Unepb/ In (b/a).
r
1
b
dr
Mr)
(2.169)
b
Section 2.14
Inhomogeneous
Analysis of Capacitors with
Dielectrics
99
Figure 2.28 Spherical capacitor
with a continuously
inhomogeneous
dielectric of
permittivity s(r), subdivided
into differentially thin
homogeneous
for
(b)
The
polarization vector in the dielectric
P{r)
-
e(r)
=
is
Example
radial, with intensity [Eqs. (2.59)
£p
D(r)
e 0 (3 b
=
and
(2.170)
bound volume charge density
of Eqs. (2.19) and (1.171), the
(2.47)]
-r)V
r 2 In (b/a)
e(r)
By means
layers;
2.1 9.
inside the
amounts to
dielectric
(2.171)
In (b/a)
while,
from Eq.
(2.23), the
bound
surface charge densities
’
on the surfaces of the
dielectric
near the inner and outer capacitor electrodes are
Ppsfl
p a +,)
P(
—
,
£0(3 b
_
—
- a)V
and
2
Ppsb
'~ p!,D
= P(b"
*
)
v
2s 0 V
=
'
a \n(b/a )
(2.172)
b\n(b/a)'
respectively.
(c)
To
verify that the total
bound charge
f
Qp
=
of the dielectric
I
dv
4Treo(3b
volume of the
— a)V
In (b/a)
In (b/a)
the
we
write
2
Sb
5a
— a)V
4neo(b
is
zero,
4na 2 +pps b 4n
dr Tppsa
Pp(r)
a
where dv
is
b
+
8 nepbV
=
0
(2.173)
,
In (b/a)
thin layer of radius r in Fig. 2.28, while
Sa and Sb are the
areas of surfaces of inner and outer capacitor electrodes, respectively.
Spherical Capacitor Half Filled with a Liquid Dielectric
Example 2.20
A spherical capacitor has a liquid dielectric occupying a half of the space between the elecThe radii of electrodes are a and b (a < b ), and the permittivity of the
Determine the capacitance of the capacitor.
trodes.
Solution
Assume
that the capacitor
charged by Q, as indicated
is
E-system, and the electric field intensity vector, E,
is
entirely radial.
dielectric
in Fig. 2.29. This
From
is e.
is
an
the boundary con-
components of the electric field intensity vector, Eq. (2.79), its magnitude
depends on the radial spherical coordinate, r, only, i.e., it is the same at all points of a sphere
dition for tangential
of radius r (a
The
<
r
<
b),
shown
D,
in Fig. 2.29.
electric flux density vector
is
Di
=
e r eoE in the dielectric
Applying the generalized Gauss’ law to the sphere of radius
and
r (Fig. 2.29),
D 2 = eoE
we
in air.
obtain
Figure 2.29 Analysis of a
spherical capacitor half filled
with a liquid dielectric; for
1
1
Dilnr + D22nr =
Q
—
->
(e r
+
2
l)£oE(r)27rr
=
Q,
(2.174)
Example 2.20.
.
100
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
and hence
^
'
+
27t(e T
l)£ 0 r 2
The capacitor voltage and capacitance then come out
b
v=
We
a
E(rHr
Jfa
2n(e T
_
+
note that this expression for
b)
\a
l)e 0
C
— 2n£
r
£oab/(b
—
r _Q_
V
~
~
and
C2 =
in Fig. 2.30(a)
is
(
2 176 )
.
2jr(s r
+
1
)e 0 ab
b-a
.
represents the equivalent total capacitance of a parallel
2n£oab/(b
—
dielectrics,
Eq. (2.159), where
a).
Coaxial Cable with a Continuously
Example 2.21
Shown
a)
2 175 )
to be
connection of two hemispherical capacitors with homogeneous
C\
(
Inhomogeneous
Dielectric
a cross section of a coaxial cable with a continuously
inhomogeneous
which is given by a function £(<p) = (3 + sin0)fo,
e [0, 2n].
The voltage between the inner and outer conductor of the cable is V. Compute (a) the
capacitance per unit length of the cable and (b) the distribution of free charges on the cable
dielectric, the permittivity of
conductors.
Solution
(a)
Let us subdivide the dielectric into thin sectors
(slices)
defined by elemental azimuthal
Each such sector can be considered as being
homogeneous, of permittivity e(</>). This way it becomes obvious that dielectric inhomogeneities in the systems in Figs. 2.29 and 2.30 are of the same type.
angles d0, as depicted in Fig. 2.30(b).
Because of symmetry, the
electric field intensity vector in the dielectric
=
and, based on the boundary condition in Eq. (2.79), Ei
[Fig. 2.30(b)].
b ),
i.e., it is
vector
is
This means that
E is
<p)
=
=
E3, E3
radial,
is
=
E4,
constant on a cylindrical surface of radius r (a
a function of the radial cylindrical coordinate,
D(r,
E2, E2
r,
only.
The
<
...
r
<
electric flux density
e(0)E(r). Generalized Gauss’ law applied to a cylinder of radius r and
height (length) h gives
r2n
I
— — = Q'h
£(<p)E(r) rd<ph
70=0'
v
-
D
dS
Qs
—
>
E{r)
=
—
q'
^
r J0
f
,
n
£((/))
(
2 177 )
.
d<p
is the area of an elemental strip of width rdcp and height h, and Q is the
charge per unit length of the cable. The voltage between the cable conductors and the
where dS
Figure 2.30 Analysis of
a coaxial cable with a
continuously inhomogeneous
dielectric of permittivity e(0):
(a) cross-sectional
structure
and
view of the
(b) subdivision of
the dielectric into differentially
thin
homogeneous
Example 2.21
sectors; for
1
'
Section 2.14
Inhomogeneous
Analysis of Capacitors with
capacitance per unit length are then obtained as
V=
E{r)dr
f
=
Q'\n(b/a)
fo*
(b)
e(<f>)
fo*
s(cp) d(p
d0
From
Eqs. (2.177) and (2.178), the electric field intensity, E(r),
filled
coaxial cable, Eq. (2.124), which
Eq.
(2.58), the free surface
(2.178)
In (b/a)'
is
expected, because this
is
6jt£q
_
In (b/a)
the
is
same
as in the air-
an E-system. Using
charge density on the inner conductor of the cable
=
Psa((p)
,,
NEV
s(4>)E(a
+v
+
)
=
is
+ sin <j>)V
eo(3
(2.179)
,
a In (b/a)
while on the outer conductor,
Psb(</>)
^
srvI.- s
-e(4>)E(b )
=
Two-Wire Line with
Example 2.22
e 0 (3
=
,
Coated Conductors
Each conductor of
a thin symmetrical two-wire transmission
the conductor axes
d and conductor
permittivity e
and thickness
a.
(d^>
ni9m
(2.1 80)
.
b In (b/a)
Dielectrically
radii a
—
+ sm<p)V
a), is
line,
with the distance between
coated by a coaxial dielectric layer of
Find the expression for the capacitance per unit length of
this
line if situated in air.
Solution Let the
line
Q per unit of
be charged by
its
length, as
shown
in Fig. 2.31. Similarly
to the analysis of a thin-wire line with bare conductors in Fig. 2.22, the electric fields
due
to
the individual charged conductors can be evaluated independently from each other, because
the line
point
is
thin.
Thus, the electric flux densities due to the conductors
1
and 2 evaluated
at the
M in Fig. 2.31 are
D2 =
and
Q'
2n(d
—x
(2.181)
)
The total electric flux density is D = D\ + D 2 To simplify the computation,
however, we take that D & D\ in the first dielectric coating (the second conductor is far
away) and D
D 2 in the second coating. Having in mind that the electric field intensity in
respectively.
.
dielectric coatings
is
E=
D/e, while
E = D/eo
in air, the voltage
between the conductors
is
given by
nd—a
V=
/
r2a
j
Didx-|
= |:ri ln2+
\_s
i(ln
sq \
rd—2a
/
(£>i
+ D2
)
dx
+
e 0 J2a
& Ja
Ja
2n
j
Edx=-
ln
d
rd—a
D 2 dx
/
£ Jd—2
-la
^_ ^)
2a
^
-
— 2a J +
Q' (
i.„2l^(
—
1
(
s
J
tc
T
7
\£
\
,
1
„
In 2
,
In
ft
—d
2a
Eq
(2.182)
[see the integration in Eq. (2.140)].
C'
=
The capacitance per
unit length of the line
d
Q!
—
= 7r[-ln2+ — —
1
.
In
eg
is
(2.183)
2a
Figure 2.31 Cross section
of a thin two-wire line with
dielectric coatings
conductors,
over
in air; for
Example 2.22.
Dielectrics
101
102
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
Example 2.23
Spherical Capacitor with a Nonlinear Dielectric
Consider a spherical capacitor with
short-circuited,
and there
steady state (Fig. 2.32).
where Po
a constant
is
field intensity
a and b (a
<
b), filled with a fer-
is
a
The
and
remanent (residual) polarization
polarization vector
r is
in the dielectric in the
radial, with
is
magnitude P(r)
=
new
Pob/r,
the distance from the capacitor center. Calculate the electric
vector in the dielectric.
Since this
Solution
Figure 2.32 Nonlinear
radii of electrodes
After being connected to a voltage generator, the capacitor electrodes are
roelectric.
is
a nonlinear capacitor (ferroelectrics are nonlinear materials), rela-
tionship in Eq. (2.112) does not hold, and there
them
a charge,
is
Q
new
and —Q, on the capacitor
state. From spherical sym-
spherical capacitor with
electrodes although the voltage between
short-circuited electrodes
metry and the given form of the vector P in the dielectric, the electric flux density vector in
the dielectric, D, has that same form (radial vector that depends on r only), as in Eq. (2.161).
This means that the capacitor in Fig. 2.32 represents a D-system. Then, using the generalized
Gauss’ law, Eq. (2.43), which is true for arbitrary (including nonlinear) media, the magnitude
of the vector D is found to be that in Eq. (2.162).
To determine the electric field intensity vector, E, from D, however, we cannot use the
relationship in Eq. (2.47), as the dielectric is not linear, but we can use the definition of the
vector D in Eq. (2.41), which yields
and remanent polarization;
for Example 2.23.
is
zero
in the
D-P
(2.184)
£o
From
the condition
V=
0,
f
E(r) dr
=
that
—
£0
J r=a
f [D{r)
—
P(r)] dr
= 0,
(2.185)
Ja
is,
P0 b In - =
a
Hence,
Q=
AtzPq In (b/a) ab 2 /(b
—
a)
E=
0.
(2.186)
and
ab
Pob
e0 r
|_
(
In (b/a)
b-a)r
(2.187)
Problems 2.42-2.56; Conceptual Questions (on Companion Website): 2.19-2.26;
:
MATLAB
2.15
Exercises (on
Companion Website).
ENERGY OF AN ELECTROSTATIC SYSTEM
Every charged capacitor and every system of charged conducting bodies contain a
amount of energy, which, by the principle of conservation of energy, equals
the work done in the process of charging the system. This energy is called the electric energy and is related to the charges and potentials of the conducting bodies in
the system. To find this relationship, we perform a numerical experiment in which
a system that was initially uncharged is being gradually charged, by bringing elementary charges to the conducting bodies by an external agent. We evaluate the
net work done against the electric forces while the charges of the bodies are being
changed from zero to their final values.
Consider first a linear capacitor of capacitance C. Let the charges of the electrodes of the capacitor be Q and —Q, and their potentials with respect to a reference
point be V\ and V2 respectively. The capacitor voltage is
certain
,
Q
V=V\-V2 = ^.
(2.188)
)
Section 2.1 5
103
Energy of an Electrostatic System
we add an elementary positive charge d Q to the first electrode and — dQ to the
second one, the change of the energy of the capacitor is equal to the work done
against the electric forces in moving d Q from the reference point (i.e., from the
zero potential level) to the first electrode (which is at the potential Vi) and — d Q
to the second electrode (at the potential V2 ). By the definition of the electric scalar
potential, the potential of a point on either one of the electrodes equals the work
done by the electric field in moving a charge from the electrode to the reference
point, or, conversely, the work done against the electric field in moving a charge
from the reference point to the electrode, divided by the charge [Eq. (1.74)]. This is
exactly what we have in our experiment, so the elementary work done while moving
If
d Q and
— dQ
is
given by
dWe = dQ El + (- d Q) V2
(2.189)
,
or, equivalently,
dWe = dQ(V -V2 = dQV = dQ^.
When
(2.190)
)
1
uncharged, the capacitor has no energy. The total energy stored
when
in the
charged with Q is therefore equal to the net work done in
changing the capacitor charge from q = 0 to q = Q, and is obtained by adding up
all elementary works dWe in Eq. (2.190):
capacitor
it
is
W
e
‘
C
(2.191)
2
Hence, the equivalent expressions for the energy of a capacitor are
We =
The
unit for the electric energy
of an isolated charged metallic
is
^
2C
l
2
the joule
body
is its
C here is the capacitance
(J).
-
CV2
(2.192)
.
energy of a capacitor
(unit: J
2
As
a special case, the electric energy
in a linear dielectric
is
C ^isolated body’
9
where
QV
^ =
(2.193)
energy of an isolated metallic
body
of the single body, given by Eq. 2.116, Violated body
and two additional
potential with respect to the reference point at infinity,
equivalent expressions, like in Eq. 2.192, can be written as well.
Note that the energy of a capacitor can also be expressed in terms of the charges
and potentials of individual electrodes as
We = -
QV =
-
Q(V
1
-V2
)
Generally, for a linear multibody system with
—
~ (Q\Vi
+ Q 2 V2 ).
(2.1
94)
N charged conducting bodies,
(2.195)
energy of a multibody
electrostatic
Finally,
tric
if
a system also includes a
between the conductors,
volume charge
distributed throughout the dielec-
in addition to surface
charges over the conductors, the
system
d
.
104
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
total electric
energy of the system
is
N
1
1
2
2
We
energy of a multibody system
with volume charge in the
l
P V dv,
(2.196)
dielectric
where p is the charge density and V the electric scalar potential in the dielectric,
and energy (p dv) V /2 of an elemental charge d Q = p d V is integrated throughout
the volume v of the dielectric.
ELECTRIC ENERGY DENSITY
2.16
Expression
in
Eq. (2.196) enables us to find the total energy associated with the
As the charges are sources of the
charge distribution of an electrostatic system.
electric field,
it
turns out that the energy of the system can be expressed also in
terms of the electric
field intensity
tion that the electric energy
is
(which can be
in the dielectric
throughout the system. This leads to an assumpand therefore
actually localized in the electric field,
air
and a vacuum) between the conductors of an
we
electrostatic system. Quantitatively, as
energy
shall see, the concentration (density) of
at specific locations in the dielectric
is
proportional to the local electric
field
intensity squared.
Let us consider
first
a simple case of a
homogeneous
electric field in the
dielectric of a parallel-plate capacitor, with plate areas 5, plate separation
is
compared
small
[Fig. 2.18(a)].
From
d
=
e
1
-
S
1
9
- e-(Ed) 2
2 d
CV 92 =
2
= -1 eE29 Sd =
2
(
dielectric permittivity e
Eqs. (2.192), (2.127), and (2.126), the energy of the capacitor
W
where
and
to the dimensions of the plates),
1
is
9
- eE2 v„
(2.197)
2
E is the electric field intensity in the dielectric and v — Sd is the volume of the
i.e., the volume of the domain where the electric field exists. We define
dielectric,
the electric energy density as
we =
W =
—
e
or,
electric
energy density
by employing Eq.
we =
1
— eE29
=
2
J/m 3 )
unit
in the
is
(2.198)
,
(2.47),
(unit:
The
1
- sE29
2
v
1
-
D
—
2
ED =
(2.199)
2e
2
J/m 3 (volume density). The energy of the capacitor can now be written
form
W =w
e
We
shall
now
energy contained
(2.200)
e v.
generalize this result to obtain a field-based expression for the
in
an arbitrary capacitor
(Fig. 2.14)
and an arbitrary electrostatic
system. Let us subdivide the domain between the conducting bodies of a system
into elementary flux tubes containing the lines of vector D, that start
nate at surfaces of the bodies.
cells of
volume
then cut each tube along
dv, such that the interfaces
pendicular to the
field,
We
field lines.
We
its
and termi-
length into small
between the neighboring
cells are per-
can metalize these interfaces, without changing the
and get an array of small parallel-plate capacitors along the tube. The entire
i.e., the domain with the electric
space between the conducting bodies of the system,
Section 2.16
field, is
thus partitioned into volume
parallel-plate capacitor with a
between the
plates.
cells,
and each
homogeneous
The energy contained
in
cell
dielectric
in all capacitors, that
is,
is
Energy Density
105
represents an elementary
and uniform electric field
each capacitor
is
dWe = w e dv,
where the energy density
Electric
(2.201)
given by Eq. (2.199).
Summing
the energies contained
dWe throughout the entire dielectric
integrating the energy
between the conducting bodies of the system (volume
v),
we
obtain the electric
energy of the entire system:
l
The expressions
ED dv.
(2.202)
and (2.202) are equivaenergy of an electrostatic system.
namely, they give the same result for the total
The latter expression, however, implies that the actual localization of electric energy
is throughout the electric field, that is, in the dielectric between the conductors of
the system (even if the dielectric is a vacuum), and not in the conductors, nor on
their surfaces. It also provides us with a
means, the energy density
in
Eq. (2.199), for
evaluating and analyzing the exact distribution of energy throughout the dielectric.
On
might
the other hand, reconsidering Eq. (2.196) and reinspecting
now come up
its
terms,
we
with an alternative viewpoint on the actual energy localization
which implies that the stored electric energy of a system resides
system charge, and not the field. Accordingly, pV / 2 might be considered
to be the volume energy density at points where p # 0 throughout the volume of
the dielectric [from the second term of the energy expression in Eq. (2.196)]. The
in electrostatics,
in the
corresponding surface energy density, equal to ps V /2, would then quantify energy
localization in the surface charge distribution over the surfaces of the conductors in
Eq. (2.196)]. Both
viewpoints have merit and are equally “correct.” Nevertheless, the assumption that
the system [constituting the
the energy
is
first
term of the energy expression
actually “contained” in the field,
turns out to be
much better
and not
in the
in
charge producing
it,
suited for energy considerations in the analysis of elec-
tromagnetic waves (to be done in a later chapter). Namely, an electromagnetic wave
consists of time-varying electric
and magnetic
fields
which travel through space
(even in a vacuum) and carry energy independent of the sources (charges and currents) that
it is
much
produced them (the sources might not even
simpler and
more
than to associate
of time. This
is
it
any more). Therefore,
fields,
which change
in
time and travel in space,
with the stationary distribution of sources at previous instants
why we have adopted
in the first place
exist
natural to describe the energy distribution of a system
with traveling waves in terms of the
the field-based energy localization approach
and the associated energy density expressions, those
in
Eq. (2.199),
to quantify the energy distribution in an arbitrary (linear) system (with static or
time-varying charges and
Example 2.24
fields).
Energy of Parallel-Plate Capacitors with
Two
Dielectric Layers
Find the energy of each of the capacitors in Fig. 2.25, neglecting the fringing effects. Express
the energy in terms of the capacitor charge in case (a) and the capacitor voltage in case (b).
Solution
Based on Eqs.
energy of an
electromagnetic system, via
for the electric energy in Eqs. (2.196)
lent,
electric
(2.202), (2.147), (2.192), (2.157),
and
(2.150), the
energy of the
capacitor in Fig. 2.25(a) can be found using either (1) the electric flux density in the dielectric,
or (2) the capacitances of two capacitors corresponding to individual dielectric layers, or
the energy density
V
106
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
(3) the equivalent total capacitance of a series
connection of capacitors with homogeneous
dielectrics:
D
—
D
—Sd
2s
2
W
e
=
Q
~+~
2
Sell
+
2e\
= Q
2
2C2
2C\
2
(1 )
(
2
{e 2 d\
C
+e\d2 )Q 2
(2.203)
2 e\e 2 S
(3)
2)
energy of the capacitor
Similarly, the
&
2
2
2.25(b) can be
in Fig.
computed by employing
two capac-
either (1) the electric field intensity in the dielectric, or (2) the capacitances of the
itors with
homogeneous
dielectrics, or (3) the total
capacitance of a parallel connection of
capacitors:
W
£\
r2
o .
= —z— S\d
e
s2
-|
E—2
Sn 2 dj
=
QV
—
2
0)
(
[see Eqs. (2.151), (2.160),
Determine
+ £ 2 S 2 )V~
(£[5]
(2.204)
2)
(3)
(2.154)].
Energy Per Unit Length of a Coaxial Cable
Example 2.25
conductor
and
CV 2
C2 V—2
1
energy density and (b) the energy per unit length of a coaxial cable, with
and b (a < b) and dielectric permittivity £, if the voltage between the cable
(a) the
radii a
conductors
V.
is
Solution
(a)
The
electric field intensity in the dielectric of the cable
is
given by Eq. (2.124), and hence
the electric energy density at a distance r from the cable axis
1
we = -
sE(r)
=
2
electric
The
electric
a transmission
J/m)
K=
We
(We)p.„.l.
W
e is
\L
/
line (unit:
where
<
(a
r
we
(2.206)
dv
the energy contained in a part of the cable of length
J/m. Because the energy density
is
W'
=
C
energy via
is,
r
and width
b
J r=a
=
/,
S
ix e
w e (r) 2nrdr
dr,
2
we adopt dS
is
in the
Eq. (1.60), and obtain
r
h
dr
txeV 2
(2.207)
2
r
In (b/a) Ja
dS
the cross-sectional
is
d SI. The unit for W'
a function of the coordinate r only,
form of an elementary ring of radius
p.u.l. electric
(2.205)
b).
—
d SI
area of the dielectric between the cable conductors, and dv
This result
<
energy per unit length of the cable can be evaluated as
energy per unit length
of
2
r
2
2r 2 ln (b/a)
2
(b)
——eV
is
In (b/a)'
of course, in agreement with the expression
WL = Ic'v 2
p.u.l.
(2.208)
,
capacitance
where C'
is
the capacitance per unit length of the cable, given in Eq. (2.123).
Example 2.26
Shown
in Fig.
2.33
Energy
is
in a
Coaxial Cable with a Dielectric Spacer
a cross section of a coaxial cable that
of relative permittivity e x
.
The
dielectric
is
in
is
partly filled with a dielectric
the form of a spacer between the cable
conductors defined by an angle a. The remaining space between the conductors is air-filled.
The conductor radii are a and b (a < b), and the voltage between the conductors is V. Find
f
1
.
Section 2.1 6
Electric
dS2
Figure 2.33 Coaxial cable with
a dielectric spacer; for
Example
2.26.
a
for
which the
energy contained in the dielectric equals a half of the total energy in
electric
the cable.
=
Wei
^e r soE2
where the
electric field intensity
between the
dielectric
and
air.
in the air-filled coaxial cable,
This
is
air part of the
^£o E
w e2 =
and
respectively,
By
and
Electric energy densities in the dielectric
Solution
cable interior are
2
(2.209)
,
E in the cable is continuous across the interface
2.21), and E is the same as
an E-system (see Example
Eq. (2.124).
the requirement in the statement of the example, the energy per unit length of the
cable contained in the dielectric spacer and that in the air are stipulated to be the same,
W'tl
= W' 2
which means that
,
f
(Fig. 2.33)
l-e T e 0 E(r)
2
ardr
=
J r=a 2
dSj
^s 0 E(r) 2 (2n - a)rdr
f
(2.210)
J r =a 2
dS2
We 2
Wel
Here, dSi and dS2 are surface areas of the parts of a thin ring of radius r and width dr
determined by angles a and 2n — a, respectively. Even without any integration, Eq. (2.210)
gives
eT a
2 .7
=2n — a
a
er
Example 2.27
dielectric
Example
(0
<
r
<
and
a)
where the
electric flux density,
=
/
w e (r)47rr2 dr+
Jo
where dv
Eq. (1.33).
is
w eo =
D,
is
D(r)
—
—
(a
<
r
<
(2.212)
oo),
2 £o
given in Eq. (2.63). Hence, the electric
is
/•
t
energy densities in the
electric
2
D(r) 2
energy of the system
W
)
air are
2e r £o
respectively,
.
2.5.
The
This system does not contain conductors.
sphere and
we =
2 211
Energy of a System with Volume Charges
Calculate the electric energy of the system from
Solution
(
+
adopted
/
OO
Weo(r)
Anr2 dr
=
Ja
in the
form of a thin spherical
p
2
—
7V
56£ 0
1
\
£r
/
(7-1
\
),
shell of radius r
(2.213)
and thickness
dr,
Energy Density
107
108
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
The above
result can also
Wc = lj P Vdv=
be obtained employing the expression
l
-l"
p ( r)VM
W
dr
^
=
(7
in
Eq. (2.196),
+ i)
(2.21 4)
,
with v being the volume of the sphere, p the volume charge density given in Example
V the electric scalar potential in Eq. (2.66).
Note
that the integration in Eq. (2.213)
2.5,
and
over both the dielectric sphere and the air,
is because the electric field exists in the
entire space, whereas the volume charge occupies the volume of the sphere only.
while that
Eq. (2.214)
in
is
is
over the sphere only. This
Problems'. 2.57-2.68; Conceptual Questions (on
MATLAB
DIELECTRIC
2.17
Companion
Website): 2.27-2.31;
Companion Website).
Exercises (on
BREAKDOWN
IN
ELECTROSTATIC
SYSTEMS
We
shall
now
analyze electrostatic systems in high-voltage applications,
uations where the electric field in the dielectric
of dielectric
down
breakdown
occurs
when
in the system.
As
so strong that there
is
i.e.,
is
in sit-
a danger
discussed in Section 2.6, dielectric break-
the largest field intensity in the dielectric reaches the critical
ECT Under
value for that particular material - dielectric strength of the material,
the influence of such strong electric fields, a dielectric material
is
.
suddenly trans-
formed from an insulator into a very good conductor, causing an intense current
to flow. While systems with gaseous and liquid dielectrics can completely recover
after a breakdown, those with solid dielectrics are most often permanently damaged by breakdown fields, as insulating properties of the dielectric are irreversibly
degraded. In this section, our goal is not to analyze local breakdown processes in
materials nor the overall behavior of systems resulting from these processes (these
phenomena are largely nonlinear and are not electrostatic), but to evaluate the systems and their performance in linear electrostatic states close to the breakdown
occurrences. In
fact,
our goal here
is
basically to determine the
maximum
extent of
values of various quantities in a system (or device) that are “permitted” for
its
safe
operation prior to an eventual breakdown.
In systems (devices) with nonuniform electric field distributions, the principal
task
is
to identify the
most vulnerable spot for
dielectric
breakdown, and
to relate
the corresponding largest electric field intensity to the charges or potentials of
conducting bodies
transmission
is
called the
line,
in the
system. In the case of a capacitor or a two-conductor
the voltage that corresponds to the critical field in the dielectric
breakdown voltage of the capacitor or
voltage that can be applied to the system (before
it
line.
This
is
the highest possible
breaks down), and
also referred to as the voltage rating of the system.
6
In
some
is
sometimes
applications,
we
the radius of the inner conductor
optimize individual parameters of a system
(e.g.,
6
vary considerably depending on the actual con-
Since the dielectric strength of a given material
ditions under which the material
is
is
may
used, as well as the
manufactured, the voltage rating and
critical
way
the particular piece of a solid dielectric
values of other quantities for a system in practice are
always defined with a certain safety factor included. For example, with a safety factor of
rating of a given capacitor equals a tenth of the voltage that
conditions and assumptions.
would lead
to a
10, the voltage
breakdown under
ideal
Section 2.1 7
of a coaxial cable) such that the
breakdown voltage
is
Dielectric
maximum.
Breakdown
It is
in Electrostatic
also of
(breakdown) values for the capacitor charge and
energy, and the corresponding values for forces on conductors (charge, energy,
and forces per unit length for transmission lines), as these are the largest possible
charge, energy, and forces for that particular system.
For systems filled with a homogeneous dielectric, the strongest electric field in
the dielectric is most frequently right next to one of the conducting bodies of the
system. In addition, this is most likely in the vicinity of sharp parts of a conducting
surface [see Eq. (1.210)]. In systems with heterogeneous dielectrics, on the other
side, the most vulnerable spots can be close to a boundary surface between dielectric
parts with different permittivities, where the normal component of the electric field
interest to evaluate the critical
intensity
is
discontinuous. In such systems, furthermore, the largest field intensity in
a given dielectric part
is
not necessarily the breakdown
field,
because the dielectric
strength vary from material to material, and only an electrostatic analysis (or an
experiment) can
tell
which of the
voltage of critical value
is
dielectric parts
would break down
first
after a
applied to the system.
For a system of conducting bodies situated in air (or any other gas), the
region, where the air is ionized by an avalanche breakdown mechanism
(see Section 2.6) and behaves like a conducting material, is usually localized only
in the immediate vicinity of “hot” spots on conductor surfaces, as the field farther
away from the conductor is not sufficiently strong to sustain the breakdown. Due to
an avalanche of impact ionization of air molecules, vast numbers of free electrons
and positive ions are created, so that the air near the conductor becomes more
and more conducting. The ionized air might even glow (at night), appearing as a
luminous “crown” around the conductor, and hence such a local discharge close
to a conductor surface is referred to as a corona discharge. A corona discharge
on a conductor is equivalent to an enlargement of the conductor, since a layer of
conducting material (ionized air) is added over the conductor surface. It substantially increases losses on power transmission lines and also emits electromagnetic
waves that can interfere with nearby communication devices and systems. On some
occasions, the field intensities in the system are so high that breakdowns occur
even away from the conductors. A continuously ionized path is formed from a
part of a conductor surface with an exceedingly high charge density to the nearest
conductor with opposite charge polarity. This path is apparent as a bright luminous
arc carrying a current of a very large intensity (typically hundreds to thousands
of amperes, sometimes as large as 100 kA). As a result, a very rapid and violent
discharge of the conductors occurs, known as arc discharge. The most apparent
and spectacular arc discharges certainly are intense cloud-to-ground lightning
discharges in the form of giant sparks in thunderstorms. Lightning strikes are, as we
know, a frequent cause of loss of human lives and property.
Analysis of electrostatic systems with fields close to breakdown levels and
predicting critical values of voltages, charges, and other quantities at breakdown
require no new theory, and is just a matter of applying what we already know in an
appropriate way and order, having in mind the above general comments. The rest
breakdown
of this section consists therefore of a
Example 2.28
Breakdown
Consider the capacitor
number
of characteristic examples.
in a Parallel-Plate
in Fig. 2.19 for a
=
1
m
Capacitor
and d
=
2 cm, and determine
its
breakdown
voltage and the corresponding charge and energy, as well as the electric pressure and forces
on capacitor electrodes
at
breakdown.
Systems
109
>
110
Chapter 2
Dielectrics,
'
Capacitance, and Electric Energy
The
Solution
breakdown
between the capacitor plates
electric field
intensity
is
approximately uniform, and
its
is
E = Ecro =
3
MV/m
(2.215)
Eq. (2.53)], from which the breakdown voltage of the capacitor,
[dielectric strength of air,
Eq. (2.126), charge of the capacitor, Eq. (2.125), and capacitor energy, Eq. (2.192), come out
to be
V = Ecrod =
2
Q = £ 0 « ^cr0 = 26.56 pC,
60 kV,
respectively. According to Eqs. (2.133) and
on electrodes at breakdown amount to
pe
Note
=
above values
that the
=
^eo^cro
W
and
e
X
=
-
QV = 0.8 J,
and
(2.132), the electric pressure
39 84 Pa
Fe =p e a 2 = 39.84
and
(2.216)
electric forces
N.
(2.217)
energy and force are quite small compared to energies and
for
forces of nonelectrical origin, and these are the largest possible values for this (quite large)
capacitor.
Maximum Breakdown
Example 2.29
Conductor
Voltage of a Coaxial Cable
= 7 mm
=
of a coaxial cable are a and 6
radii
(a
with a homogeneous dielectric of relative permittivity e r
ECI =
MV/m
20
maximum,
(b)
(polystyrene), (a) Find a for which the
What
is
the
maximum breakdown
<
b).
The cable
filled
is
2.56 and dielectric strength
breakdown voltage of the cable
is
voltage of the cable?
Solution
(a)
Assume
between the cable conductors
that the voltage
in the cable is
given by Eq. (2.124). Obviously, the
V.
is
The
electric field intensity
field is the strongest right
next to the
inner conductor of the cable.
maximum
electric field of
V
E(a + )
a
(2.218)
a \n(b/a)
coaxial cable
meaning
that dielectric
breakdown occurs when
this field intensity
reaches the
critical
value - dielectric strength,
E(a + )
For a given radius
withstand,
is
a,
breakdown
the
= ECI
voltage,
(2.219)
.
the largest voltage that the cable can
i.e.,
therefore
VCT fa) =
(2.220)
Eci-aln -.
a
To
find the optimal radius a, for
VCT
which the voltage
is
the largest,
we impose
the
condition
dVcr
=
—
0
(2
da
The
solution
- =
maximum breakdown
a
e
=
2.718
—
a opt
»
Because the second derivative of Vo-fa) for a
Ecr (a) has indeed
The maximum breakdown
function
(b)
221
)
is
coaxial cable radii ratio for a
voltage
.
a
maximum
=
—
a 0 pt
is
2.57
^cr(^opt)
mm.
negative (equals
(and not a minimum)
voltage of the cable
(^cr)max
= — =
(2.222)
-ECT /a opl
),
the
at that point.
is
— Ecr
fl
Q pt
—
51.5 kV.
(2.223)
Section 2.1 7
Dielectric
Breakdown
in Electrostatic
Systems
111
HISTORICAL ASIDE
The
rod
lightning
was
Stream
invented around the mid-
electricity,
and proposed that upright
of
the
eighteenth
sharp-pointed iron rods be mounted on roofs
century
by
dle
Benjamin
of buildings and connected by conducting wires
Franklin (1706-1790), an
to the iron bars buried in the earth (grounding
American statesman and
electrodes).
tor in the area of electricity
principle of
(e.g.,
conservation
labeling
of charge,
and
positive
negative charges, investigation of lightning, etc.), as well as in
other areas of scientific inquiry
(e.g.,
One
of the
first
lightning rods with
grounding conductors based on his design was
that installed in 1752 on the Pennsylvania State
House (now Independence Hall), in Philadelphia.
With several improvements by Franklin and others in decades to follow, grounded lightning rods
soon proved to be an efficient way of protecting
buildings from lightning damage. (Portrait: Library of
inven-
scientist, a prolific
sses,
chart, etc.). Franklin speculated that light-
ning was
numerous
bifocal gla-
Congress)
Franklin stove, improved printing press, Gulf
Breakdown Voltage
Example 2.30
of a Thin
Two-Wire Line
Consider a symmetrical two-wire line with conductor radii a = 1 mm and the distance
between conductor axes d = 0.5 m. The line is situated in air. Compute (a) the breakdown
voltage of the line and (b) the largest possible forces on line conductors per unit length.
Solution
(a)
Let Q' > 0 and
The
—Q
be the charges per unit length of the line conductors, as in
electric field intensity of the line
of the conductors.
At
Fig. 2.22.
the largest very close to the surface of each
is
these points, the total field (due to both charged conductors),
Eq. (2.138), can approximately be evaluated as the field due to a single isolated charged
wire conductor, because the other conductor is far away (d^> a). Thus, by means of
Eq. (1.196),
Q!
(2.224)
2tteq a
—
The breakdown voltage of the
vcr =
line
—C — —
0cr
In
where
C
is
- = 2aEcr0 In - =
37.29 kV,
(2.226)
a
computed
and
as in Eqs. (1.223)
Oa
=
=
2tt £o d
the largest possible force for this
Example 2.31
A
(2.225)
the corresponding intensity of electric forces on line conductors per unit
FL
is
nC/m.
the capacitance per unit length of the line [Eq. (2.141)].
«d,
length can be
which
167
to
a
jreo
(b) Since a
cr
amounts
Q'a
= 2neoaECXQ =
Q'
=
= 5 mm,
is
1
and the
result
is
mN/m,
(2.227)
line.
Grounded Wire Conductor
wire conductor, of radius a
earth, at a height h
(1.224),
electric field of
thin two-wire line
and, at breakdown, the critical charge per unit length of the line turns out to be
Emax = Ecr0 = 3 MV/m
maximum
as a Lightning Arrester
positioned horizontally above the surface of the
6 m, as shown in Fig. 2.34. The conductor
is
grounded.
A
uniform
a
.
112
Chapter 2
'
Dielectrics,
Capacitance, and Electric Energy
atmospheric electric
field
of intensity Eq, due to a large charged cloud, exists above the
and directed upward. 7 Assuming that Eq is known
conductor per unit of its length and (b) the
electric field intensity on the surface of the conductor, (c) Consider a cloud discharge
to the ground and discuss the protection that the wire conductor provides to the space
below it.
earth’s surface.
(£o >
The vector Eo
vertical
is
induced
0), find (a) the charge
in the
h
Solution
•2
fi
0
:3
I
1
V=0
(a)
The
earth’s surface represents a conducting plane. Let the potential of the plane be zero.
Since the wire conductor
is
connected to the earth, a charge of opposite polarity to that of
is, of positive polarity, is induced in the conductor (pulled
the lower part of the cloud, that
out from the earth) under the influence of the atmospheric
Figure 2.34 Grounded wire
conductor in a uniform
atmospheric electric field;
for Example 2.31
charge per unit length of the conductor
(
Q
>
'
of the conductor, V\, can be expressed as the
0), as
sum
first
plane, as in Fig. 1.49, and thus using Eq. (1.119) with
ri
=
2 h (since
The
this
potential
r\
—
its
image
in the
as
conducting
a (distance of the point
at
computed form the axis of the original wire equals the wire radius)
2 h 3> a). The other component of the potential can be found from
is
Eo from the conductor to the earth's surface along the
uniform, the voltage is simply — Eq times the length
potential of the conductor is given by
Eq. (1.90), by integrating the
straight, vertical path.
As
of the path, and the total
field
this field is
V =
In
X
——E
the condition that the conductor
Vi
=
is
(2.228)
0 h.
a
lit Eq
From
Q designate
component can be obtained
the potential at the surface of the wire due to both Q' and
and
Let
of the potential due to charge Q' and
the potential produced by the external field Eo- This
which the potential
field.
indicated in Fig. 2.34.
grounded, we
Q'
0
=
now have
27t£oEoh
(2.229)
ln(2 h/a)
(b)
The
electric field intensity
on the surface of the wire conductor (point
1
in Fig. 2.34)
is
approximately [Eq. (2.224)]
Q'
E0 h
2neoa
a ln(2 h/a)
Ei
(c)
On
the other hand, having in
conductor
Fig. 2.34)
at a height
amounts
mind Eq.
= 154£0
(2.230)
.
(2.139), the electric field intensity
below the
H — 2 m, for instance, with respect to the ground level (point 2 in
to
N
£2 =
+ -i--^fi
)+£o = 0.71£
2h — r
2tieq
\
r
()
(2.231)
)
(E2 is directed upward), where r = h — H = 4 m. Similarly, the field intensity below the
conductor close to the earth’s surface (point 3) turns out to be £3 = —Q'/(neoh) + £0 =
0.743£0 (r = h).
As the atmospheric field becomes stronger and stronger, in a thunderstorm, the field
intensity E\ reaches the
breakdown value [Eq.
E\
The
air ionizes
7
= £cro-
and becomes conductive, so
cloud flows through this conducting
air
(2.225)]
(2.232)
that a portion of the negative charge of the
channel to the wire conductor and
down
to the
and cause eventual lightning discharges, there is a positive
in its base. The bottom buildup induces another
positively charged area at the earth's surface, and these two charge layers, like a giant charged capacitor,
generate an electric field above the earth that is directed vertically upward, as in Fig. 2.34.
In cold clouds, that
charge buildup
at the
contain water and
ice,
top of a cloud and a negative one
—
2
Section 2.1 7
ground. The
E2
field intensities
£2 = 0. 71 E„
=—£, =
Dielectric
and £3, however, are much lower than
—
—
£3 = 0.743E, =
and
£,
Breakdown
in Electrostatic
Systems
113
critical,
= fe,
(2.233)
meaning that a structure or a person that may be underneath the conductor is safe from
breakdown and cloud discharge (or lightning strike). The grounded conductor protects
8
therefore the space below it and serves principally as a lightning arrester.
Breakdown
Example 2.32
The
in a Spherical
Capacitor with
dielectric of a spherical capacitor consists of
two concentric
Two
Dielectrics
layers.
The
relative permit-
= 2.5 and its dielectric strength £cr = 50 MV/m. For the outer
layer, e 2 = 5 and £ cr2 = 30 MV /m. Electrode radii are a = 3 cm and c — 8 cm, and the
radius of the boundary surface between the layers is b = 5 cm. Calculate (a) the breakdown
tivity
of the inner layer
e ri
is
i
X
voltage and (b) the corresponding energy of the capacitor.
Solution
(a)
Let
be the charge of the capacitor. The electric flux density in the dielectric
Q
by Eq.
(2.162).
With reference
is
given
to Fig. 2.35, the electric field intensities in individual
layers are
—-—
E\(r)
=
E2 (r)
= —4ne
-
=£l( fl+ )-2’
47re r ieo ^1
<
a
r
<
(2.234)
b,
Figure 2.35 Evaluation of
= E2
-
T 2£or
2
+
(b
)^j,
b
<
r
<
(2.235)
c,
r
spherical capacitor with a
where r is the radial spherical coordinate. Note that E\{a + ) is the largest field inten+
sity in the inner layer, while E2 (b ) represents the strongest field in the outer layer.
Combining Eqs. (2.234) and (2.235), we arrive to the following relationship between
these field intensities:
£ Tl a
We
do not know
in
voltage of critical value
possibility of a
2
E
1
(a
+
advance which
is
breakdown
)
=
the
same
time,
in the inner layer,
from Eq.
As E2 {b + ) < £cr2, we conclude
Note
that this
is
would break down
first
first
after a
check the
which implies that
= £cr i.
(2.237)
gri «
1
£r2 b
2
2
= 9 MV/m.
(2.238)
that the electric field intensity at
lower than the
critical
assumption of a breakdown occurring
8
(2.236)
).
(2.236),
)
is
E2 (b +
dielectric layer
E2 (b + = £cr
dielectric layer
2
applied across the capacitor electrodes. Let us
£i(a+)
At
e r2 b
all
points in the outer
value for that dielectric. This means that the
in the inner dielectric layer
the principle of operation of lightning arresters, in general.
is
correct.
Grounded
lightning rods,
placed as high as possible, protect exposed objects (buildings or other structures, as well as
humans and
animals) and their immediate surroundings from violent atmospheric electrical discharges by enforcing
breakdown leading to a lightning strike is “induced” near the surface of the
some other part of the object. Then, the rod routes the lightning discharge through an
that an eventual dielectric
rod,
and not
at
insulated wire conductor to a grounding electrode, rather than through the object. In addition to simple
vertical metallic rods with pointed ends, designs of lightning arresters include rods with spherical metallic
tops or umbrella-like wire extensions, which have capacity for carrying
charge.
the breakdown voltage of a
much
larger
amounts of induced
two-layer dielectric; for
Example 2.32.
114
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
The other
which assumes that an eventual breakdown occurs
possibility,
in the outer
layer, gives
E2 (b + = £cr2
)
and E\(a + )
larger than
(2.239)
,
= Ecr2 £ 2 b /(e ia = 166.7 MV /m. This is impossible, as E\ cannot be
Ecx under the assumption that the inner layer is in a normal regime, while
2
2
)
T
t
\
the outer layer breaks down.
With the
the capacitor
field-intensity values in Eqs. (2.237)
+ E2 (b +
+ 2
Ei(a )a f
Tcr =
and
(2.238), the
breakdown voltage of
is
b
)
=
2
r-
Ja
f
rL
Jh
e T \a
2
Ecrl
=
(
\£ r ab
769 kV.
e x2 bc J
i
(2.240)
This voltage can also be obtained by considering the critical value of the capac-
charge for breakdown. Let us denote by
itor
an eventual breakdown
in
Q
and
??’
the charge in the case of
the inner and outer dielectric layer, respectively. Based
on
Eqs. (2.234), (2.235), (2.237), and (2.239),
= 4ne
Qct
As
0
becomes
and
larger
T ie 0
a
2
ECTl
larger, the
the two charges in Eqs. (2.241).
The
0® = 4ne
and
x2 eob
breakdown occurs when
critical
2
Ecr2
it
(2.241)
.
reaches the smaller of
charge of the capacitor
is
thus
Q cr = min (0^,0®).
=
1
For the given numerical data,
Q^x < 0c? [0^
occurs in the inner dielectric layer. Hence,
0 cr =
The corresponding voltage
C=
where
The
pF
16.28
largest possible
is
=
1
],
meaning
that the
2.242)
breakdown
12.52 fiC.
(2.243)
769 kV,
(2.244)
is
Tcr =
(b)
0<J>
(
0.30c?
|
^=
j
the capacitance of the capacitor [Eq. (2.164)].
energy of the capacitor
W
t
=
l
is
CVl =
-
4.81
(2.245)
J.
j
Problems 2.69-2.81; Conceptual Questions (on Companion Website): 2.32-2.34;
:
MATLAB
Exercises (on
Companion Website).
Problems
2.1.
Nonuniformly
A
lelepiped.
lelepiped
is
polarized
dielectric
rectangular
dielectric
situated in air in the
first
the Cartesian coordinate system
(z/c)
paral-
densities of
octant of
(x, y, z
>
edges,
of lengths a, b
to coordinate axes x, y,
The
is
,
and
and
z,
c,
parallel
respectively.
polarization vector in the parallelepiped
given by POc, y, z)
—
Po[{x/a) x
+
(y/b) y
+
z],
of the parallelepiped, (b)
Show
that the total
bound charge of the parallelepiped
0),
with one vertex at the coordinate origin, and
the
where Pq is a constant, (a) Find the
volume and surface bound charge
paral-
2.2.
is
zero.
Uniformly polarized disk on a conducting
plane.
A
uniformly polarized dielectric disk
surrounded by
as
shown
air
is
lying at a conducting plane,
in Fig. 2.36.
The
polarization vector
115
Problems
in the disk is
P = Pi,
the disk radius
and
is a,
thickness d. Calculate the electric field intensity
vector along the disk axis normal to the con-
ducting plane (z-axis).
Figure 2.38 Very thin
dielectric disk with a
nonuniform
polarization; for
Problem
and
free space
The
is
2.4.
lying at a conducting plane.
polarization vector
is
P,
and
it is
normal
to the plane, as depicted in Fig. 2.39. Find (a)
bound surface charge density
the
Figure 2.36 Dielectric disk with a uniform polarization
lying at a
.
and
and
(b)
the electric field intensity vector at the center of
conducting plane; for Problem 2.2.
the
2.3
at the flat
spherical surfaces of the hemisphere
Uniformly polarized hollow dielectric cylinder.
A hollow dielectric cylinder of radii a and b,
and height 2h, is uniformly polarized and situ-
flat
that
surface (point
it is
on the
O in the figure), assuming
dielectric side of the
boundary
surface (dielectric-conductor), very close to the
surface.
ated in free space. The polarization vector, of
magnitude P,
as
shown
is
parallel to the cylinder axis,
in Fig. 2.37.
Find the electric
field
intensity vector at the center of the cylinder
(point O).
Figure 2.39 Dielectric hemisphere with a uniform
polarization lying at a conducting plane; for
Problem
2.6.
2.5.
Nonuniformly polarized large
dielectric slab.
An infinitely large dielectric slab of thickness
d = 2a, shown in Fig. 2.40, is polarized such that
the polarization vector is P = Pqx2 x/a 2 where
Figure 2.37 Hollow
dielectric cylinder
,
with a uniform
The medium outside the slab
is air. Compute (a) the distribution of volume
and surface bound charge of the slab, (b) the
Po
polarization; for
Problem
2.3.
a constant.
.
A
thickness
d (d <£
a), is
polarized throughout
shown
in Fig. 2.38, the polarization vector
(c) the voltage
of the slab.
its
volume. In the cylindrical coordinate system
everywhere, and
between the boundary surfaces
electric field intensity vector
2 4 Nonuniformly polarized thin dielectric disk.
very thin dielectric disk, of radius a and
.
is
2.7. Electric flux
density vector. Find the electric
flux density vector,
is
P = Pori /a, where
The medium around the disk
D,
(a) at the center of the
defined by the expression
polarized dielectric sphere in Fig. 2.7 and (b)
P0
along the axis of the polarized dielectric disk
is
a constant.
Determine (a) the distribution of bound
charge and (b) the electric field intensity vector
(z-axis) in Fig. 2.36.
is air.
along the z-axis.
.
hemisphere of radius a
Total (free plus bound) volume charge density.
2 5 . Uniformly polarized dielectric hemisphere.
dielectric
2.8.
is
A
situated in
The
electric field
dielectric, E,
is
a
intensity vector in a
known
function of spatial
coordinates, (a) Prove that the total (free plus
116
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
a closed surface
S situated entirely inside the
body.
2.12. Total
enclosed bound and free charge. Con-
S
sider an imaginary closed surface
homogeneous
Infinitely
large dielectric slab
charge enclosed by S
is
is
homogeneous medium. Prove that
homogeneous linear medium with no
volume charge, there is no bound volume
2.13. Charge-free
with a nonuniform
in
a
polarization; for
Problem
The
Q$. What
the total bound charge Q p s enclosed by S?
total free
Figure 2.40
inside a
dielectric of permittivity s.
free
2.6.
charge either.
bound) volume charge density
tric,
£oV
ptot
=
P
+ Pp,
in the dielec-
can be obtained as p t0
E. (b) Specifically, find p to t for
•
E
1
=
given
as the following function of Cartesian coor-
dinates:
(2 x
2.9,
2
-f z
y
Uniform
E(jc, y, z)
3
)
z]
=
[4xyz
V/m (x, y, z
x
in
field in a dielectric.
+
(2jc
2
z
— y3
)
y
+
m).
There
is
a uniform
The
volume charge density is p. Determine the
bound volume charge density, p p
free
.
Closed surface
in a
uniform
field.
Consider-
ing an arbitrary closed surface in a uniform
= 0) region
prove
the
following
vector
identity:
(Fig. 2.41),
electric field, in a charge-free (p tot
fs
dS
volume charge.
A
very long homogeneous dielectric cylinder, of
radius a and relative permittivity e T
,
is
charged
uniformly with free charge density p throughout its volume. The cylinder is surrounded by
air.
electric field in a certain dielectric region.
2.10.
2.14. Dielectric cylinder with free
(a) Calculate the voltage
and the surface of the
bound charge
between the
axis
cylinder, (b) Find the
distribution of the cylinder.
volume charge distribution.
Repeat Example 2.6 but for a model of a pn
junction given by the volume charge density
p{x ) = pq{x/o) e~ |j:|/a where po and a are posi-
2.15. Linear-exponential
,
tive constants, as
shown
in Fig. 2.42.
= 0.
Figure 2.42 Model of a pn junction with a
linear-exponential charge distribution; for Problem 2.15.
boundary
2.16, Dielectric-dielectric
conditions.
Assume that the plane z = 0 separates medium
1 (z > 0) and medium 2 (z < 0), with relative
permittivities e x \
Figure 2.41 Closed surface
electric field
in a
region with a uniform
and no volume charge;
for
Problem 2.10.
The
volume charge
2.11. Flux of the electric field intensity vector.
polarization vector, P, and free
density, p, are
tric
known
at
every point of a dielec-
body. Find the expression for the flux of
the electric field intensity vector,
through
The
=
4 and e T2
=
2, respectively.
electric field intensity vector in
medium
near the boundary (for z = 0 + ) is Ei =
(4x — 2y + 5 z) V/m. Find the electric field
1
medium
2 near the boundno free charge exists
on the boundary (ps = 0) and (b) there is a sur2
face charge of density p s = 53.12 pC/m on the
intensity vector in
ary (for z
boundary.
= 0_ E2
),
,
if
(a)
117
Problems
boundary conditions.
and conductor-
2.17. Conductor-dielectric
Obtain
boundary conditions, Eqs. (1.186),
and (1.190), from Eqs. (2.84) and (2.85).
free space
(2.58),
boundary. Sketch the field
emerging from water (s r = 80) into air,
2.18. Water-air
“incident” angle (in water)
2.19.
2.20.
a
is
=
2 23
.
.
ductors,
2.43
total
2 . 24 .
.
=
5 cm,
and assume that
a nonuniform free volume charge whose denthe function p(r) given in Eq. (1.32), with
3
Po = 3 C/m exists between the electrodes, for
a < r < b, where the permittivity is £q. In addi,
tion, let the potential of the inner electrode
(sphere) with respect to the outer one (shell) be
Vo
=
10 V.
Under
these circumstances, com-
pute (a) the electric potential and (b) the electric field
2.22.
computer program -
direct solution.
vector at an arbitrary point between
number
of
same quantities as in the previous
problem, and compare the results obtained by
the two programs.
2 25
.
.
Capacitance of the earth. Find the capacitance
of the earth assuming that it is a conducting
sphere of radius
R=
6378
km
(grounds and
waters are conducting media).
2 . 26 . Capacitance of a person.
C
that the capacitance
ducting body
is
in
It
can be shown
of an arbitrary con-
between the capacitance of
body and that of
the sphere inscribed in the
the sphere overscribed about the body, that
is
[Eq.
where
i?
spheres.
(2.121)],
47r£ 0
m n and R m ax
j
Based on
R m in < C < 4^£ 0 R m ax,
are the radii of the two
that, estimate the capaci-
the electrodes.
tance of an average
Application of Laplace’s equation in spherical
sues are conducting media).
coordinates. Repeat the previous problem but
As
plot the
.
enclosed by a shell) as the one in Fig. 1.41,
N=
which equals the number of unknown potentials], The system of equations, in which known
potentials at nodes on the surface of conductors appear on the right-hand side of equations,
should be solved by the Gaussian elimination
method (or by matrix inversion). Compute and
Vacuum
concentric metallic electrodes (a solid sphere
and
nodes in Fig. 2.13(b) as unknowns [applying
Eq. (2.106) to each interior grid node, we get
coordinates. Consider a system of two spherical
sity is
= a/N
12, respectively.
equations with the potentials at interior grid
anode
and b
FD
and
an alternative to the iterative technique based
on Eq. (2.107), write a computer program for
the FD analysis of a square coaxial cable by
directly solving the system of linear algebraic
Ps2^
= 1cm
and the tolerance of
V. (b) Compute the
a set of simultaneous equations the
2 20
electric
charge per unit length of the inner and
2, 3, 5, 7, 9, 10,
Application of Poisson’s equation in spherical
with a
= a/10
= 0.01
the outer conductor, taking d
diode; for Problem
2.21.
and the
space between the con-
and the surface charge density on
spacing to be d
V=V0
d
V,
1
V. (a) Plot the results for the
the potential 8y
,
Figure 2.43
(2.107).
the surfaces of conductors, taking the grid
0 < x < d,
be described as V(x) = Vo (x/d)
where d is the distance between the electrodes.
Find (a) the volume charge density in the diode,
(b) the surface charge density on the cathode,
(c) the surface charge density on the anode, and
(d) the total charge of the diode.
0
— —1
on Eq.
= 3 cm, Va =
cm, b
1
distribution of the potential
4/ 3
cathode
Vt,
=
that a
field intensity in the
shows a vacuum
two flat electrodes,
the cathode and the anode, and a charge
distribution in a vacuum between them. Let
the potential of the cathode be zero and the
potential of the anode Vo (Vo > 0). The dis-
Psl
computer program - iterative solution.
Write a computer program for the finite-
and
diode, which consists of
V=0
b).
FD
Assume
45°.
tribution of the potential in the diode can
<
r
difference analysis of a coaxial cable of square
inhomogeneous media.
(a) Derive Poisson’s equation for an inhomogeneous medium, (b) Write Laplace’s equation
for an inhomogeneous medium.
diode. Fig.
<
cross section (Fig. 2.13) based
the
Poisson’s equation for
Vacuum
0 (no volume charge) between the
electrodes (for a
lines
if
—
for p(r)
conductor-dielectric
human body (human
tis-
,
118
2.27.
Chapter 2
Capacitance, and Electric Energy
Dielectrics,
Capacitance of a metallic cube, computed by
the MoM. Find the capacitance of the metallic cube numerically analyzed by the method
of
moments
in
Problem
plates in Fig. 2.19 are at potentials V\
and
metallic spheres, respectively: (a) the sphere
inscribed in the cube, (b) the sphere over-
radius
mean
the arithmetic
is
spheres
in (a)
and
,
whose
.
of the radii of
(b), (d) the
sphere having
same surface as the cube, and (e) the sphere
with the same volume as the cube.
the
2.28.
RG-55/U
ial
6
coaxial
A
cable.
RG-55/U
cable has conductor radii a
=
3
2.25).
mm. The
dielectric
is
=
0.5
mm
Determine the capacitance per
.
.
,
-
-
coax-
and
polyethylene (s r
V
on the two plates are equal in magnitude and
opposite in polarity. With this, the unknowns
in the procedure are charge densities p s i, p s 2
ps /v on the upper plate only, and matching
points, at which the potentials are computed,
are centers of the same patches; however, these
potentials are due to pairs of patches on both
plates, with charge densities p s; and — p S( (/ =
1,2,..., N). Using the MoM program, find C
fora = 1 m and the following d/a ratios: (i) 0.1
(ii) 0.5, (iii) 1, (iv) 2, and (v) 10, and compare
the result with capacitances of the following
scribed about the cube, (c) the sphere
1
V, respectively, as well as that
the charge densities of pairs of corresponding
patches that are right above/below each other
and compare
1.85,
=
V2 = —1
=
the results with the corresponding
unit
C
,
values
obtained from Eq. (2.127), which neglects the
length of the cable.
fringing effects.
2.29.
Capacitance
FD
analysis.
p.u.l.
of a square coaxial cable,
Compute
the capacitance per unit
2.32.
line.
Derive the
length of the coaxial cable of square cross sec-
expression for the capacitance per unit length
by a finite-difference
technique in Problems 2.23 and 2.24, and compare the result (using the grid spacing of d =
of a nonsymmetrical thin two-wire transmis-
tion numerically analyzed
sion line in
a (standard) coaxial cable (of circular cross
section) having the
radii
(
b/a
=
3)
and
same
2.33.
conductor
as the square
cloud.
A
model of
its
a thunder-
electrical properties
are concerned, as a parallel-plate capacitor
vertical separation
d—
=
km 2
and
km. Assume that the
charge Q = 300 C and
with horizontal plates of area S
15
conductors are
Thick symmetrical two-wire line. Consider a
symmetrical two-wire line with conductors of
is
— 1}. Take d/a to be 3, 5, 10, 20, and
using this expression and
and calculate
s/[d/(2a)] 2
1
100,
C
upper plate has a total
the lower plate an equal amount of negative
the expression in Eq. (2.141), obtained for thin
charge. Neglecting the fringing effects, find (a)
two-wire lines. Compare the two sets of results
and evaluate the error due to thin-wire approx-
the capacitance of this capacitor, (b) the volt-
age between the top and bottom of the cloud,
imation of the line for individual distance to
and
radius ratios.
|
(c) the electric field intensity in the cloud.
MoM
numerical analysis of a parallel-plate
2.34.
small metallic spheres in
radii, a,
and write a computer program
based on the method of moments to evaluate
its capacitance (C). In specific, subdivide each
of the plates into N = 10 x 10 = 100 square
patches, and assume constant charge densities on individual patches (as in Fig. 1.46).
Additionally, assume that the upper and lower
Two
air.
A
capacitor
consists of two small metallic spheres of equal
capacitor. Consider the parallel-plate capacitor in Fig. 2.19,
I
not
necessarily small compared to the distance
between wire axes, d) in air. Using a version of image theory for line charges in the
vicinity of conducting cylinders, it can be
shown that the capacitance per unit length
of this line is given by C' — ttsq/ \n{d/(2a) +
thundercloud can be approximately
represented, as far as
radii of the
b),
arbitrary thickness (radius of wires, a,
cable.
2.30. Parallel-plate capacitor
The
and the distance between the
^
axes of conductors is d (d ~S> a, b).
ratio of
dielectric (air)
air.
a and b (a
a/ 10) with the per-unit-length capacitance of
2.31.
Nonsymmetrical thin two-wire
placed
tance d (d
»
in air at a
a).
center-to-center dis-
Find the capacitance of
this
capacitor.
2.35.
Four
parallel wires in air.
shown
The transmission
line
two pairs of
gal-
in Fig. 2.44 consists of
vanically interconnected thin wire conductors,
situated in
air.
The distance between
the axes
l
£
119
Problems
of adjacent wires
radii are a
=
1
is
d
=
200
mm
mm. Compute
per unit length of the
and the wire
O
the capacitance
oo
Qd
Qb Qc
Qa
line.
Cab
Figure 2.46
C-cd
Equivalent circuit for
+ Vcd
+ vab
+
the system
Fig.
1/
1
.41
in
for
;
Problem 2.38.
and solving an equivalent circuit with three
Repeat Problem 1.78 using the
same circuit and the corresponding set of con-
Figure 2.44 Cross section
of a transmission line
consisting of
two
capacitors, (b)
pairs of
short-circuited wires in
air;
for
ditions.
Problem 2.35.
2.40.
2.36.
Two
wires at the
same
A short-circuited two-wire
large
flat
metallic
foil,
and a
potential
line
is
foil.
parallel to a
Equivalent circuit with parallel-plate capacitors.
Repeat Example 1.28 by solving the equivshown in Fig. 2.47.
alent circuit
as portrayed in Fig. 2.45.
= 1 mm, h =
mm, and d = 30 mm, and the medium is air.
(a) What is the capacitance per unit length of
Q
The geometry parameters are a
20
-Q\ Qi\
Q\
2
l
O on the
II
4
II
3
c3
c2
c'1
+
at the central
-Q\
|
[
whose one
conductor is the two-wire line and the other
conductor is the foil? (b) If the voltage between
the line and the foil is V = 20 V, find the
point
—02 03
#
a two-conductor transmission line
induced surface charge density
'
V
Figure 2.47 Equivalent circuit for the system
foil.
Fig.
1
.42; for
in
Problem 2.40.
2.41. Equivalent circuit with cylindrical capacitors.
Figure 2.45
Repeat Problem
Short-circuited
an equivalent
by generating and solving
1.79
circuit
with capacitors.
horizontal two-wire
line
metallic
foil;
for
layer
Consider the
system
composed of
a
wire parallel to a corner screen in Fig. 1.57,
and assume that a
—
2
mm
and h
—
10 cm.
Calculate the capacitance per unit length of this
system (transmission
2.38.
line).
Equivalent circuit with two spherical capacitors. (a) Consider the system described in
Example
1.27, and show that it can be replaced
by the equivalent circuit given in Fig. 2.46.
(b)
What
are the capacitances of the capaci-
is
the outer layer
is oil (
r2
= 2.3). The
geometri-
—
8 cm, and c =
connected
to a source
is
—
of voltage V
100 V. The source is then disconnected, and the oil is drained from the
capacitor. Find the voltage between the elec-
cal
parameters are a
16 cm.
=
2 cm, b
The capacitor
new
trodes of the capacitor in the
electrostatic
state.
without disconnecting the source.
2.43. Oil drain
If
the
oil in
problem
is
the capacitor from the previous
drained without disconnecting the
Obtain the
voltage source, determine the flow of electricity
expression for the potential at the center of the
through the source circuit (that is, the difference in the charge of the capacitor) between
tors in this schematic
diagram?
(c)
structure in Fig. 1.41, Eq. (1.202), that
Example
2.39.
and assume that the inner dielectric
made from mica (e r i = 5.4), whereas
Fig. 2.27
Capacitance per unit length of a wire-corner
line.
Consider the spherical capacitor in
dielectric.
Problem 2.36.
2.37.
capacitor with a solid and liquid
2.42. Spherical
over a large
1.27,
by solving
is,
repeat
this circuit.
Equivalent circuit with three spherical capacitors. (a)
Repeat Problem
1.77
by generating
the two electrostatic states.
2.44. Metallic
sphere
with
dielectric
metallic sphere of radius a
=
1
coating.
cm
is
A
covered
120
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
with a concentric dielectric layer of relative permittivity £ r
=
situated in
air,
tial
—
4 and thickness b
as
shown
a
=
cm and
2
The poten-
in Fig. 2.48.
of the sphere with respect to the reference
point at infinity
is
V=
1
kV. Compute
(a) the
cable with two coaxial dielectric layers.
The
geometry parameters are a — 1 mm, b = 2 mm,
and c — 4 mm. The dielectric parameters are
£ r = 5 and e T 2 = 2. Calculate the capacitance
i
per unit length of the cable.
capacitance of the sphere, (b) the free surface
charge density on the sphere surface,
(c) the
bound volume charge density in the dielectric,
and (d) the bound surface charge densities on
dielectric interfaces.
Figure 2.50 Cross
section of a coaxial
cable with
Figure 2.48 Metallic sphere
with a dielectric coating
for
two
dielectric layers; for
Problem 2.48.
in air;
Problem 2.44.
2.49. Coaxial cable with a radial variation of permit-
2.45.
Charge densities in a
half-filled spherical
capac-
Consider the half-filled spherical capacitor
from Fig. 2.29, and assume that a — 2 cm, b —
10 cm, £ r = 3, and Q — 10 nC. Find the distributions of (a) free surface charges on metallic surfaces and (b) bound surface charges on
itor.
Empty and
air-filled
radii a
=
half-filled spherical capacitor.
3
cm and
trostatic state
An
= 15 cm is connected to
V = 15 kV. After an elec-
is
established, the source
is
dis-
is
then half
filled
with a liquid dielectric of relative permittivity
2.5.
What
is
the
new
voltage between the
A
sphere half
embedded
metallic sphere of radius a
given by the following function
=
<r <
r/a (a
=
b ), where a and b
the cable conductor radii.
(a) the capacitance
If
and
(b) the
5a are
the potential difis
V, find
per unit length of the cable
bound charge
distribution of the
dielectric.
2.50. Coaxial cable with four dielectric sectors.
is
in a dielectric.
pressed into a
up
volume, as shown in
medium in the upper half-space
Fig. 2.49.
The
(a)
Find
its
is air.
the capacitance of the sphere, (b)
If
to a
the sphere
A
with a piece-wise homogeneous dielectric composed of four 90° sectoral
coaxial cable
is filled
parts with different permittivities, as
Fig. 2.51.
shown
in
Let the relative permittivities of the
=
6, e r 2
=
10, the radii of the cable
dielectric half-space of permittivity £
half of
£ r (r)
sectors be e r \
electrodes of the capacitor?
2.47. Metallic
is
of the radial distance r from the cable axis:
b
connected. The capacitor
=
permittivity
spherical capacitor with conductor
a source of voltage
er
Consider a coaxial cable with a continu-
ously inhomogeneous dielectric, whose relative
ference between the cable conductors
dielectric surfaces.
2.46.
tivity.
2, £ r 3
=
1,
and
conductors a
=
£ r4
2
=
mm
and b = 1 mm, and the potential of the outer
conductor with respect to the inner conductor V = 25 V. Compute (a) the capacitance per
unit length of the cable and (b) the free charge
density at an arbitrary point on the surface of
the inner conductor.
charged with Q, determine what portion of
this charge is located on the upper half of the
sphere surface.
is
£q
Q
e
Figure 2.49 Charged
Figure 2.51
metallic sphere half
Cross section
embedded
of a coaxial
in a dielectric
half-space; for
Problem
2.47.
cable with
four 90°
dielectric
2.48. Coaxial cable with
ers. Fig. 2.50
shows
two coaxial
dielectric lay-
a cross section of a coaxial
sectors; for
Problem 2.50.
121
Problems
2.51.
Charge distribution for two coated
Consider the two-wire
ings in Fig. 2.31,
of thickness d.
and assume that a = 1 mm,
4, and V = 10 V. Under these
permittivity of the dielec-
= 2(1 + 3x/d)eo (0 < x < d).
Neglecting fringing, calculate the capacitance
charge and (b) bound charge in the system.
of this capacitor.
=
25
mm, £ r =
^-coordinate: e(x)
Two
metallic spheres with dielectric coating. If
2.55. Permittivity
two
identical metallic spheres with dielectric
plates.
gradient
Assume
dielectric
are placed in air so that the distance between
y-coordinate, e(y)
is
d
=
1
y <
m, determine the capaci-
tance of such a capacitor.
Two
5
mm
itor.
spheres half
metallic
dielectric.
Two
are half
embedded
in
a
metallic spheres of radii a
=
embedded
2.56.
= 4,
as
air.
Fig.
capacitor
2.53
is
a function of the
= 2[1 + 3sin(7ry/b)]eo
<
Neglect fringing.
shows a
is
layer.
parallel-plate capacitor that
half filled with a ferroelectric.
tor
(0
find the capacitance of the capac-
part of the capacitor
The distance between the centers
is d = 30 cm. The upper medium
to
Capacitor with a nonlinear dielectric
is
shown
of the spheres
is
and
Fig. 2.54
in a dielectric half-
space of relative permittivity e r
in Fig. 2.52.
b),
in
parallel
that the permittivity of the
coating as the one described in Problem 2.44
their centers
2.53.
The
given by the following function of the
tric is
circumstances, find the distribution of (a) free
d
2.52.
b and continuously inhomogeneous dielectric
wires.
line with dielectric coat-
is
air-filled.
The other
The capaci-
charged by being connected to a voltage
The source is then disconnected, and
source.
The spheres are charged with charges
the capacitor electrodes are short-circuited. In
between the spheres
new electrostatic state, there is a remanent
uniform polarization throughout the volume
200 V. For such a capacitor, find (a) the
of the dielectric, with the polarization vector
capacitance, (b) the distribution of free charges
being normal to the capacitor plates and its
magnitude being P. Determine the electric field
intensity vector between the capacitor plates
in (a) air and (b) dielectric. Fringing can be
of equal magnitudes and opposite polarities.
The
is
potential difference
V=
over conductors, and
(c) distribution of
the
bound
charges in the dielectric.
d
neglected.
£o
e
Figure 2.54
Short-circuited
Figure 2.52
Two charged
metallic spheres
parallel-plate
pressed into a dielectric half-space; for
capacitor containing
Problem 2.53.
2.54. Permittivity
a nonlinear dielectric
gradient
plates. Fig. 2.53
shows
normal
layer with a uniform
to
capacitor
remanent
a parallel-plate capaci-
tor with rectangular plates of dimensions a
polarization; for
and
Problem 2.56.
2.57.
Energy of a spherical capacitor with two layers.
For the spherical capacitor with two concentric
dielectric layers
from Example
2.18,
compute
the electric energy stored in each of the layers
by
(a) integrating the electric
energy density
Figure 2.53
over the volume of each layer and (b) rep-
Parallel-plate
resenting the capacitor as a series connection
capacitor with a
of
continuously
inhomogeneous
dielectric; for
Problem 2.54.
two spherical capacitors with homogeneous
dielectrics.
2.58.
Change
in
energy of a spherical capacitor. For
the spherical capacitor with solid and liquid
122
Chapter 2
Dielectrics,
Capacitance, and Electric Energy
from Problem
dielectric layers
electrodes from Example 1.28
two ways: (a) using the electric field intensities between electrodes (found
in Example 1.28) and the corresponding energy
densities and (b) using the equivalent circuit in
Fig. 2.47 and the involved capacitances, respec-
2.42, find the
parallel large
energy of the capacitor between the
electrostatic state with the capacitor connected
change
to the voltage source
and the
final electrostatic
state.
2.59.
Energy of a coated metallic sphere. For the
charged metallic sphere with a dielectric coating from Problem 2.44, determine the radius b
such that 1/2 of the total energy of the system is
tively.
2.67.
Energy of a coaxial cable with two coaxial
layers. For the coaxial cable with two coaxial dielectric layers from Problem 2.48, assume
that the voltage between the cable conductors
is V == 100 V and find the per-unit-length electric energy contained in each of the layers by
energy density over
the cross section of each layer and (b) representing the cable as a series connection of
(a) integrating the electric
from Problem
2.68.
Example
elec-
pn junction from
in the
assuming that the area of the
2.6,
junction cross section perpendicular to the xaxis
is S.
For
liq-
surface.
2.69.
Example
energy stored
is
2.20, calculate the
in the liquid, if the
2.70.
charge
Q.
Problem
total
electric
stored
y
=
what percentage of the
energy of the capacitor, if charged, is
the lower half of the dielectric, from
in
0 toy
=
=
5
10
cm
(a
<
b ).
is
It
filled
and
2.71.
dielectric strength
voltage of the capacitor
is
the
ECT =
Find a for which the breakis
maximum,
maximum breakdown
the capacitor? (c)
What
is
(b)
voltage of
the energy of the
breakdown?
Breakdown in a wire-plane transmission line.
(a) Compute the breakdown voltage of the
wire-plane transmission line
assuming that a
is
the
maximum
—
1
cm and h
2.24(a),
Fig.
in
=
1
m. (b) What
energy per unit length of
this
Determine the largest possible elecforce on the wire conductor per unit of its
line? (c)
For the parallel-plate capacitor with
from
=
(a)
capacitor at
Energy of a capacitor with an inhomogeneous
a permittivity variation parallel to plates
mittivity e r
25MV/m.
down
What
Energy of a capacitor with a variable permittivConsider the parallel-plate capacitor with
a permittivity variation normal to plates from
Problem 2.54. Determine what percentage of
the total electric energy of the capacitor, when
charged, is contained in the first half of the
dielectric, from x = 0 to x — d/2.
voltage of a spherical
spherical capacitor has electrodes
with a homogeneous dielectric of relative per-
sectors.
ity-
2.55, find
A
of radii a and b
Energy of a coaxial cable with four
dielectric.
Maximum breakdown
capacitor.
For the coaxial cable with four dielectric sectors from Problem 2.50, find the per-unit-length
electric energy contained in each of the sectors.
2.65.
energy contained
half-filled spherical capacitor.
Energy of a
of the capacitor
2.64.
2.14.
the spherical capacitor half filled with a
electric
2.63.
volume charge.
Breakdown charge and energy of the earth.
Determine the maximum possible charge and
electric energy that could be stored on the
earth, as described in Problem 2.25, and in the
electric field around it, limited by an eventual
dielectric breakdown of air near the earth’s
uid dielectric from
2.62.
free
Energy of a pn junction. Calculate the
tric
two coaxial cylindrical capacitors with homogeneous dielectrics.
2.61.
Energy of a system with
Find the electric energy per unit length of the
system with a volume free charge distribution
stored in the coating.
2.60.
flat
in the following
in
tric
length.
2.72.
Grounded
arrester.
metallic
sphere
Repeat Example
as
2.31
a
lightning
but
for
a
grounded small metallic sphere (instead of the
b/ 2.
Energy of a system of spherical conductors.
wire) in a uniform atmospheric electric field
Calculate the energy of the system of three
above the surface of the earth. Adopt the same
radius and the same height of the conductor.
spherical conductors from
2.66. Energy of a system of
flat
Problem
1.80.
electrodes.
Compute
the electric energy stored in the system of five
two dielectric layConsider the parallel-plate capacitor with
2.73. Parallel-plate capacitor with
ers.
123
Problems
two
dielectric layers in Fig. 2.25(a),
that
=
s r2
d\
—
Ecr2 =
11
mm, d2 =
4
mm,
and assume
— 3, and
2.78.
Ecr = 20 MV /m
two concentric
are
MV/m,
respectively. Find the break-
\
,
down voltage of the capacitor. Neglect fringing.
2 . 74
.
dielectric layers in Fig. 2.27.
Let the dielectric strengths of the inner and
outer layer be Ecr \ and Ecr2 respectively. Find
the relationship between the parameters of
and
layers
Simultaneous breakdown in two spherical
Consider the spherical capacitor with
layers.
er\
well as that the dielectric strengths
5, as
the
for
2
this capacitor
two dielectric secRepeat the previous problem but for the
parallel-plate capacitor with two dielectric sectors from Fig. 2.25(b) and d = d\ + d 2
Parallel-plate capacitor with
(
a b,
,
e r \, e r 2
c,
,
Ecr \,
and
FC 2 )
r
such that, for sufficiently large capacitor volt-
tors.
age, dielectric
breakdown
will
occur in both
dielectric layers simultaneously.
-
2 . 75 .
Breakdown
in a spherical capacitor with
two
2.79.
(Fig. 2.50).
tric layers
state
(outer layer
The
is air).
2.80. Metallic
dielectric strengths for
mica and oil are ECT \ = 200
15
/m, respectively.
dielectric.
MV /m and E 2 =
.
Breakdown
potential
of a
sphere, (a) Determine the
ct
.
coated metallic
breakdown potential
maximum potential
of this sphere such that
the
from Problem 2.44
breakdown will not occur after it is
removed from the liquid and raised high above
maximum
.
A
half
of the metallic sphere with a dielectric coating
the coating
2 77
sphere half immersed in a liquid
metallic sphere of radius a — 2 cm
immersed in a liquid dielectric, as in
Fig. 2.49. The relative permittivity of the dielectric is £ r = 3 and its dielectric strength is E cx =
20 MV/m. The upper medium is air. Calculate
is
MV
2 . 76
lay-
Repeat the previous problem but for a
coaxial cable with two coaxial dielectric layers
Find the breakdown voltage of the
spherical capacitor with two concentric dielec-
layers.
from Problem 2.42 in (a) the first
(outer layer is oil) and (b) the second state
Simultaneous breakdown in two coaxial
ers.
if
Ecr —
the dielectric strength for
30
energy of
Breakdown
MV /m.
(b)
What
is
the
the interface.
this structure?
in a coaxial cable with
two
layers,
Find the breakdown voltage of the coax-
(a)
ial
is
cable with two coaxial dielectric layers from
Problem 2.48. The dielectric strengths of the
inner and outer layer are ECI 1 = 40 MV /m and
ECT2 =
the
20
MV /m,
maximum
per unit of
its
respectively,
energy that
length.
dielectric
(b)
this cable
Calculate
can store
2.81.
Breakdown
in a coaxial cable with a dielectric
spacer. Consider the coaxial cable with a dielectric
2
spacer in Fig. 2.33, and assume that a
mm,
b
= 6 mm,
a
—
60°,
and
er
=
as that the dielectric strength for the spacer
Ect =
200
=
5, as well
is
MV/m. Find (a) the breakdown volt-
age of the cable and (b) the electric energy per
unit length of the cable at
breakdown.
Steady Electric Currents
Introduction:
S
o
far,
we have
dealt with electrostatic fields,
associated with time-invariant charges at rest.
We now
consider the charges in an organized
macroscopic motion, which constitute an electric
current. Our focus in this chapter is on the steady
flow of free charges in conducting materials, i.e.,
on steady (time-invariant) electric currents, whose
macroscopic characteristics (like the amount of
current through a wire conductor) do not vary
with time. Steady currents are also called direct
currents, abbreviated dc. The subject of steady
electric currents links field theory to several important concepts of circuit theory, such as Ohm’s law,
Kirchhoff’s current and voltage laws, Joule’s law,
resistance, conductance, voltage and current generators, and power balance in a circuit. Discussions
regime, and help us further
develop and understand the concept of a transmisin a time-invariant (dc)
sion line as a circuit with distributed parameters.
We
shall derive
and discuss
integral
and
dif-
ferential field equations for steady electric currents
and
their electric field, along with the correspond-
boundary conditions. We shall also study the
mechanism of conduction for various materials,
introduce models of energy sources, and derive
expressions for power and energy calculations.
These equations and concepts will enable us to
develop and demonstrate techniques for analysis
ing
of several general configurations with steady currents,
such as resistors of various composition and
shapes, capacitors with imperfect
inhomogeneous
dielectrics, transmission lines with
imperfect con-
of steady electric currents will bring us to the field
ductors and imperfect inhomogeneous dielectrics,
and
and grounding electrodes buried
124
circuit analysis of transmission lines with losses
in
the earth.
Section 3.1
Current Density Vector and Current Intensity
125
CURRENT DENSITY VECTOR AND CURRENT INTENSITY
3.1
In the absence of an externally applied electric
random
field,
free charges in a conduc-
motion due to their thermal energy. This
is so-called thermal motion of charges. The corresponding velocity is the thermal
velocity, denoted as v t Generally, v is rather large. In metallic conductors, free elec5
trons move at thermal velocities on the order of v ~ 10 m/s at room temperature,
between collisions with the atoms and with one another. Because of the entirely
random nature of thermal motion of charges (without any external electric field
applied), there is no net macroscopic motion in any given direction, i.e., the macroscopic average vectorial resultant of thermal velocities of individual charges at any
point in the conductor is zero. For an electric current, defined as a macroscopic net
flow of free charges, to exist, there must be a nonzero macroscopic average velocity
of microscopic velocities of charges in some direction in a conductor. This can be
achieved, as we shall see, by establishing and maintaining an external (i.e., externally
tor are in a state of
(chaotic)
.
t
t
1
+
I
V
applied) electric field within a conductor.
Consider a conducting body whose two ends are connected to a generator
of voltage V, as shown in Fig. 3.1. Because a potential difference is maintained
between the conductor ends, there is an electric field of intensity E inside the conductor [the line integral of E, Eq. (1.90), through the conductor is nonzero]. Note
from the situation in Fig. 1.38, where the
transient redistribution of charge occurred and electrostatic equilibrium with zero
field inside the conductor was established. Here, the conductor is not isolated but
wired to a source of electromotive force (generator), providing a mechanism that
forces the free charges to move and prevents them from piling up, which would tend
to reduce the field in the conductor. Assume, for simplicity, that the free charge carthat this situation
riers in the
is
conductor are electrons only (as in metallic conductors). The electric
on each charge
force
essentially different
is
thus [Eq. (1.23)]
Fe = -eE,
(3.1)
e (e > 0) is given in Eq. (1.3). This force compels the
move through the conductor, between its ends. However, since the elec-
where the charge amount
electrons to
trons are not in free space, they cannot accelerate indefinitely under the influence
of the electric field. Before they can acquire any appreciable speed, the electrons collide with the atomic lattice
the field
E
A tc ~
-14
ture),
in a systematic
is
velocities
.
manner. This relatively
slight systematic drift of the free electrons
the basis of electric conduction (current). After a brief initial transient, the elec-
trons acquire a steady-state average velocity, determined by the balance
the accelerating force of the applied field,
lisions
by v d
1
vt Since
s (typical average time interval between collisions at room temperacan change the random thermal velocities of the electrons only slightly, but
10
it
and acquire new random
essentially has to start accelerating the electrons all over again every
The
with the
lattice.
This velocity
is
Fe and
,
between
the scattering effect of the col-
termed the
drift velocity
and symbolized
.
free electrons in a metallic conductor are so-called conduction-band electrons (or valence elec-
trons),
which are very loosely bound to their atoms and are essentially free to move about the crystalline
atom of copper has 29 electrons, 28 of which are bound
structure of the metal. For example, each
electrons (tightly
bound
in their shells), while the
outermost one
is
a free electron.
Figure 3.1 Conductor
whose two ends
maintained at
difference.
are
a potential
126
Chapter 3
Steady
Electric
Currents
In general, v d <£ v t
,
makes only a slight change in the
was applied, and v d is a macro-
since the electric field
velocity distribution that existed before the field
scopic resultant of microscopic velocities of free charges in a direction along the
-4
electric field lines. In most cases, its magnitude is not larger than v d ~ 10
m/s in
amounts of current carried
metals, for reasonable
The
(as
we
proportional to the electric
drift velocity is linearly
Vd
drift velocity
=
shall see in
an example).
field intensity vector,
-/Z e E,
(3.2)
where the constant pt e is the so-called mobility of electrons in the given material.
The mobility is measured in the units of m 2 /(Vs) and is positive by definition.
For electrons, the direction of v d as well as the direction of Fe is opposite to the
,
,
Good conductors have high
now say that charge carriers in
direction of E.
We
can
mobility.
the conductor
with the macroscopic average velocity v d This
.
is
move through
volume
its
an organized, directive motion of
charges, which constitute an electric current throughout the conductor volume.
introduce then a
density vector,
J.
new field quantity,
By definition,
J
current density vector (unit:
A/m
We
to describe the current at a point: the current
= Ny (-e )\ d
(3.3)
,
2
)
where
/Vv
or per
1
is
m
3
the concentration of charge carriers,
(the unit
is
m~ 3
The current density can
sity, /,
which, in turn,
surface
current intensity
current (unit:
or,
(e.g.,
is
i.e.,
their
number per
unit
volume
2
).
alternatively be defined by
defined as a rate of
movement
means of the current
inten-
of charge passing through a
cross section of a cylindrical conductor). That
is,
simply,
(3.4)
A)
amount of charge that flows through the surface
dr. The unit for current intensity, which is
current, is ampere or amp (A), equal to C/s. The cur-
In other words, I equals the total
during an elementary time
usually referred to
as,
rent density vector
is
the current
current
current density
vs.
lines. If
dr,
simply,
divided by
directed along the macroscopic motion of charges,
i.e., along
perpendicular
elementary
surface
of
area
dS
to the
set an
3.2(a), the magnitude of the current density vector is given by
we
lines, as in Fig.
intensity
(3.5)
where d I is the current flowing through dS. We see that the unit for J is A/m 2
which means that it actually represents a surface density of a volume current.
,
Figure 3.2 (a) Definition of the
means of the
current density by
current intensity, (b) Evaluation
of the total current
through
a
surface.
2
For example, the concentration of conduction electrons
in
copper
is
Nv = 8.45
x
number of copper atoms per unit volume, since copper has one conduction-band
The number of atoms per unit volume is approximately the same for all solids.
the
It)
28
m -3
.
It
equals
electron per atom.
1
To show
equivalent,
that the definitions of current density, J, in Eqs. (3.3)
we
amount
realize that the total
time interval d t
127
Current Density Vector and Current Intensity
Section 3.1
of charge that crosses
and (3.5) are
dS during the
is
dQ =
dSVd dr (— e)
(3.6)
Av
time interval
(in the
volume Av
dr, a
charge moves a distance v^dr, and
all
charges in the
Ny times Av, pass
=
dSVd dr, the number of which is the concentration
through dS). Using the definition of J in Eq. (3.3), this becomes
dQ=JdSdt.
Dividing by
dr,
we
is
dS
obtain the current intensity d I through
the
same
terms of /:
in
= JdS,
dI
which indeed
(3.7)
(3.8)
as in Eq. (3.5).
In the case where the current density vector
element, the current through the element
dl
is
not perpendicular to the surface
is
— J dS n = JdScosa =
J
•
dS,
(3.9)
where dSn is the projection of dS on the plane normal to J and a is the angle
between J and dS (see Fig. 1.30). Hence, the total current through an arbitrary
surface S, Fig. 3.2(b), equals the flux of the current density vector through the
surface,
1
If
=
J
J
dS.
(3.10)
total current
through
a surface
there are several types of free charge carriers in a conductor drifting with
different average velocities, the resultant current density vector
is
a vector
sum
of
current densities in Eq. (3.3),
M
J
—
'y
'
(3.11)
i=
that correspond to individual types of carriers (e.g., electrons
conductor). Equivalently, the net charge flow
current intensity in Eq. (3.4). Positive charges
is
and holes
move
(3.10) are therefore valid for
dl
in the direction of E, negative
charges in the opposite direction, but both add to the total current. Eqs.
and
in a semi-
taken in evaluating the resultant
(3.4), (3.5),
any conductor and any combination of charge
carriers.
In
many
situations, current flow
infinitely thin) film
current, described
N
localized in a very thin (theoretically
3.3.
This
is
by the surface current density vector, J s which
,
Js
where
is
over a surface, as shown in Fig.
= N q\d
s
,
so-called surface
is
Figure 3.3 Surface current
density vector (J s ).
defined as
(3.12)
surface current density vector
(unit:
A/m)
the surface concentration of charge carriers (number of carriers per
-2
unit surface area, in
). Note that the surface current density vector is somes
is
m
times denoted as K. In terms of the current intensity, the surface current density
is
given by
(3.13)
surface current density
current intensity
vs.
128
Chapter
3
Steady
Electric
Currents
where d I
the current flowing across a line element d / set normal to the current
is
The unit for J s which represents
A/m. For example, the surface current density
flow (Fig. 3.3).
is
with a current
I/w and J s
,
Example
=
/
10 A.
directed parallel to the strip axis.
3.1
Electron Drift along a 1-km
of length
The current
which the electrons
Solution
As
=
/
km
1
and radius a
tv,
strip (foil)
equals 7S
drift
=
3
mm carries a steady current of intensity
the current
is
distributed uniformly across the cross section (S) of the wire, the
is
[Eq. (3.10)]
=
= -4z =
(3.3), the drift velocity
of electrons
v<j
3.54 x 10
A/m 2
5
na
S
=
—
=
(3.14)
.
is
2.62 x 10
-5
m/s,
(3.1 5)
/Vv e
/V v
=
28
8.45 x 10
m -3
is
the concentration of conduction electrons in copper and e
the absolute value of the charge of an electron, Eq. (1.3).
along the wire
drift
is
in
along the wire.
J
where
=
Copper Wire
uniformly distributed across the wire cross section. Find the time
is
current density in the wire
From Eq.
aluminum
of a very thin
uniformly distributed across the strip width,
I that is
is
A copper wire
a line density of a surface current,
,
The time
it
is
takes for an electron to
hence
t
=
— =3.82 x 10
7
(3.16)
s,
Vd
which
is
3.2
approximately 442 days. 3
OHM S LAW IN
CONDUCTIVITY AND
LOCAL FORM
Consider again a metallic conductor (the charge is carried by electrons), with the
J. Substituting Eq. (3.2) into Eq. (3.3), we obtain
current density
J
current density in a metallic
Ny efx e E.
(3.17)
=
ctE,
(3.18)
N^ene,
(3.19)
conductor
This equation can be rewritten as
Ohm
's
law
in local
J
form
where the proportionality constant,
o
conductivity of metallic
conductors (unit:
S/m)
is
a
=
macroscopic parameter of the medium called conductivity.
It is
always positive,
conduct electric
reciprocal of a
meter
(S/m).
The
current. The unit for conductivity is siemens per
x meter (£2m).
resistivity.
The
unit
is
ohm
is denoted by the symbol p and termed
and represents,
3
a
We know
in general, a
we do
that, of course,
1-km long transmission
line.
measure of the
ability of materials to
not need to wait 442 days to receive a communication signal sent via
We
shall see in a later
chapter that time-varying signals traveling along
transmission lines propagate as electromagnetic waves outside the conductors that constitute the
line,
motion of electrons within the conductors. The conductors actually serve as
guides for the waves along the line. That is why signals travel at the velocity of electromagnetic waves in
the medium surrounding the line conductors. If the medium is air, the velocity is 3 x 10* m/s (speed of
and not
light in a
via the drifting
vacuum), and the
travel time
is
only 3.33 ps for a 1-km
line.
J
J
J
Conductivity and
Section 3.2
Using
resistivity,
Ohm's Law
in
Local
129
Form
Eq. (3.18) becomes
E=
1
pJ,
p
(3.20)
o
p -
resistivity (unit:
fimj
Both Eqs. (3.18) and (3.20) are known as Ohm’s law in local or point form.
Note that Eq. (3.18) is one of the three general electromagnetic constitutive
equations for characterization of materials [another one being Eq. (2.46)]. It can be
written in the following form:
J
=
J(E),
(3.21)
constitutive equation for J, for
an arbitrary material
to
encompass
all
possible conducting properties of materials.
However,
in
terms of
most conducting materials are linear and isotropic, i.e., J(E) =
independent of electric field intensity and current density (the property of linearity), and is the same for all directions (isotropy). In homogeneous
conductors, a does not change from point to point in the region being considered. For inhomogeneous conductors, on the other hand, a is a function of spatial
their conductivity,
crE,
where a
is
[e.g., a = <j(x, y, z) in the region].
Almost always, the conductivity is a function of temperature,
few exceptions is an alloy called constantan (55% copper, 45%
coordinates
conductivity
is
practically constant in a temperature range
One
of the
nickel),
whose
T.
0— 100° C. For
metallic
conductors, the mobility of electrons decreases with an increase in temperature
(because the average time interval between collisions with vibrating atoms,
decreases). Hence, the conductivity decreases
perature
rise.
and
resistivity increases
Around a room temperature of Tq = 293
and we can write
K
A tc
,
with a tem-
(20°C), the resistivity
varies almost linearly with T,
P(T)
where po
=
= po[l + a(T-
(3.22)
To)],
p( 7o). For most metals (copper, aluminum, silver,
is approximately 0.4% per kelvin.
etc.),
the temperature
coefficient of resistivity, a,
For some materials the
resistivity
drops abruptly to zero below a certain
temperature:
p( J)
=0
for
J < Jcr
,
(3.23)
where Jcr is called the critical temperature of the material. This property, discovered
by Kamerlingh Onnes in 1911, is called superconductivity, and the materials are
said to behave like superconductors. Most superconductors are metallic elements
that exhibit transition into superconducting states at a temperature of a few kelvin.
Examples are aluminum ( cr = 1.2 K), lead ( cr = 7.2 K), and niobium ( cr =
9.2 K), as well as their alloys and compounds. More recently, new ceramic materials were discovered that become superconducting at considerably higher (and
thus less expensive to produce and maintain) temperatures. For example, yttriumbarium-copper oxide (YBa 2 Cu 3 C> 7 ), discovered in 1986, has Jcr = 80 K, so its
superconductivity can be utilized by cooling with liquid nitrogen. Interestingly,
some of the best of the normal conductors, such as silver and copper, cannot become
superconducting at any temperature, while the ceramic superconductors are normally good insulators - when they are not at low enough temperatures to be in a
superconducting state.
Ohm’s law in local form holds also for conductors with more than one type of
charge carriers [see Eq. (3.11)]. In plasmas and gases, the charge carriers are electrons and positive ions (electron-deficient atoms or molecules). In liquid conductors,
called electrolytes, the charge is carried by positive and negative ions. In all cases,
superconductors
1
30
Chapter
3
Steady
Electric
Currents
both positively and negatively charged particles (ions and electrons) contribute to
the conductivity.
and germanium), vacancies in the atomic crysby electrons, called holes, can move from atom to atom and behave
like positive charge carriers, each hole carrying charge e. The conductivity of a
semiconductor is therefore
In semiconductors (e.g., silicon
tal lattice left
conductivity of
a
=
Nwe ep, e +Nyh en h
(3.24)
,
semiconductors
4
where the first term represents the contribution to the conductivity from electrons,
moving opposite to the field E, while the second term represents the contribution
from holes, which move with E. The concentrations /Vve and Nv h rapidly increase
with an increase in temperature (temperature rise accelerates generation of free
electrons and holes). Consequently, the conductivity of semiconductors increases
with increasing the temperature, which
is
opposite to the temperature behavior of
metallic conductors.
By adding very
small amounts of impurities to pure (intrinsic) semiconduc-
may be increased dramatically. Impurities called donors (e.g.,
phosphorus) provide additional electrons and form n-type semiconductors, while
acceptors (e.g., boron) introduce extra holes, forming p-type semiconductors. This
procedure is known as doping of semiconductors. Note that the boundary between
p-type and n-type parts of a single semiconductor crystal forms a junction region,
called pn junction (see Fig. 2.9), which is utilized in semiconductor devices (diodes
tors,
and
the conductivity
transistors).
Unlike the relative permittivity (e r ), shown in Table 2.1, the conductivity of
materials varies over an extremely wide range of values, as we go from the best
insulators to semiconductors, to the finest conductors. In S/m, o (at room temperature) ranges from around 10
water, and 2.2 for
from quartz
ductivity
it
germanium
goes to
infinity for
to silver
-17
for fused quartz, 10
to 6.17 x 10
is
7
for silver.
-9
We
as large as 25 orders of
superconductors. Table 3.1
lists
for bakelite, 10
-2
for fresh
see that the range in con-
magnitude (10 25 ), and then
values of the conductivity of
selected materials.
Copper (Cu),
the most
commonly used metallic conductor,
conductivity of copper, at
cr
Cu
=
58
has a conductivity of
MS/m.
(3.25)
20° C
many
In
electric
applications,
we
consider copper and other metallic conductors as perfect
conductors (PEC), with
(3.26)
oo.
perfect electric conductor
(PEC)
Of
course, superconductors also
(3.26),
no electric
body
field inside
we conclude
fall
under
this category.
From
Eqs. (3.18) and
that
a PEC
(3.27)
a
4
Generally, the current density in semiconductor devices
=
is
composed of two components:
a drift current
which depends on the gradient of the concentration
of charge carriers in a material and, therefore, does not satisfy Ohm’s law in local form. Consequently,
the relationship between the total current density vector and the electric field intensity vector, Eq. (3.21),
density. J
in
crE,
and
a diffusion current density, Jjjf,
semiconductor devices
is,
in
general, nonlinear.
Conductivity and
Section 3.2
a (S/m)
Material
- 17
~ io
~ io - 17
~ 1(T 16
~ 10~ 15
~ io - 15
~ 1(T 15
~ 1CT 15
~ KT 14
2 x KT 13
~ IO" 12
Quartz (fused)
Wax
Polystyrene
Sulfur
Mica
Paraffin
Rubber (hard)
Porcelain
Carbon (diamond)
Glass
Polyethylene
1.5
io
Bakelite
~ KT
Marble
10-8
-
12
IO
"8
9
6
1(T
10“ 4
Granite
soil
KT 4
2 x
Distilled water
Silicon (intrinsic)
4.4
Clay
5
x 1(T
1(T
Fresh water
~
soil
x 1(T
4
3
2
io
Material
o (S/m)
Carbon
7.14 x 10 4
(graphite)
Bismuth
8.70
Cast iron
~
-2
10- 2
1.04 x 10
x 10
Stainless steel
1.1
Silicon steel
2 x 106
Titanium
2.09 x 10 6
Constantan (45% Ni)
2.26 x 10 6
German
6
3 x 10
silver
Lead
4.56 x 10 6
Solder
7 x 10 6
Niobium
6
8.06 x 10
Tin
8.7
Platinum
9.52 x 10
x 10 6
Bronze
10
Iron
1.03
x 10 7
Nickel
1.45
x 107
Brass (30% Zn)
1.5
Zinc
7
1.67 x 10
Tungsten
7
1.83 x 10
7
2.17 x 10
x 10 7
0.4
Aluminum
3.5 x 10
Animal blood**
0.7
Gold
4.1
Germanium
7
5.8 x 10
(||
to fiber)**
2.24 x 10
3 x 10
2.2
Copper
Seawater
3-5
Silver
Ferrite
10 2
Mercury
(at
<4.1 K)
DO
Tellurium
~
Niobium
(at
<9.2 K)
OO
(intrinsic)
5
x 102
1.18 x 10
Silicon (doped)
For dc or low-frequency currents,
at
7
7
7
x 10 7
7
6.17 x 10
YBa 2 Cu 3 0 7
3
6
7
0.22
8 x IO"
(_L to fiber)**
2
6
6
Animal muscle
**
131
x 10 5
Animal, body (average)**
4 x
fat**
Animal muscle
*
Form
10 6
Sodium
Magnesium
Duralumin
Animal
Local
106
Nichrome
Mercury (liquid)
x 1(T
- 11
Wood
Wet
in
Conductivity of selected materials*
Table 3.1.
Dry
Ohm's Law
(at
<80 K)
OO
room temperature.
Also for humans.
i.e.,
the electric field
is
always zero in perfect conductors. This, in turn, implies that
the voltage between any two points of a perfect conductor [Eq. (1.90)] is zero.
Finally, so-called convection currents, which are the result of the motion of posin a vacuum or rarefied gas (where a = 0),
Examples are electron beams in a cathode ray tube
and a violent motion of charged particles in a thunderstorm. The convection current
density is given by
itively or negatively
charged particles
are not governed by
Ohm’s
law.
J
=
p\,
(3.28)
where v is the velocity of particles and p is the volume density of charges (charge per
unit volume) in the vacuum or rarefied gas. Noting that p = Nw q, Ny being the concentration of particles and q an elementary charge, we observe the equivalency of
the definitions of convection and conduction current densities given by Eqs. (3.28)
convection current density
1
32
Chapter 3
Steady
Electric
and
Currents
(3.3), respectively.
(vj) of charges
and Eq.
However, the velocity v
(3.2)
in
Eq. (3.28)
is
not a drift velocity
not satisfied.
is
Conceptual Questions (on Companion Website):
3.1
and
3.2.
HISTORICAL ASIDE
Kamerlingh Onnes (1853-1926), Dutch
and professor at the University of
Leyden, was awarded a Nobel Prize in Physics
which was a fascinating result at that time.
the first to produce liquid helium (in 1908).
Onnes demonstrated in 1911 that the resistivity
of mercury absolutely disappears at temperatures
below about 4 K, and thus discovered supercon-
Heike
0.9 K,
physicist
He was
in
1913 for his investigations of the properties
of matter at extremely low
temperatures. His
experiments led him as close to absolute zero as
ductivity.
The SI unit of power, the watt, was named in honor
of James Watt (1736-1819), a Scottish mechanical
engineer and inventor, who is famous for his rev-
the 1760s, which led to great advancements in the
olutionary improvements of the steam engine in
his
Industrial Revolution,
ing the “horsepower”
steam engines.
AND JOULE'S LAW
LOSSES IN CONDUCTORS
3.3
and is also known for devis- to measure the power of
IN
LOCAL
FORM
now consider the current flow in a conductor from
As we know, free charge carriers (e.g., electrons) are
Let us
the energy point of
view.
accelerated on their
paths between collisions with vibrating atoms, and at every collision they lose
their acquired kinetic energy.
Energy
is
thus transmitted from the electric
field,
E,
and ultimately
resulting in a higher temperature of the conductor. This means that in a conductor
with electric current, electric energy is constantly converted into heat. We wish to
derive the expression for the rate (power) of this energy transformation.
We start with the electric force on a charge d Q given in Eq. (3.7), which equals
dQ E. The work done by this force in moving d Q a distance d / along the field lines
via charge carriers to the atoms, enhancing their thermal vibration
is
[Eq. (1.72)]
dWe = dQEdl = JdSdtEdl = JEdvdt
where dv
to heat,
is
an elementary volume
known
power of Joule’s
volume dv is
called the
in the
The volume
Joule's
law
ohmic power
W/m 3 )
form; p\
density (unit:
in local
density of this
rate of this conversion,
losses or
dPi
power
Pi
=
conductor. This work
in the
The
as Joule’s heat.
(dv=dSdl),
ohmic
=
d
W
e
losses.
is
(3.29)
converted
dWe /dr (J/s),
is
(lost)
power,
Thus, the power of Joule’s losses
=JEdv.
(3.30)
is
dPj
/2
= JE=
— =oE
(3.31)
dv
and this is known as Joule’s law in local (point) form. The unit for power
(W), and hence the unit for power density is W/m 3
.
is
watt
Section 3.4
Continuity Equation
133
The total power of Joule’s losses (the electric power that is lost to heat) in a
domain of volume v (e.g., in the entire conducting body) is obtained by integration
of power dPj throughout v:
Pj
=
l
JEdv.
(3.32)
power of joule's (ohmic)
losses (unit:
W)
dPj
Conceptual Questions (on Companion Website):
3.3.
CONTINUITY EQUATION
3.4
We now
consider one of the fundamental principles of electromagnetics - the
continuity equation, which
is
the mathematical expression of the principle of con-
and cannot be lost or created. It can
nowhere nor disappear. Let
an arbitrary closed surface, S, enclosing volume v, be situated in a region with timevarying currents, as shown in Fig. 3.4(a), and let Qs and Qs + d Qs denote the net
charges in v at instants of time t and t + dr, respectively. The change in charge, d Qs,
cannot be created in v, but only brought in from the domain outside the surface S,
and it is brought by the current flowing through S. So, the charge dQs passes the
surface S during the time dr, and this is exactly what we have in the definition of
servation of charge. Charge
move from
is
indestructible,
place to place, but can never appear from
current intensity in Eq. (3.4). Therefore,
dQs
(3.33)
dr
is
the intensity of the current flowing across the surface
current crossing the surface in the opposite direction,
region,
is
S
i.e.,
into the region
v.
The
the current leaving the
hence
Jout
—
dQs
~hi
(3.34)
‘
dr
On
(b)
the other hand, by virtue of Eq. (3.10), the current leaving v across S equals
the total outward flux of the current density vector through S (which is a closed
Figure 3.4 Arbitrary closed
surface), that
surface
is,
in a
region with
currents: (a) general case
/out
= ^J-dS.
(3.35)
Combining the two preceding equations, we obtain
(f
Js
This
is
J-dS =
node.
--^.
(3.36)
dr
the continuity equation (in integral form).
It tells
is
us that the outward flux of
equal to the negative of the
derivative in time of the total charge enclosed by that surface.
By
expressing the charge in terms of the volume charge density, p, the
becomes
continuity equation, for
currents of any time
dependence
the current density vector through any closed surface
continuity equation
and (b) current flow
through wires meeting
at a
1
1
34
Chapter 3
Steady
Electric
If
Currents
S does not change
the surface
the
volume
time derivative can be
in time, the
moved
inside
integral, yielding
continuity equation in terms
(3.38)
of the volume charge density
The ordinary
derivative
is
replaced by a partial derivative because p, generally,
is
a
multivariable function of time and space coordinates.
By applying the divergence theorem, Eq. (1.173), to Eq. (3.38) or simply by
analogy to the differential form of the generalized Gauss’ law, Eq. (2.45), we get the
differential
form of the continuity equation:
V
continuity equation in
differential
J
•
= - dp
(3.39)
dt'
form
It tells
us that the divergence of J at a given point equals the negative of the time
rate of variation in charge density at that point,
and
is
also called the continuity
equation at a point.
For steady (time-invariant) currents, the charge density is constant
= 0, and the integral form of the continuity equation reduces to
in time,
dp/dt
=
dS
continuity equation for steady
(3.40)
0.
currents, integral form
The
differential equation of continuity of steady currents
is
given by
differential continuity
V
equation for steady currents
J
=
(3.41)
0.
Thus, time-invariant current density vector has zero divergence everywhere, and
under
all
We
circumstances.
solenoidal.
say that steady electric currents are divergenceless or
The zero divergence
of a vector field indicates that there are no sources
means
upon themselves (steady currents must
or sinks in the field for the lines of flux to originate from or terminate on. This
that the streamlines of steady currents close
flow in closed loops), unlike the streamlines of the electrostatic field intensity, which
originate and end
If
on charges.
the steady current
is
carried into and out of the
volume
v
by a number
(TV)
of wire conductors meeting at a point, as indicated in Fig. 3.4(b), then Eq. (3.40)
implies that the algebraic
sum
of
all
the currents leaving the junction
is
zero.
N
Kirchhoff's current
y>=°.
law
(3.42)
k=
For the situation and notation
in Fig. 3.4(b), I\
—h—h+h —
0-
Eq. (3.42),
if
applied to a node in a dc circuit, represents Kirchhoff’s circuital law for currents.
Like Kirchhoff’s voltage law, Eq. (1.92), Kirchhoff’s current law
also applies to time-varying situations
assumptions, which
will
and ac
be discussed later
In studying steady current fields,
circuits,
meaning
conservative nature of
field
E
in
in
mind
field,
that time-invariant
which
is
a conservative
that the line integral (circulation) of the electric field intensity vector,
E, along an arbitrary contour (closed path)
steady current
above form
in this text.
we always have
currents in a conductor are produced by a static electric
field,
in the
with certain restrictions and
l
E
dl
is
=
zero,
0.
(3.43)
'
Section 3.4
We
also
have
in
mind
that J
and
E
135
Continuity Equation
are related at any point in the conductor by
the constitutive equation for the current density, Eq. (3.21) [or Eq. (3.18) for linear
media].
Example 3.2
Element Law for a Capacitor
Prove that the current through a capacitor equals the product of the capacitor capacitance
and the time rate of change of the voltage drop across the capacitor.
Solution Fig. 3.5 shows a capacitor of capacitance C connected to a time-varying voltage 5
v. We assume that the capacitor is ideal, i.e., its dielectric is perfect (nonconducting), as well
as that there is no excess charge along the connecting conductors in the circuit (charge is
localized only on the capacitor electrodes). To relate the current through the capacitor, i,
to the capacitor charge, Q, we apply the continuity equation for time-varying currents in
integral form, Eq. (3.36), to a surface S enclosing completely only the upper electrode of
the capacitor (Fig. 3.5).
The
total current leaving the
0
c
—
Q
c
enclosed domain (left-hand side of the
,
)
-Q
equation) equals —i, whereas the total enclosed charge (appearing on the right-hand side of
the equation) equals the charge of the upper electrode,
i.e.,
the capacitor charge:
Qs = Q.
Thus, Eq. (3.36) becomes
0
(3.44)
time-varying current; for
Example
Substituting
Q=
Cv
=C
dv
(3.45)
dt
see that
i
is
linearly proportional to the rate of
proportionality constant. This
itor,
which
is
3.2.
[Eq. (2.112)] yields
i
We
1
Figure 3.5 Capacitor with a
is
element law (current-voltage
characteristic) for
change of v
in time, with
C
as the
a capacitor
Wlul a time-varying current
the element law (current-voltage characteristic) for a capac-
widely used in circuit theory, in conjunction with Kirchhoff’s laws and other
element laws.
Spherical Capacitor with an Imperfect Dielectric
A
spherical capacitor is filled with an imperfect homogeneous dielectric of conductivity a.
The radius of the inner electrode is a and the inner radius of the outer electrode is b (b > a).
The electrodes have a conductivity that is much larger than a, so that they can be considered
as perfect conductors. The capacitor is connected to a generator of time-invariant voltage V.
Find (a) the current density vector in the dielectric and (b) the power of Joule’s losses in the
capacitor.
Solution
(a)
Since a
0,
cated in Fig.
there exists a steady current of intensity / in the capacitor circuit, as indi3.6.
To determine
the current density vector in the dielectric,
we apply
the
continuity equation for steady currents in integral form, Eq. (3.40). In general, appli-
cation of the continuity equation
law, Eq. 2.43. This
current density vector
is
is
analogous to application of the generalized Gauss’
one with spherical symmetry (see Example
of the form
problem
is
J
where
r is
We
The surface S
for applying the continuity equation
and the
(3.46)
the radial coordinate in the spherical coordinate system and
vector (Fig. 3.6).
5
= J{r) r,
1.18),
is
f is
the radial unit
a spherical surface
use lowercase (small letter) symbols for the voltage and current here to emphasize that those are
time-varying quantities.
volume current distribution
with spherical symmetry
1
36
Chapter
3
Steady
Electric
Currents
Figure 3.6 Spherical capacitor
with an imperfect
homogeneous
dielectric
and
time-invariant current; for
Example
3.3.
of radius r (a
<
r
<
b) centered at the origin (the
surface in electrostatics,
outward direction
is
e.g.,
that in Fig. 2.16).
zero, that
The
same
as the corresponding
total current flowing
Gaussian
through S
in the
is.
JOAnr2 -1 = 0.
(3.47)
This means that the outward flux of J through S equals the current / through the capacitor
terminals. Hence,
J{r)
From Eq.
=
(
a
<r <
b )•
3.18, the electric field intensity in the dielectric
(3-48)
is
I
(3.49)
Anar2
The voltage between
the electrodes
given by
is
v = v,-n =
where
Va
because
li
Using Eq.
power of Joule’s
r
"
[
dv=
,
Pj
in
power
equipotential
=
aabV
(b — a)r 2
losses in the dielectric
— Wdr=-
"-'
J(r)
2
.
°
9
r
dv
(3.51)
,
Anaa l2 b 2l V 2z
(b
-
ry2
a)
r
b
/
Ja
is
— = AnaabV
—
dr
r
2
b
2
(3.52)
a
the form of a thin spherical shell of radius r and thickness dr
[Eq. (1.33)]. There are
overall loss
6
J r=a
Jv
with dv adopted
is
»
(3.32), the
Pj=
no
losses in the electrodes (perfect conductors), so this
is
the
in the capacitor.
Conceptual Questions (on Companion Website): 3.4-3. 6;
Companion
(3.50)
^(l-l}
and Vb are the potentials of the electrodes (each electrode
a). Combining Eqs. (3.48) and (3.50), we get
cr e ie C trodes
7(r)
(b)
E *r =
Website).
MATLAB
Exercises (on
Section 3.5
3.5
Boundary Conditions
137
for Steady Currents
BOUNDARY CONDITIONS FOR STEADY CURRENTS
Applications of steady current fields involve considerations of interfaces between
conducting media of different conductivity. In this section, we shall formulate the
boundary conditions that govern the manner in which the current density vector, J,
and the electric field intensity vector, E, behave across such interfaces.
Comparing the integral form of the continuity equation for time-varying currents, Eq. (3.38), to the integral form of the generalized Gauss’ law, Eq. (2.44), we
conclude that the boundary condition for normal components of the vector J is of
the same form as the boundary condition for normal components of the vector D,
Eq. (2.85). The only difference is on the right-hand side of the equation, where p s
(the surface charge density that may exist on the surface) is replaced by —dp s /dt.
With this.
n
•
—n
Ji
dps
=
J2
•
(3.53)
dt
boundary condition
for J n/
time-varying currents
is the normal unit vector on the surface, directed from region 2 to region 1.
For steady currents, —dps /dt = 0 in Eq. (3.53), and the boundary condition that
corresponds to Eq. (3.43) is that in Eq. (2.84) [Eq. (3.43) is the same as for the
electrostatic field], so the complete set of boundary conditions for steady currents is
given by
where n
n x Ei
n
•
—n
Ji
E2 =
x
0
— n-J 2 =
E\ t
or
0
or
J\ n
— E2
1
,
=hn-
E
(3.54)
boundary condition
for
(3.55)
boundary condition
for J n/
t
regime
We see that in the steady current field, the tangential component of the electric field
intensity vector
and the normal component of the current density vector,
are both continuous across the boundary.
boundary condition
for
E
t
in
As we
Eq. (3.54) has
this
E and J n
t
,
shall see in a later chapter, the
same form
for fields of any time
variation.
By analogy to the procedure
of deriving Eq. (2.87) for the electrostatic
field,
we
obtain the law of refraction of the current density lines at a boundary between two
linear conducting
media of conductivities
oq
and
tan oq
oq
tan oq
a2
oq:
(3.56)
law of refraction of current
streamlines
where aq and 012 are the angles that current lines in region 1 and region 2 make with
the normal to the interface, as shown in Fig. 3.7.
Note that if medium 2 is a good conductor and medium 1 is a low-loss dielectric,
then <72
oq, i.e., oq/oq ~ 0, and Eq. (3.56) gives tanoq « 0 for any 012 Therefore,
»
.
oq
meaning
«0
(oq
» oq),
(3.57)
that the current lines always leave (or enter) a
good conductor
at a right
angle to the boundary (zero angle to the normal on the boundary).
Note
tan
012
also that
if
medium
1 is
a perfect dielectric
(e.g., air),
then oq
=
0,
and
-> 00 from which
,
oq
= 90°
(oq
=
(3.58)
0).
means that the lines of current flow are always parallel to the surface of a
conductor surrounded by a nonconducting medium.
This
Figure 3.7 Refraction of
steady current lines at a
conductor-conductor
interface.
Conceptual Questions (on Companion Website): 3.7 and
3.8.
dc
a
138
Chapter 3
Steady
Electric
Currents
DISTRIBUTION OF CHARGE IN A STEADY CURRENT
3.6
FIELD
The
electric field intensity vector, E, in a steady current field
is
produced by
station-
ary excess charges in the system. In the general case, these charges exist not only
on the surfaces of conductors, but also inside their volume. Distribution of charges
in the system is conditioned by the distribution of currents, and can be determined
only after the current distribution is determined first.
The distribution of the current density vector, J, inside conductors can be
obtained by solving the basic equations that govern steady current fields, which we
summarize here:
Maxwell's
first
equation for
§c E
static fields
J
dc continuity equation
J
constitutive equation for J
Once
the solution for J
is
=0
dS = 0
•
=
dl
•
J(E)
(3.59)
=
[J
known, the charge
crE]
distribution can be obtained
from the
generalized Gauss’ law in conjunction with the constitutive equation for the electric
flux density vector,
D:
[&D-dS = Q S
D = D(E) [D = eE]
Maxwell's third equation
constitutive equation for
D
(3.60)
|
Note
that,
assuming that the medium is linear, both D and J are linearly
As a result, we have a linear relationship between D and J:
proportional to E.
duality of
D
D = £E = - J.
and J
(3.61)
a
This duality relationship will be used on
many
occasions.
Starting from the differential form of the generalized Gauss’ law, Eq. (2.45),
and using Eqs.
volume charge
in
(3.61)
and
(3.41), the
volume charge density
in the
conductor
p-V.D-V.(£j)-[v(i)].J+i(V.J) = J.v(i).
a steady
is
6
(3.62)
current field
We
see that
V
=
J
0 does not imply that p
=
Volume charge
0.
inhomogeneous conducting media where e/a
in
portional to the gradient of e/o.
On
cannot be volume excess charges (p
^
const,
we
the other hand,
= 0)
inside
and
is nonzero
magnitude is pro-
density
its
also conclude that there
homogeneous media
(
and
e
do
not vary with position) with steady currents.
The corresponding boundary condition
D, Eq.
for
(2.85), in
combination with
Eqs. (3.61) and (3.55), gives the surface charge density on the interface between
medium (with parameters o\ and £j) and medium 2 (with cr2 and £ 2 ):
1
surface charge in a steady
Ps
=
r.
Dl - n D?
n
•
=
—n
*1
o\
current field
^Having
in
mind the
•
(/a)
=
(V/)
a
J,
1
—n
02
•
J2
--W
=
V^i
rule for calculating the derivative of a product of
gence operator, which
V
«
is
+ /(V
(3.63)
two functions, we apply the diverand a vector function, and get
a differential operator, to a product of a scalar
a).
a2 J
0
Section 3.7
Relaxation
Time
n 3\ — n J2 is the normal component of the current density vector
boundary (n is directed as in Figs. 2.10 and 3.7). Note that ps — 0 only for
the special case of £\/o\ — 82 / 02 If both media are metallic conductors, we have
approximately s\ = e 2 = £o> so that Eq. (3.63) becomes
where Jn
=
•
•
across the
-
(3.64)
Ps
Finally, the distribution of bound (polarization) charge in the dielectric can
be found by computing first the polarization vector, P, from Eq. (2.59), and then
bound volume and surface charge densities, p p and pps using Eqs. (2.19) and (2.89),
,
respectively.
Problems'. 3.1; Conceptual Questions (on
Companion Website):
3.9
and
3.10;
MATLAB Exercises (on Companion Website).
3.7
RELAXATION TIME
We know
from
electrostatics that charge placed in the interior of a
move to the conductor surface and redistribute itself in such
way
a
conductor
that
E=
will
0 in the
conductor under electrostatic equilibrium conditions. With the continuity equation
(for time-varying currents) now in hand, we can quantitatively analyze this nonelectrostatic transitional process and calculate the time it takes to reach an equilibrium.
Consider a homogeneous conducting
medium
of conductivity 0 and permittiv-
The current density vector, J, and electric flux density vector, D, in the medium
interconnected by the duality relationship in Eq. (3.61). Combining the differ-
ity e.
are
ential
form of the generalized Gauss’
equation, Eq. (3.39),
law,
Eq.
£
£
= V.D = V- (-j)\ = — V-J =
\o /
a
/
/0
where the
ratio £ /
—
£ dp
o
(3.65)
dt
can be brought outside the divergence sign because it is a conRewriting, we have that the charge density, p,
-\
dt
is
—
0
dp_
which
and that of the continuity
medium is homogeneous).
medium satisfies the equation
stant (the
in the
(2.45),
we can write
£
P
— 0,
(3.66)
a first-order partial differential equation in time,
p
where po
is
=
po
e'
~(a/e)t
Poe'
e.g.,
Its
solution
is
-t/T
p
—
given by
(3.67)
the initial value of the charge density at
function of the space coordinates as well,
t.
t
—
0.
In general, p
p(x,y,z,t). Eq. (3.67)
is
redistribution of charge
a
tells
us that the charge density at a given location in the conductor decreases with
time exponentially, completely independent of any applied electric
field.
The time
constant
(3.68)
is s) and it equals the time required
any point to decay to 1/e (36.8%) of its initial value. It
can easily be shown that p decreases to 1% of po after approximately 4.6r, while at
_5
t = lOr, p % 4.5 x 10
p 0 he., p « 0.
is
referred to as the relaxation time (the unit
for the charge density at
,
relaxation time
139
140
Chapter
3
Steady
Electric
Currents
For metallic conductors, r is so short that it can hardly be measured or observed.
For example, copper has rc u % 10 -19 s. Even much poorer conductors than metals
-5
have very short relaxation times (e.g., rn 2 o % 10
s for distilled water). On the
other hand, the relaxation time for good dielectrics (insulators)
(e.g., r
minute, rm
1
g ass
i
i
ca
~
15 hours, and rquar tz
%
is
relatively long
50 days).
devoted to time-invariant currents and fields, we note
is also used to determine the electrical
nature (in terms of conducting and dielectric properties) of materials for timevarying currents and fields. Namely, whether a material of parameters o and s
is considered a good conductor or a good dielectric is decided on the basis of
While
chapter
this
is
here that the concept of relaxation time
the relaxation time, as
compared
to times of interest in a given application. Thus,
for a time-harmonic (sinusoidal) field of frequency /, the relaxation time, given
by Eq.
is
(3.68),
classified as
said to be a perfect electric conductor (PEC).
considered a good dielectric (insulator)
(lossless) dielectrics (a
rial is classified as
=
0).
For
all
a quasi-conductor.
that are considered as
good
= \/f. If x
T, the medium
= 0 (a -> oo), the material is
compared to the time period, 7 T
a good conductor. In particular, if r
is
if
On
the other hand, the material
other (intermediate) values of
We
can
good conductors
now understand
that
at certain frequencies
dielectrics at sufficiently higher frequencies
(i.e.,
is
oo for perfect
T. In a limit, r
r
r,
some
the mate-
materials
may behave
like
shorter time periods). For
= 10 MHz, and fi, = 30 GHz, average rural
kHz,
S/m, assuming no change in the parameters as a
function of frequency) behaves like (1) a good conductor, (2) a quasi-conductor,
and (3) a good dielectric, respectively. Further discussions of conducting and
example, at frequencies
ground
(e r
=
14 and a
f\
=
=
10
1
-2
dielectric properties of materials at different frequencies, in a context of general
Maxwell’s equations and electromagnetic wave propagation, are provided
in later
chapters.
3.8
RESISTANCE, OHM'S LAW,
AND JOULE'S LAW
Consider an arbitrarily shaped conductor made from a linear (generally inhomogeneous) material of conductivity a, as shown in Fig. 3.8. Let the end surfaces 5 a and
5 b of the conductor be coated with perfectly conducting material (or with material of conductivity much greater than a). If the voltage V is applied between the
conductor ends, the current, of density J, in the conductor flows normal to the end
surfaces [Eq. (3.57)] and parallel to the sides of the resistor [Eq. (3.58)].
of the continuity equation, the
same
section of the conductor. This current
total current
is
By
virtue
must pass through every cross
given by
(3.69)
where E
is
any path
On
the other hand,
same
vector, E, along
the electric field intensity vector in the conductor.
the voltage between
5 a and 5 b equals the
line integral of the
conductor connecting these surfaces [Eq.
in the
(1.90)], that
is,
B
(3.70)
7
The time period
(
T)
is
defined as an interval after which a time-harmonic (or other time-periodic)
over time, whereas the frequency (/)
per unit time (one second), so that / times T equals unity.
function repeats
itself
is
the
number
of repetitions of the function
Section 3.8
Resistance,
Ohm's
Law, and Joule's
Law
141
Figure 3.8 Arbitrarily shaped
conductor -
Now, we note
that
not change shape, but
and, by
means
if
resistor.
V is increased (for some reason), the electric field lines do
E proportionally increases everywhere within the conductor,
of Eq. (3.69), so does the current
I.
The
ratio of
V
and
I
is
thus a
constant,
(3.71)
called the resistance of the conductor.
stantial) resistance
voltage, current,
R
is
O)
A conductor with two terminals and a (subThe relation between
Ohm’s law:
usually referred to as a resistor.
and resistance of a
resistance (unit:
resistor
is
known
as
V — RI.
the
(3.72)
Ohm's law
The resistance is always nonnegative (R > 0), and the unit is the ohm (£2), equal to
V/A. The value of R depends on the shape and size of the conductor (resistor), and
on the conductivity a (or resistivity p) of the material.
HISTORICAL ASIDE
Georg
Ohm
analogies between the flow of electricity and flow
German
of heat. In a series of papers in 1825 and 1826, he
gave a mathematical description of conduction in
Simon
(1789-1854),
a
and mathematiwas a professor
physicist
cian,
at
University
the
Munich. As a
received
a
and
matical
tific
child,
fine
of
Ohm
mathescien-
education from his
father,
who was
a lock-
smith
and
entirely
an
man. In 1811, he received a doctorate
mathematics from the University of Erlangen.
He was also interested in physics, where he studied
self-taught
in
modeled after Fourier’s study of
He assumed that, just as the rate
at which heat flowed between two points depended
on the temperature difference between the points
and on the ease with which heat was conducted
by the material between the points, the electric
current between two points should depend on the
difference in electric potential between the points
and on the electric conductivity of the material
between the points. By experimenting with wires
of different lengths and thicknesses, he found that
electric circuits
heat conduction.
142
Chapter 3
Steady
Currents
Electric
the current intensity through a wire, for a given
Analyzed Mathematically), published in
first, Ohm’s work was received
with little enthusiasm, and a full acknowledgement
and recognition of his results, including Ohm’s law,
did not come until 1841, when he was awarded the
Copley Medal of the Royal Society and soon after
became member of several European academies.
Only in 1852, two years before his death, Ohm
Circuit,
between the wire ends, was
inversely proportional to the length and directly
1827 in Berlin. At
potential difference
proportional to the cross-sectional area of the wire.
As
very knowledgeable of both mathematics and
Ohm
physics.
was able to deduce mathematical
on the experimental evidence
relationships based
that he
a wire
existed
had tabulated. He defined the resistance of
and showed, in 1827, that a simple relation
among
(voltage),
and current
now known
achieved his lifelong ambition of being appointed
the resistance, potential difference
Ohm’s
as
intensity of a wire. This
law.
The
fully
to the chair of physics at the University of
developed
ohm
presentation of his theory of electric conduction
lished the
appeared in his famous book “Die Galvanische
Kette, Mathematisch Bearbeitet” (The Galvanic
Edgar Fahs Smith
Libraries)
The siemens
the mechanic Johann
adopted as the SI unit for
honor
of brothers Werner and
conductance in
Wilhelm von Siemens, German engineers and
inventors. Ernst Werner von Siemens (1816-1892)
is
contributed to then
new
the “Siemens
Company”
(ohm
ter.
Georg Halske (1814—1890),
Halske Telegraph Construction
in Berlin, later
He
later
as “Siemens.”
furnace and electric pyrome-
became
British subject, Sir Charles
is
called the conductance
the siemens (S), where S
is
=
1
I
R
V
= A/V.
Note
and symbolized by G.
that
sometimes the
shows the circuit-theory representation of a resistor.
assumed that the resistances are located only in the resistors
In circuit theory,
in a circuit,
the interconnecting conductors are considered as perfectly conducting.
necting conductor has therefore zero resistance and
The only voltage drops in the
Ohm’s law, Eq. (3.72), is valid also
circuit).
that
of a resistor (v
Figure 3.9 Circuit-theory
representation of a
resistor.
elements
=
mho
used instead of the siemens.
Fig. 3.9
it is
known
William Siemens.
of resistance
spelled backwards)
&
the regenerative
(unit: S)
Its unit is
(Portrait:
Collection, University of Pennsylvania
mechanical engineer by training and successful
businessman. His most important inventions are
discipline of electrical
phy (needle telegraph), cable transmission (large
undersea cables), and energy generation, and was
one of the first great entrepreneurs in electrical industry. In 1847, he founded, together with
conductance
as the unit of resistance.
Karl Wilhelm von Siemens (1823-1883) was a
engineering with several inventions in telegra-
The reciprocal
From Ohm’s law,
Munich.
Ohm’s name was further immortalized in 1881,
when the International Electrical Congress estab-
is
is
while
Each con-
equipotential (acts as a short
occur across circuit elements. Note
for time-varying voltage (v) and current ( i )
circuit
It represents the element law for a resistor, as one of the basic
theory [another element law being Eq. (3.45) for a capacitor].
Ri).
in circuit
For linear resistors, R remains constant for different voltages and currents, V
and / For nonlinear resistors, however, the conductivity of the material depends
on the applied electric field, E, and R therefore depends on the applied voltage (or
.
current),
nonlinear resistor
R=
R(V).
(3.74)
'
Section 3.8
Resistance,
Ohm's Law, and
Joule's
Law
143
Figure 3.10 Evaluation of the
power
of Joule's (ohmic) losses
a resistor of general shape.
in
Examples are semiconductor (pn ) diodes, whose current-voltage
characteristics, I
=
I(V), are nonlinear functions.
The
total
power of
Eq. (3.32), where v
is
Joule’s or
ohmic
losses in the resistor in Fig. 3.8
the volume of the resistor.
we first cut v into thin slices with bases chosen to be
dicular to current lines).
We
then subdivide each
small tubular cells of volume dv
=
is
given by
To carry out the volume integration,
equipotential surfaces (perpen-
slice
d S d /, as depicted
along the current lines into
in Fig. 3.10,
and write
JEdv — f fjEdSdl.
J A Js
dv
—
Note
same
that
E dl
equals the voltage
dV between
(3.75)
the bases of the
cell,
which is the
between
for all cells within a slice (because of the equipotentiality of interfaces
adjacent
Therefore,
slices).
dV
is
a constant for cross-sectional integration over S,
and can be taken out of that integral. We have thus separated the overall integration
throughout v into two independent integrals:
-Us JdSEdl
dV
Pi
(3.76)
Figure 3.1
1
which equal, respectively, the voltage V [Eq. (3.70)] and the current I [Eq. (3.69)]
of the resistor. Employing Eqs. (3.72) and (3.73), the equivalent expressions for the
in a
power of Joule’s
section; for
losses in a resistor are
conductor of a uniform cross
Example 3.4.
2
Pj
This
is
known
I
= VI = RI2
G'
as Joule’s law.
Current Uniformity
Example 3.4
A steady current
is
in
Conductors with Uniform Cross Section
flowing through a long conductor
made
of a
homogeneous material and
having a uniform cross section of an arbitrary shape. Prove that the current density
same
(3.77)
in the entire
is
the
conductor.
With reference to Fig. 3.11, the current density vector in the conductor has a
= Jz (x, y, z) z. From the continuity equation for steady currents in differential form, Eq. (3.41), and the expression for divergence in the Cartesian coordinate system,
Eq. (1.167), we have
Solution
2 -component only, J
V
which means that Jz
is
not a function of
J
=
z, i.e.,
(3.78)
it
does not vary along the conductor.
Steady current
homogeneous
Joule's
law
144
Chapter 3
Steady
Electric
Currents
HISTORICAL ASIDE
Janies
Prescott
Joule
was
scientist,
a
produced by a current
that the heat
(1818-1889), an English
tric circuit
mem-
in
an elec-
over a certain interval of time equals
the square of the current intensity multiplied by
ber of the Royal Society.
the resistance of the circuit and time. This
As
to be called Joule’s law.
a son of a wealthy
He measured
came
the heat
and was especially
produced by every process he could think of, and
studied the relationship between the amount of
work entering the system and the amount of heat
exiting the system. In his famous “paddle wheel”
experiment in 1847, he used a falling weight to
spin a paddle wheel in an insulated barrel of water
and measured very precisely the increase in the
water temperature produced by the friction of the
wheel - to determine the mechanical equivalent
interested in the efficiency of electric motors. After
of the heat dissipated in the water. His general
several attempts to design a superefficient electric
conclusion was that heat was only one of
would produce infinite power (this possibility had been suggested in previous papers) and
thus offer an ideal alternative to steam engines,
he realized that this goal was not achievable and
became interested in studying the heat gener-
forms of energy, which can be converted from one
form to another but the total energy of a closed
system remains constant. With this, he contributed
fundamentally to the discovery and recognition of
ated by electricity. Joule lacked in mathematical
Joule also by using joule as the unit for
brewer.
Salford
Joule
was educated by private
tutors, including the famous English chemist and
physicist
John Dalton
(1766-1844). Later, while
working
in
the
family
brewery, he studied in his spare time the subject
of electricity
motor
(new
at that time)
that
rigor, but
was a
fanatic experimentalist.
the principle of conservation of energy.
Based on
energy.
(Portrait:
many
We
honor
work and
National Bureau of Standards Archives,
extensive measurements of heat in electric motors
courtesy AIP Emilio Segre Visual Archives, E. Scott Barr
and other
Collection)
electric circuitry,
he discovered,
To prove
in 1840,
that J does not
change
Eq. (3.43) to a rectangular contour
parallel to current lines (Fig. 3.11).
<t
C
Using also Eq.
E dl=
Jc
where a
the
is
the conductivity of the
medium
is
homogeneous ( a
—
in a cross section of the
-
(f
dl
dl
1
8),
(f
we can
J-dl
=
write
0,
(3.79)
° Jc
1/cr
can be taken out of the integral because
Hence,
const).
•
(3.
=-
Jc °
medium, and
J
conductor either, we apply
placed inside the conductor, with sides of length a set
= J\a — J2 a =
0,
(3.80)
with J\ and Jj standing for the current densities along the two sides of the contour parallel to
J.
We
conclude that these current densities are the same:
J\=J2
Note
that
C can
be translated to any position
in
.
(3.81)
the conductor, so that J\ and Ji can be associ-
depend
on x and y, namely, it is the same in the entire cross section of the conductor. Combined with
our previous conclusion about the current uniformity in the z-direction, we have that
ated to any pair of points
dc current uniformity
in
in
the conductor cross section. This implies that J does not
a
J
homogeneous conductor of a
uniform cross section
everywhere inside the conductor.
=
const
(3.82)
.
Section 3.8
Example
Resistance,
Ohm's Law, and
Joule's
145
Law
Resistance of a Resistor of a Uniform Cross Section
3.5
Determine the resistance of a homogeneous resistor with a uniform cross section of an
arbitrary shape and surface area S. The length of the resistor is / and the conductivity of the
material is a
Eq. (3.82)
Solution
tells
us that the current
current density in the resistor
is
distributed uniformly in the resistor.
is
The
thus
J
=
(3.83)
S'
where
I
is
the current intensity through the resistor, Eq. (3.69). Using Eq. (3.70), the voltage
across the resistor
is
J/
//
11
a
cro
V S El=-l=—.
(3.84)
Hence, the resistance of the resistor comes out to be
(3.85)
oS
resistance of a
resistor with
homogeneous
a uniform cross
section
Example 3.6
Fig. 3.12(a)
Two
shows a
Resistors in Series
two parts of lengths
cylindrical resistor of radius a consisting of
Conductivities of the parts are ay and 02 (ay
^ 02
).
The voltage
across the resistor
l\
is
and
l2
.
V. Find
the current through the resistor.
Solution
The current through the two
parts of the resistor
represented as two homogeneous resistors connected in
total resistance of the
connection
is
the same, and they can be
series, as
shown
in Fig. 3.12(b).
The
is
R=
R\
+ R2
(3.86)
,
equivalent resistance of two
resistors in series
where, using Eq. (3.85), the resistances of individual resistors amount to
Ri
h
=
R„ 2
and
G\na^
respectively.
The current through the
1
Two
g2 ttcij2a
resistor in Fig. 3.12(a)
=
R\
Example 3.7
h
-
is
(3.87)
’
hence
2
V
+ Ri
nax a2 a V
vih
(3.88)
+ <f\h
Resistors in Parallel
A cylindrical resistor of length
/ consists of two coaxial layers of different conductivities, oy
and G2 as shown in Fig. 3.13(a). The cross-sectional surface areas of the layers are S\ and S2
The current through the resistor is I. Compute (a) the current density in the resistor and (b)
+
Vi
,+
V2
.
,
^WW\
the voltage across the resistor.
R2
/?!
it
Solution
(a)
Eq. (3.54)
tells
us that the electric field intensities in the two layers are the same:
E
X
=E
2
=
E.
V
(3.89)
Hence, the current densities in the layers are not the same, as
total current through the resistor is given by
J\
= g\E
and J2
= g2 E.
The
I
from which
= Jx Si + J2 S2 =
(giS\
+ g2 S2 )E,
(b)
Figure 3.12 Cylindrical
resistor
(3.90)
with two parts of
different conductivities:
(a)
(b)
for
geometry and
network representation;
Example 3.6.
p
146
Chapter 3
Steady
Electric
Currents
and
o\I
=
J\
CTlSl
(b)
The voltage
across the resistor
and
+ (7252
J2
<
El
5
i
(3.92)
+ O2 S 2
II
=
ctj5i
V can
7i
is
V=
Note
ail
=
(3.93)
+ 02S2
found using the equivalent circuit shown in Fig. 3.13(b). Namely,
two layers of the resistor is the same, they can be represented as two homogeneous resistors connected in parallel. The total conductance of the
connection turns out to be
that
also be
since the voltage across the
G—
equivalent conductance of
two
resistors in parallel
that
[Eq. (3.85)],
is
+ G 2 ,\
(3.94)
— —
(3.95)
G\
G= 01S1
The voltage of the
resistors
is
V=
Conductance of
Example 3.8
+
a2 Si
I/G.
a Spherical Capacitor
Find the conductance of the spherical capacitor with an imperfect (conducting) dielectric (of
conductivity
cr)
from
Fig. 3.6.
Solution Using Eqs. (3.73) and
by definition,
(3.50), the
conductance of the nonideal spherical capacitor
is,
I
leakage conductance of a
V
nonideal spherical capacitor
This, actually,
is
=
4 naab
(3.96)
b-a'
the so-called leakage conductance of the capacitor.
(An
ideal capacitor has
a perfect dielectric and zero leakage conductance.) Note, however, that the system in Fig. 3.6
may
also represent a spherical resistor, with resistance
J_
_ P(b-a)
G
(3.97)
4ji ab
where p = \/a is the resistivity of a (resistive) material between the electrodes.
Note also that, employing the conductance (or resistance) and Joule’s law, Eq.
the power of Joule’s losses in the capacitor (resistor) can now easily be found as
PjJ
which, of course,
Problems
:
MATLAB
3.9
is
the
3. 2-3. 9;
same
= GV 2 =
result as in
Eq.
—R
AnaabV 2
b — a
(3.98)
(3.52).
Conceptual Questions (on Companion Website):
Exercises (on
(3.77),
3.1 1-3.17;
Companion Website).
DUALITY BETWEEN CONDUCTANCE AND
CAPACITANCE
(b)
Figure 3.13 Resistor with
two
coaxial layers (a),
which
can be represented as two
homogeneous
resistors
connected in parallel
for Example 3.7.
(b);
Consider a pair of metallic bodies (electrodes) placed in a homogeneous conducting
medium of conductivity o and permittivity e, as shown in Fig. 3.14. The conductivity
of electrodes is much larger than cr. Let the voltage between the electrodes be V.
From Eq. (3.62), there is no volume charges ( = 0) in the medium (a and e are
constants). Having in mind Eq. (3.57), the current (and field) lines are normal to
the surfaces of electrodes. We now use the duality relationship between the current
Section 3.9
Duality
between Conductance and Capacitance
/
/
\
\
Figure 3.14
Two
metallic
homogeneous
conducting medium.
electrodes
in a
D, in the medium, Eq. (3.61), to
and capacitance, C, between the electrodes.
Applying the continuity equation for steady currents, Eq. (3.40), and generalized Gauss’ law, Eqs. (3.60), to an arbitrary surface S completely enclosing the
density vector,
J,
and
electric flux density vector,
relate the conductance, G,
positive electrode (Fig. 3.14) gives
(3.99)
where
I
is
the total current leaving the positive electrode through the conducting
medium (and
entering the negative electrode),
Q is the total charge of the positive
electrode (the total charge of the negative electrode
outside the integral sign because
By
dividing this equation by V,
it is
we
a constant,
is
i.e.,
— Q), and o/e can be brought
medium is homogeneous. 8
the
obtain
L=
V
°_Q
£
(3.100)
v'
which, by means of Eqs. (3.73) and (2.113), yields the following duality relationship
G and
between
C:
(3.101)
£
Note
charges
that, since the definition of C,
Q
Eq. (2.113), depends on the existence of
is independent of whether or not a cur-
and — Q on the electrodes and
medium (of permittivity e), G and C in Eq. (3.101)
do not necessarily represent the conductance and capacitance of a system with
an imperfect dielectric of parameters a and e. They can also be associated with
two independent dual systems, representing the capacitance (C) between a pair of
rent also exists in the dielectric
electrodes
when placed
in a perfect dielectric (of permittivity e, including the case
= £o,
and zero conductivity) and the conductance (G) between the same electrodes when placed in a conducting medium (of conductivity o and totally arbitrary
£
permittivity), respectively.
8
Note, however, that this
a/e
=
const.
is
true also for
inhomogeneous media
for
which functions a and
e are
such that
duality of
G
and
C
147
148
Chapter
3
Steady
Currents
Electric
R between
In terms of the resistance,
the electrodes, Eq. (3.101) can be
,
rewritten as
R-C analogy
(3.102)
where
C
Q
given by Eq. (3.68),
is the relaxation time of the material between the elecshows the circuit-theory representation of the system in Fig. 3.14.
a first-order capacitive circuit, whose time constant (r c = RC) is equal, by
r,
trodes. Fig. 3.15
-Q
This
is
virtue of Eq. (3.102), to the relaxation time of the material (r
The
and (3.102) are very useful
relations in Eqs. (3.101)
in
=
e/o).
deriving expressions
conductance and resistance of electrode configurations for which we already
have the capacitance, or vice versa. For example, from the expression for the capacitance of a spherical capacitor with dielectric permittivity e (and zero conductivity)
and electrode radii a and b (a < b), given in Eq. (2.119), we can immediately determine the expression for the conductance of the same capacitor if filled with a
conducting medium of conductivity a\
for
Figure 3.15 Network
representation of the system
in Fig.
3.14.
4ncrab
G= —C=
b
e
[the
same
as in Eq. (3.96)].
Self-Discharging of a Nonideal Capacitor
Example 3.9
A
(3.103)
—a
parallel-plate capacitor is filled with an imperfect dielectric of relative permittivity e r
-14
=
6
and conductivity a — 10
S/m. The capacitor was connected to an ideal battery and fully
charged to a voltage of 20 V between its plates, and then disconnected from the battery. Find
the time after which the voltage of the capacitor decays to 1 V.
Solution The capacitor can be represented by the equivalent network in Fig. 3.15, where C
and R are the capacitance and resistance between the capacitor plates, respectively. With no
battery in the circuit, the current discharging
C is
also the current across R, which, with help
of Eqs. (3.45) and (3.72), gives
—dv—
v
=0,
I
dt
where the time constant rc
equation
= RC is found
from Eq.
(3.104)
rc
(3.102).
The
solution of this differential
is
self-discharging of a capacitor
v(/)
=
v(0) e
_r/(/?C)
=
V (0)e~
(<7/S) '.
(3.105)
with an imperfect dielectric
Finally,
we
we
take the natural logarithms of both sides of the equation, and obtain the time
seek:
t
=—
where
v(0)
We
=
20
V
and
v(f)
=
1
=
In
er
265 minutes,
(3.1
06)
v(0)
V.
charged capacitor with homogeneous imperfect dielecis completely independent of its shape (geometry)
and size, and completely analogous to redistribution of charge in a conductor described by
Eq. (3.67). The capacitor voltage, v(t), and charge, Q(t) = Cv(r), decrease exponentially with
time, at a rate that is set by the relaxation time of the lossy dielectric filling the space between
tric
that
realize that discharging of a
is left
to itself (open-circuited)
the capacitor electrodes.
Conceptual Questions (on CD): 3.18-3.21;
Website).
MATLAB
Exercises (on
Companion
External Electric Energy
Section 3.10
3.10
149
Volume Sources and Generators
EXTERNAL ELECTRIC ENERGY VOLUME SOURCES
I
AND GENERATORS
We
observed in connection with
chemical battery)
is
Fig. 3.1 that
an external voltage source
(e.g.,
a
required to maintain a steady current in a conductor. This
source creates a rise in potential in the circuit and a potential difference between
the ends of the conductor. From the energy standpoint, an external source of
energy is necessary to maintain steady electric current flow through the circuit
by continuously supplying the energy that is then dissipated in the conductor as
Joule’s heat. In concrete situations, various forms of the external energy (chemical energy,
mechanical energy, thermal energy,
converted
light energy, etc.) are
and ultimately lost to heat.
analogy to circuit-theory generators, we use two field-theory mod-
to the energy of the electric field in conductors
Generally, in
els
of volume-distributed energy sources capable of transmitting energy to electric
and sources analogous to current
charges: sources analogous to voltage generators
generators.
Consider a source region of volume v shown in Fig. 3.16(a) in which a nontermed the impressed force, acts on charge carriers (e.g.,
electric external force, Fj,
and separates positive and negative excess charges. An example is the
on electrons in a metallic wire moving in a magnetic field (as we shall see in
a later chapter). We can formally divide F, by the charge of a carrier, q (q = — e for
electrons), and what we get is a quantity expressed in V/m:
electrons)
force
Ej
=
—
(3.107)
,
q
which we
call
the impressed electric field intensity vector. This
although of the same dimension and unit as E,
due to charges). Since there
force on q is
is
is
in the
domain
v,
A
however,
not a true electric field (a
due to charges
also a field
field,
(b)
field
the total
Figure 3.16 (a) External
electric
Ftot
=
F;
+ qE =
(/(Ej
+ E),
(3.
08)
energy sources
modeled by an impressed
electric field, of intensity Ej.
and
compels the charges to move through the region. The
the region is, therefore, given by
this is the force that
current density vector in
J
= <r(E + Ej),
(3.109)
(b) Voltage generator in
circuit theory.
Ej
- impressed
electric field
intensity vector (unit:
where a
is
the conductivity of the material in
v. It is
very important to note that Ej
does not depend on J.
The model with a volume distribution (throughout v) of energy sources modeled by an impressed electric field, of intensity Ej, can be used in many different
electromagnetic situations, as we shall see later in this text. However, we note here
that the particular situation in Fig. 3.16(a) actually represents a voltage generator in
circuit theory.
The generator
is
connected to a resistor of resistance R. Combining
Eqs. (1.90) and (3.109), the voltage between the ends
B and
A of the source region
(generator), can be written as
dl.
Reversing the order of integration
FB a =
limits,
we have
B
f
/
JA
r
Ej
-
(3.110)
dl
-
/
B
JA
-i
<r
dl.
(3.111)
’V/m)
150
Chapter
3
Steady
Electric
Currents
The
first term on the right-hand side of
motive force (emf) of the generator:
equation
this
is,
by definition, the electro-
B
electromotive force
(unit:
£
- emf
V)
= f
JA
Ej-dl.
(3.112)
that £ equals the work done by the external force, Fj, in moving the charge q
through the generator, from its negative terminal (A) to its positive terminal (B),
divided by q. The unit for emf is V. The second integral in Eq. (3.111) is linearly
Note
proportional to the current
/ in
the circuit:
J
= Rg I
dl
a
where the constant of proportionality, R g
generator. Hence, Eq. (3.111) becomes
,
is
FBA = £ -
voltage generator
(3.113)
,
called the internal resistance of the
RgI.
(3.114)
shows the equivalent circuit-theory representation of the generator.
the generator terminals A and B are open-circuited (R —» oo), no
current flows through the generator (/ = 0), and we can write [Fig. 3.16(b)]
Fig. 3.16(b)
Note
that
if
(^BA) open-circuited
Thus, the
An
no
its
emf of
= ^CA =£
the generator also equals
ideal voltage generator
is
its
voltage
(1
=
(3.1
0).
when
internal losses, implying that the voltage of such a generator
it.
That
is
(
Rg =
for
all
values of
I
impressed electric
and
all
i.e.,
is,
(3.116)
values of R. This, essentially,
field intensity, Ej,
0),
always equal to
Vba=S
ideal voltage generator
5)
open-circuited.
one with a zero internal resistance
emf, irrespective of the current flowing through
1
is
a
consequence of the
being completely independent of the current
density, J, in the generator.
By Ohm’s
law,
Eq. (3.72), on the other hand, the voltage
Tba
in Fig- 3.16
can
be written as
F BA =
RI,
(3.117)
which, combined with Eq. (3.114), gives
£
= Rg I +
RI.
(3.118)
many emf
sources), we have
In general, for a closed path in a circuit with
(including the internal resistances of the
Kirchhoff's voltage
terms of emf's
law
sources and resistors
M
N
—
$
Rkhyi
in
and voltage
y=l
(3.119)
k=l
drops Rf
This equation
is
an expression of Kirchhoff’s voltage law, Eq. (1.92).
It
states that
sum of the emf’s (voltage rises) around a closed path in a circuit equals
algebraic sum of the voltage drops (RI) across the resistances around the path.
Multiplying both sides of Eq. (3.1 18) by /, we obtain
the algebraic
the
£1
= Rg I 2 +
RI 2
.
(3.120)
Section 3.10
By
Joule’s law, Eq. (3.77), the expressions appearing
External Electric Energy
on the right-hand
Volume Sources and Generators
151
side of this
equation represent the power of Joule’s losses in the generator and that in the resistor of resistance R, respectively. Both these powers are lost to heat. Therefore, by
the principle of conservation of energy, the expression on the left-hand side of the
equation,
= £I,
Pi
(3.121)
must be the power generated by the emf of the generator. In local form, the volume
3
density (in W/m ) of the power of the impressed electric field of intensity Ej is [see
Eq. (3.31)]
(3
dv
The
power of sources
total
in the
domain v
-
122 )
thus [see Eq. (3.32)]
is
J dv.
(3.123)
power generated by
impressed
Let
now
field
the charges in a source region be carried by an impressed nonconduc-
tion current of density
as depicted in Fig. 3.17(a),
Ji,
independent of the
electric
model of energy sources we use in
9
electromagnetics. The current density Jj does not enter Ohm’s law in local form,
and the total current density vector in the region is given by 10
This
field intensity, E.
is
the second general
J
— E + Jj,
(3.124)
cr
Jj
- impressed current density
vector (unit:
A/m 2 )
which, in scalar notation [for the situation in Fig. 3.17(a)], becomes
J
This
is
= —oE + J[.
analogous to a current generator
(3.125)
in circuit theory,
shown
in Fig. 3.17(b)
and
described by
I
where
Ig
and
respectively.
circuited
0),
is
(V
= -G g V + Ig
=
0).
that /g equals the current of the generator with
its
terminals short-
An ideal current generator has a zero internal conductance (G g =
or an infinite internal resistance. This
means
that the current of such a generator
always constant,
i.e., it is
A
= h.
(3.127)
independent of the generator terminal voltage (and of the conductance G).
example
classical
is
so-called
Van de Graaff
generator, where the impressed current consists of
charges transported mechanically on a moving dielectric belt, so that the velocity at which the belt
moves
actually represents an equivalent drift velocity (vd) of charges. Here, however, the velocity of
charges
(i.e.,
the velocity of the belt) obviously does not
E, and Eq. (3.2) does not
10
current generator
G g are the current intensity and internal conductance of the generator,
Note
I
9
(3.126)
,
Note
make
depend on the
electric field intensity vector,
sense.
that both Eq. (3.109), for a region with energy sources
modeled by an impressed
electric field,
and
Eq. (3.124), for a region with an impressed current, can be regarded as versions of the general constitutive
equation for the current density, Eq. (3.21). In other words, Eq. (3.21) generally includes models of vol-
ume energy
of materials.
sources based on impressed fields and impressed currents, along with conduction properties
ideal current generator
5
152
Chapter 3
Steady
Electric
Currents
The power density of sources represented by an impressed current of density
amounts to
Pi
= —E
•
Jj
(3.128)
Jj,
where the minus sign is necessary because vectors E (the electric field intensity
vector, due to positive and negative charges) and J; (the impressed current density
vector, which, like any other current density vector, carries positive charges in
direction
and negative charges
For the situation
in the
= EJ
in Fig. 3.17(a), p\
its
opposite direction) are directed oppositely.
.
X
The
total
E
power generated by
Jj
power of sources
in v
is
dv.
(3.129)
impressed current
Note
that the
power generated by
7g in Fig. 3.17(b)
Pi
This power
is
=
VIg
is
given by
(3.130)
.
delivered to the rest of the circuit and dissipated to heat in the two
resistors, the total
power of Joule’s
losses being
G g V 2 + GV 2
.
ANALYSIS OF CAPACITORS WITH IMPERFECT
3.1 1
INHOMOGENEOUS DIELECTRICS
We now
(a)
/
deal with steady current fields in capacitors containing piece-wise
homo-
geneous and continuously inhomogeneous imperfect (lossy) dielectrics. Generally,
charges in these systems exist not only on the surfaces of capacitor electrodes, but
also on the boundary surfaces between homogeneous dielectric layers and over
the volume of continuously inhomogeneous dielectrics. The analysis starts with the
evaluation of the current distribution, i.e., the computation of the current density
vector, J, in the dielectric. The electric field intensity vector, E, is then determined
by Ohm's law in local form. By integrating E through the dielectric, we find the
voltage between the electrodes and the conductance (and resistance) of the capacitor. Finally, the distribution of charge in the system can be found from the electric
flux density vector, D, using the generalized Gauss’ law in differential form and the
corresponding boundary condition.
As an illustration, consider a parallel-plate capacitor with two imperfect dielectric layers shown in Fig. 3.18(a). Let the permittivities of the layers be e\ and £ 2
their conductivities o\ and 02 and thicknesses d\ and d 2 respectively. The surface
area of each of the plates is S. The plates can be considered as perfect conductors.
Let us analyze this capacitor, assuming that it is connected to an ideal generator of
,
,
,
(b)
Figure 3.17 (a) External
electric
energy sources
modeled by an impressed
electric current, of density
Jj.
(b) Current generator in
circuit theory.
time-invariant voltage V.
The current density vector in the dielectric is normal to the plates [Eq. (3.57)]
and uniform in each dielectric layer [Eq. (3.82)]. From the boundary condition
for the normal components of J, Eq. (3.55), applied to the interface between the
two dielectrics and the continuity equation for steady currents in integral form,
Eq. (3.40), applied to a rectangular closed surface enclosing the positive plate
[Fig. 3.18(a)], with the right-hand side positioned in either one of the dielectrics,
we have
.
JX
where
/ is
-J2 =J =
the current through the capacitor.
~,
The
(3.131)
electric field intensities in the
dielectrics are
FEl =
L = ±_
/
a,
/
a,
and
E2 =
-L_.
— = —.
(72
(72
S
(3.132)
Section 3.1
1
Analysis of Capacitors with Imperfect
Inhomogeneous
153
Dielectrics
Figure 3.18 Analysis of a
capacitor with a two-layer
imperfect dielectric:
geometry
(a)
of the structure
with vectors
J,
and D
and (b)
E,
individual layers
equivalent circuit of the
(b)
(a)
The voltage across the capacitor
V=
is
given by
E\d\
S\o
so that the conductance of the capacitor
G_
The
11
V
(3.1 33)
02 /
to
be
crio2 S
_
J_
+
i
comes out
(3.134)
o2 di+oid2
electric flux densities in the layers are
D = e\E\ =
system.
s\GV
and
i
12
£2
D 2 = s 2 E2 =
GV
(3.135)
o2 S
CTlS
Eq. (3.62) tells us that, since the dielectric layers are homogeneous, there is no volume charge in them. Using Eq. (3.63), the surface charge on the boundary surface
between the layers amounts
Psi2
to
— D 2 - Di
( £2
V° 2
r
_
£l \
04 /
j
_
(£204
~
&\d2
ei<*2)V
-f-
o2 d\
(3.136)
where we take Jn = —J in Eq. (3.63) because the unit vector n in the corresponding
boundary condition is directed from medium 2 to medium 1. Note that if £\/o\ —
£ 2 /o2 Psl2 = 0. From Eq. (2.58), the surface charge densities on the plates are
,
Psi
=D\ =
The system
£\Q2V
o\d2
+ o2 d\
and
ps2
= -D 2 = -
£204
0\d2
F
+ CT2^1
(3.137)
can be explained and analyzed also invoking the
and concepts. The interface between the dielectric layers
is equipotential, and can, therefore, be metalized. We thus get two nonideal capacitors in series, each of them being represented by a parallel connection of an ideal
in Fig. 3.18(a)
circuit-theory point of view
11
Note that the current
field in the resistor of Fig. 3.12(a)
can be analyzed in the same way.
Note that D\ ^ D 2 contrary to Eq. (2.147) for the same capacitor with two perfect dielectric layers,
shown in Fig. 2.25(a). The capacitor with a perfect dielectric represents an electrostatic system (with
no current), and the analysis starts with the generalized Gauss’ law. The capacitor with an imperfect
12
,
dielectric represents a
system with steady current, and the analysis
starts with the continuity equation.
in
,
154
Chapter
3
Steady
Electric
Currents
capacitor and an ideal resistor, as
shown
in the equivalent circuit in Fig. 3.18(b).
The
characteristics of the elements in the circuit are:
Ci
eiS
=
r
C2 =
£2 S
=
d2
r2 =
aiS’
A,
CT2
(3.138)
S
where the resistances R and R 2 can be obtained either from the corresponding
capacitances, C\ and C2 [also see Eqs. (2.157)], using the relationship in Eq. (3.102),
or from Eq. (3.85).
Using the expression for the equivalent total resistance of a series connection
i
of resistors, Eq. (3.86), the conductance of the system
1
G=
R\
The charges
+ R2
is
=
cr 2 d\
(3.139)
+ o\d2
of the capacitors in Fig. 3.18(b) are
computed
as
C\R\I
= C\R\GV =
— GV,
Qi = C2 V2 = C2 R 2 I
= C2 R 2 GV =
—
GV
a
01
=
Ci V\
=
(3.140)
(3.141)
,
2
with V\ and
V2 standing for the voltages across individual dielectric layers.
13
Finally,
the surface charge densities in Fig. 3.18(a) are given by the expressions
0i
Psi
= y.
Psi2
which give the same results as
Example 3.10
——
= 02-0i
in Eqs. (3.136)
A
and
i
and
-Qi
=—
—
ps2
(3.137).
Spherical Capacitor with a Continuously
Inhomogeneous Imperfect
Dielectric
A
is filled with a continuously inhomogeneous imperfect dielectric. The
and conductivity of the dielectric depend on the distance r from the capacitor
center and are given by the expressions e(r) = 3eo b/r and o(r) — aob 2 /r (a < r < b), where
a and b are radii of the inner and outer capacitor electrodes, and cto is a (positive) constant.
The capacitor is connected to a time-invariant voltage V. Find: (a) the conductance of the
spherical capacitor
permittivity
bound charge
capacitor, (b) the free charge distribution of the capacitor, (c) the
distribution
of the dielectric, and (d) the total free charge in the capacitor.
Solution
(a)
Because of spherical symmetry, the current density vector
given by Eq. (3.46).
From
the continuity equation, J(r)
that the electric field intensity vector in the dielectric
E=
J
r
^Trj
with
/
13
Note
same
that the voltages of the elements in Fig. 3.18(b) are
= VR 2 /(R\ +
r
47r<ro b
in the entire dielectric
in the dielectric is of the
the
same
form
as in Eq. (3.48), so
(Fig. 3.19)
=
4na(r)r2
being the current intensity of the capacitor.
turns out to be the
is
is
2
=
Er,
(3.143)
We note that the electric field intensity
E = const). The voltage between the
(
determined by the
resistors: V\
=
VR\/(R\
+
Ri) - resistive voltage divider. In the case of the same system with perfect
dielectric layers [Figs. 2.25(a) and 2.26(a)], which is an electrostatic system, the voltages are determined
by the capacitors: V\ = VCi/{C\ -I- C2 ) and Vj = VC\/(C\ -I- C2 ) - capacitive voltage divider.
Rl) and
a
Section 3.1
Analysis of Capacitors with Imperfect
1
Inhomogeneous
/
Figure 3.19 Spherical
capacitor with a continuously
inhomogeneous
dielectric
with losses; for
Example 3.10.
electrodes and the capacitor conductance
amount
b
V=
—
Edr = E(b -a)
[
to
14
G
Ja
(b)
The
electric flux density vector in the dielectric
1
4 na0 b 2
=
(3.144)
=V ~b^
given by
is
D = <•(/•)£?,
where E = V/(b —
a).
By means
density inside the dielectric
volume charge
of Eqs. (3.62) and (1.171), the free
is
E
The
(3.145)
3 e 0 bV
d
(3146)
on the surfaces of the inner and outer capacitor
free surface charge density
electrodes are
Psa
= D(a + =
)
e(a
+ )E
=
a(b
-
and
ps
t,
= —D(b
-
)
a)
3g V
— s{b )E = — 0
b — a'
(3.147)
respectively.
(c)
From Eq.
(2.59), the polarization vector in the dielectric is given
P=
Using Eq.
[g(r)
bound volume charge
(2.19), the
- e 0 ]Er.
(3.148)
density over the
ldr ?„1
„
by
volume of the
dielectric
is
— 2 r)
b (30
£oU(3
l
(b
(3.149)
- a)
while Eq. (2.23) gives us the bound surface charge density on the surfaces of the
dielectric:
Ppsa
(d)
The
—
total free
‘
r
Q=
f
P{&
+.
charge
)
_
—
- a)V
a(b-a)
eo(3 b
and
pp sft
=
7
)
=
2e 0 V
(3.150)
b-a
Q in the capacitor is zero. To prove that, we write
b
p(r) 471^ dr
j
P(b
v
+psa Sa + psb Sb = \2neobV +
llnsoabV
12tt£o b
b
“r
2
—a
V
-
0
,
dv
(3.151)
14
Note
that, since e(r)/a(r)
capacitance
C is found
in
^ const, G
Example
2.19.
is
not proportional to C,
i.e.,
Eq. (3.101)
is
not
satisfied.
The
Dielectrics
155
156
Chapter 3
Steady
Electric
Currents
where dv is adopted to be the volume of a spherical shell of radius r and thickness dr [see
2
Fig. 1.9 and Eq. (1.33)], Sa = 4jra (surface area of the inner electrode), and Sb = 4nb 2
(surface area of the outer electrode).
Problems: 3.10—3.16; Conceptual Questions (on Companion Website): 3.22 and
MATLAB Exercises (on Companion Website).
3.23:
ANALYSIS OF LOSSY TRANSMISSION LINES WITH
3.12
STEADY CURRENTS
In this section,
we analyze
transmission lines with losses in a time-invariant (dc)
regime. Consider, as an example, a coaxial cable with imperfect conductors of
(finite)
conductivity
oc and an imperfect
dielectric of (nonzero) conductivity 04.
radius of the inner conductor of the cable
is a,
The
the inner and outer radii of the outer
conductor are b and c, respectively (a < b < c), and the length of the cable is /. The
cable is fed by a time-invariant (dc) voltage generator of electromotive force £ and
internal resistance R g as shown in Fig. 3.20. A load of resistance R l is connected to
,
the other end of the cable.
Let
I designate the current flowing
The same current returns through
cable
is
imperfect, there
the cable conductors.
is
between
symmetry of the structure, this current is
and its density, J, depends only on the radial
a leakage (stray) current through the dielectric,
Due
to cylindrical
radial with respect to the cable axis
distance r from the axis.
through the inner conductor of the cable.
the outer conductor. Since the dielectric in the
The current leakage, in turn, causes a continual decrease
meaning that I is a function of the coordi-
of the current intensity / along the cable,
nate z along the cable, I
cylindrical closed surface
—
I(z). Applying the continuity equation, Eq.
S of radius r and length A z (Fig. 3.20) gives
I(z
+
Az) —
a/
[currents I(z
in
+
Az
current leakage in a
imperfect dielectric
Figure 3.20 Coaxial cable with
dc regime.
0
(3.152)
rd
Az) and /^Az leave 5, while /(z) enters it], where A 1 is the change
A z, and I'd is the leakage current per unit length (p.u.l.)
A/m). Hence,
(^d)p.u.l.>
transmission line with an
imperfect conductors and
=
current I along a distance
of the cable (in
dielectric in a
I{z) +72flrr
(3.40), to a
(3.153)
/
/
Section 3.1 2
157
Analysis of Lossy Transmission Lines with Steady Currents
and
J
A.
=
(3.154)
2rtr
Using Ohm’s law
the dielectric
is
Eq.
in local form,
computed
(3.18), the electric field intensity vector in
as
—=
E=
I'
(3.155)
r,
2no&r
cr(j
so that the voltage between the inner and outer conductor of the cable amounts to
V
^ln».
a
27rad
Edr =
-S'
(3.156)
By definition, the conductance per unit length of a transmission
conductance for one meter (unit length) of the line divided by 1 m, is
,
G'
line, i.e.,
the
/j
= Gpul = ^.
(3.157)
,
conductance per unit length
of a transmission line (unit:
also referred to as the p.u.l. leakage conductance.
It is
cable,
from Eq.
The
unit
is
S/m. For a coaxial
S/m)
(3.156),
27T<7d
=
G'
(3.158)
In (b/a)
conductance
p.u.l.
of a
coaxial cable
The drop of the current
/ along a distance
AG =
is
Az
equals
A I = A GV, where
G'Az
(3.159)
the conductance through the dielectric at the length Az.
As Az approaches
zero, the expression
becomes the derivative of
and (3.157), we have
on the left-hand side of Eq. (3.153)
combining Eqs. (3.153)
/ with respect to z. Therefore,
d
=
-G'V.
(3.160)
dz
This
is
transmission line with a lossy
dielectric
a first-order differential equation in z for the current and voltage of a trans-
mission line with a lossy dielectric.
along the line
is
It tells
us that the rate of change of the current
proportional to the negative of the line voltage, with the conduc-
tance per unit length of the line as the proportionality constant. In the case of a
transmission line with a perfect dielectric (no losses in the dielectric), G'
that dl
dz
=
0,
=
0,
so
yielding
I
Because of losses
varies along the cable,
in the
i.e.,
—
conductors
V=
V(z).
const.
(l/crc
(3.161)
/
From Eq.
0),
the voltage between conductors
transmission line with a
perfect dielectric
(3.85), the total resistance (for time-
invariant currents) of the cable turns out to be
R = R 1+ R 2 =
-L + -1-,
(3.162)
where R\ and R 2 are the resistances of the inner and outer conductor, respectively,
and 5] and S 2 are cross-sectional surface areas of the conductors. The resistance per
unit length of the line is given by
R'
= R p.u.l.
R
1
(3.163)
resistance per unit length of a
transmission line (unit: £2/ m)
<
158
Chapter 3
Steady
Electric
Currents
(the unit
dc resistance
p.u.l.
is
£2/m), so that
of a
R'
{
1
=
The
resistance along the length
Az
+
\a
ttoc
coaxial cable
1
1
of the line
AR =
(3.164)
ci~^bi
is
R'Az,
(3.165)
and the voltage drop across that resistance
-
V(z)
+
V(z
= A RI.
Az)
(3.166)
This equation can be rewritten as
V(z
+
Az)
-
V(z)
=
-R'AzI,
(3.167)
AV
which, as
Az approaches
zero,
becomes
transmission line with lossy
(3.168)
conductors
We
see that the rate of change of the voltage along a transmission line with lossy
conductors
is
proportional to the negative of the line current, with the resistance
per unit length of the line being the proportionality constant. For a transmission
=
conductors (PEC), R'
line with perfect electric
V
transmission line with perfect
0 and
dK/ dz =
0,
so that
- const.
(3.169)
conductors
Eqs. (3.167), (3.153), and (3.157)
of length
Az can be represented by
resistance
R'Az and a shunt
(parallel) resistor of
transmission line can be replaced by
in Fig. 3.21.
us that each section of a transmission line
tell
a circuit cell consisting of a series resistor of
Thus, a transmission line
conductance G' Az, and the entire
many cascaded
is
equal small
cells,
as indicated
said to be a circuit with distributed
(parameters per unit length), and Eqs. (3.160) and (3.168) are
Kirchhoff’s laws and Ohm’s law for the cells of the circuit.
ters
in fact
paramebased on
Eqs. (3.160) and (3.168) are called transmission-line equations or telegrapher’s
equations for time-invariant currents and voltages on transmission
transmission line with both R!
^
0 and G'
lines.
For a
these two equations are coupled dif-
0,
two unknowns: /(z) and V(z). By taking the derivative with
respect to z of one equation and substituting its right-hand side by the corresponding expression from the other equation, we can eliminate one unknown (current
ferential equations in
l(z-Az)
o
A,A/\
R'Az
/(z+Az)
/(z)
—+ >-VVv
—+ rvA
t
R'Az
V(z-Az)
V(z+Az.)
V{z)
G'Az
o
<
G'Az.
—
>
c>
Figure 3.21 Circuit-theory
representation of a transmission
line
with losses
in
conductors
and
dielectric in a
dc regime.
r
1
R'Az
z+Az
z-Az
Az
—L
G'Az
^
1
Section 3.12
159
Analysis of Lossy Transmission Lines with Steady Currents
or voltage) and obtain the following second-order differential equations with only
voltage and only current as unknowns:
—Vt -R'G'V =
d
2
d 1
2
dz z
—
and
0
*
dzl
- R'G'I =
0
(R'
0,
G'
# 0).
(3.170)
Their solutions are exponential functions for the voltage and current along the
R!
If
=
Eq. (3.160) is
and R'
0,1
line.
V=
const along the line [Eq. (3.169)], and the solution to
/ 0,
a linear function for the current along the line. Conversely, if G' = 0
0 and G'
=
const along the line [Eq. (3.161)], and the solution to Eq. (3.168)
is
=
0,
a linear function for the voltage along the line. Finally,
if
both G'
—0
and R'
both current and voltage do not vary along the line.
Note that using the duality relationship in Eq. (3.101), the leakage conductance
per unit length of a transmission line with homogeneous imperfect dielectric can be
found from the capacitance per unit length of the line as
(3.171)
where
e
is
duality of
G
1
and
C
the permittivity of the dielectric [for a coaxial cable, see Eqs. (3.158) and
Note also that R' ^ 1/G' in Eqs. (3.170).
The power of Joule’s losses per unit length of a transmission line, P), can be
obtained as the power of Joule’s losses in one cell in Fig. 3.21, divided by A z. The
unit is W/m. Specifically, in conductors of the line (series resistor in Fig. 3.21),
(2.123)].
(
A RI2
=
Az
pj) c =
R'l
2
(3.172)
.
joule's or
line
ohmic
losses p.u.l. in
conductors (unit:
W/m)
In the dielectric (shunt resistor in Fig. 3.21),
A GV2
Az
«)d
The
total
power of Joule’s
3.1
(3.173)
= R'l2 + G'V2
is
hence
(3.174)
.
Transmission Line with Perfect Conductors and Imperfect Dielectric
A transmission line of length
and conductance per unit length G is connected to an ideal
voltage generator of time-invariant emf £. The other end of the line is terminated in a load
of resistance Pl- The losses in the conductors of the line can be neglected. Find: (a) the
1
l
distribution of current along the line conductors, (b) the total
line,
and
(c) the
power of
Joule’s losses in the
power of the generator.
Solution
(a)
With reference to
Fig. 3.22,
voltage along the line
is
15
since there are
15
no
losses in the line conductors, R'
= 0, the
[Eq. (3.169)]
V(z)
Although resembles a two-wire
line, a
=£
(0
<z<l).
pair of parallel horizontal lines in Fig. 3.22
(3.175)
is
a symbolic repre-
sentation of an arbitrary two-conductor transmission line, with conductors of completely arbitrary cross
section
and generally inhomogeneous
dielectric.
Joule's (ohmic) losses p.u.l. in
line dielectric (unit:
losses per unit length of the line
P'j
Example
G'V2
W/m)
.
160
Chapter
3
Steady
Electric
Currents
m
Figure 3.22 Transmission
m
j
line
with lossless conductors and
lossy dielectric in a
for
Example
3.1
dc regime;
1
Eq. (3.160) becomes
2L = -<*,
(3.176)
az
and by integrating
it,
we
obtain
— -G'£z + In
I(z)
(0
<z<l).
the condition /(/) = V//?l — £/Rl (the line current
through the load), the integration constant is found to be /0
(3.177)
From
at z
=
=
G' £1
/
equals the current
+ £/Rl
[/o
=
7(0)
is
the current of the generator], so that
=
I(z)
G'£(l
~
R
-z) +
(R'
=
0,
a£
(3.178)
0).
L
(b)
Based on Eq.
(3.173), the
P'j
As
=
power of Joule’s
(Pj)
= G'V 2 = G’£ 2 =
d
fj does not vary along the
the losses in the load,
losses per unit length of the line
line,
comes out
Pi=
the total
const
(0
<
power of Joule’s
z
<
is
given by
(3.179)
/).
losses in the line, including
to be
^l +
C=
v
line
G/+
(
^)^
2
(3
'
'
180)
’
load
(c)
Eq. (3.121)
tells
us that the
power of the generator
is
P =£1(0) =e(g'£1+
(3.181)
{
Note
that
Pi
which, of course,
is
in
=P
3
(3.182)
,
agreement with the principle of conservation of energy and power;
namely, the power generated by the generator
and dissipated
is
delivered to the rest of the circuit,
to heat in the lossy dielectric of the transmission line
and
in
the resistive
load.
Example 3.12
Thin Two-Wire Line with Losses
Determine the resistance and conductance per unit length of a thin symmetrical two-wire
line with lossy conductors and a lossy dielectric. The conductivities of both conductors and of
the dielectric are erc and oj, respectively. The conductor radii are a and the distance between
their axes is d (d
a).
The
dc resistance
two-wire
p.u.l. of
resistance per unit length of the line
R'
a thin
=
[see also Eq. (3.164)].
twice that of a single wire:
1
2
ac na 2
line
is
(3.183)
—
Analysis of Lossy Transmission Lines with Steady Currents
Section 3.1 2
161
Figure 3.23 Cross section of a
thin symmetrical two-wire line
with imperfect dielectric
coatings immersed
conducting
Example
for
From
—
— C' —
n<Td
Conductors of a thin two-wire
o \ The
of conductivity
(3.1
'
.
is
is a,
Conducting Liquid
the thickness of coatings
» a). The line
d (d
in a
coated with thin coaxial layers of imperfect dielectric
line are
radius of each wire
distance between wire axes
84)
immersed
is
is
also a
,
and the
a2
in a liquid of conductivity
.
the conductance per unit length of this line?
The evaluation of the (leakage) conductance per unit length of the line, whose
is shown in Fig. 3.23, can be performed completely in parallel to the evaluation
of the capacitance per unit length of the line in Fig. 2.31. Here, instead of Q' we have I'd - the
Solution
cross section
current that leaks from conductor
ductor 2 of the
conductors,
line,
i.e.,
per unit of
its
1
of the line through the coatings and the liquid to con-
length.
The leakage current
densities
the current density due to the leakage from conductor
leakage into conductor
2,
due to the individual
and that due to the
1
M in Fig. 3.23, are
h = id_-
evaluated at the point
_
V
_k_
Ji
2.71
and
T'
2n(d
x
— x)
,
(3.185)
The total current density along the x-axis between the conductors is J = J\ + J2
The electric field intensity is E — J/o\ in the coatings and E = J/a2 in the liquid. The voltage
between the conductors, V, is calculated as in Eq. (2.182), and the conductance per unit length
of the line comes out to be
respectively.
.
G'
Note
—
V
=7z (
In
2
+
\ a\
—
a2
that this expression for G' can be obtained also
ln^-^)
2 a)
(3.186)
.
by substituting the permittivities
in the
expression for C' in Eq. (2.183) by the corresponding conductivities. Namely, the inhomogeneity in terms of o in the system in Fig. 3.23
in
terms of e in the system in
Fig. 2.31,
and a
is
of the
same form as the inhomogeneity
between the conductance and
sort of duality
capacitance of the two systems can be exploited.
Example 3.14
Planar Line with
Two
Imperfect Dielectric Layers
shows a planar transmission line consisting of two parallel metallic strips of width
and a two-layer dielectric between the strips. The strips are perfectly conducting, whereas
both dielectric layers are imperfect, with conductivities o\ and a2 The thicknesses of layers
are d\ and d2 and the length of the line is /. The line is connected at one end to an ideal
voltage generator of time-invariant emf £. The other end of the line is open. Calculate (a)
the conductance per unit length of the line and (b) the current along the strips.
Fig. 3.24
w
.
,
Solution
(a)
An
application of the continuity equation to a rectangular closed surface
Fig. 3.24,
much
like that in
S portrayed
in
Eq. (3.152), yields
JwAz =
I(z)
-
I(z
conductance
two-wire line
Two-Wire Line with Imperfect Coatings
Example 3.13
is
In (d/ a)
e
is
3.1 3.
Eqs. (3.171) and (2.141), the conductance per unit length of the line
G'
What
in a
liquid;
+ A z) =
I'
d
Az
(3.187)
p.u.t.
of a thin
162
Chapter
3
Steady
Electric
Currents
Figure 3.24 Planar
transmission line with a
two-layer imperfect dielectric;
for
Example 3.14.
( I'd is
the leakage current per unit length of the line), from which, the current density in
both dielectric layers
is
/'
/=
A
w
(3.188)
Having in mind Eqs. (3.133) and (3.157), the voltage between the
conductance per unit length of the line are obtained as
1/
= —di +
a1
(b)
—d
2
=
rs2
The current along
W
the line
(
d
A
+ ?l)
V
CTl
is
— w (— +
02 )
V<Tl
given by Eq. (3.178) with
I(z)
=
G'£(l
w£{l
-z) =
d\/o\
—
and the
1
.
o2 'j
)
°o (the line
-
strips
is
(3.189)
open):
z)
+d2 /o2
(3.190)
’
Problems 3.17-3.22; Conceptual Questions (on Companion Website): 3.24-3.26;
:
MATLAB Exercises (on Companion Website).
3.1 3
GROUNDING ELECTRODES
Consider a metallic body (electrode) of arbitrary shape buried under the flat surface
body may represent a grounding electrode for some electrical or
16
electronic device (residing on the earth’s surface).
Assume that the conductivity
of the earth is the same, cr, in the entire lower half-space. Let a time-invariant current of intensity / flow from the electrode into the earth (this current is supplied to
the electrode through a thin insulated wire), as shown in Fig. 3.25(a). As the conductivity of the electrode is much larger than a, Eq. (3.57) tells us that the current
lines leaving the electrode are perpendicular to its surface. From Eq. (3.58), on the
of the earth. This
other side, the current lines
(parallel) to the
l6
Grounding
boundary
in the
earth near to
its
boundary with
air are tangential
surface.
electrodes, in general, are used for draining static electricity induced on parts of a device,
for equalizing the electric potential of a device housing with the neighboring conducting objects, for
reducing the so-called conduction interference (undesired currents induced
“regular" currents
in
in a device, interfering
with
the device and with neighboring devices), for routing (together with lightning rods)
the charge flow in lightning to the ground, etc.
Section 3.1 3
Grounding Electrodes
163
Figure 3.25 Grounding
electrode with a steady current
(a)
and the equivalent system
with the earth-air boundary
removed
(b)
(a)
Consider now the system shown in Fig. 3.25(b), obtained by taking an image
of the lower half-space of Fig. 3.25(a) in the earth-air boundary plane. Namely, the
grounding electrode in the lower half-space is the same as in the original system
and a symmetrical electrode (image electrode) with respect to the boundary plane
is introduced in the upper half-space, while the entire space around the electrodes is
filled with the medium of conductivity a. The current of the image electrode is the
same
as that of the original
By symmetry,
image).
one
(/),
and
it
flows also out of the electrode (positive
the total current density vector (due to both electrodes) at
an arbitrary point of the plane of symmetry in the system in
izontal
component
only. This
to the plane of symmetry,
in the
system
means
and
in Fig. 3.25(a).
Fig. 3.25(b)
has a hor-
that the resultant current lines are tangential
plane corresponds to the ground-air interface
conclude that the distributions of current in the
this
We
lower half-space in the systems in Figs. 3.25(a) and (b) are the same. This is the soimage theory (or theorem) for steady currents, which states that an arbitrary
called
steady-current source
(e.g.,
a grounding electrode with a steady current flowing into
the ground) situated in a conducting half-space near a
flat
boundary surface with a
nonconducting medium (e.g., earth-air interface) can be replaced by a new system
of sources in an infinite conducting medium. The new system consists of the original source configuration itself and its positive image in the boundary plane. The
image electrode (i.e., the charge on its surface) actually represents an equivalent
for the surface charge on the boundary between two half-spaces in the original system. There is no direct way to determine this charge and take its contribution to the
total electric field into account in the original system, and that is why the equivalent
(homogenized) system, with the boundary removed, is generally much simpler for
analysis.
After the current density vector,
J, in the earth for a given current I of a grounddetermined by analyzing the equivalent system in Fig. 3.25(b), all
other quantities of interest for a specific application can be found - in the original
system in Fig. 3.25(a). The electric field intensity vector in the ground and on its surface is E = i/a. An important parameter of a grounding electrode is its grounding
ing electrode
is
(b)
- image theory
steady currents.
for
164
Chapter
3
Steady
Electric
Currents
resistance, defined as
R gr —
grounding resistance
^gr
(3.191)
1
’
where Egr is the potential of the electrode with respect
infinity. According to Eq. (1.74), this potential is given by
to the reference point at
infinity
potential of a grounding
V,gr
-I
E
•
(3.192)
dl.
electrode
electrode (reference point at
infinity)
By
integrating
E
along the
field lines
on the earth’s surface between two points
that are a distance equal to an average person’s step apart,
we
get the so-called
voltage of a step, Estep, around the grounding electrode. This voltage corresponds to
the potential difference between the feet of a person walking in the direction along
the field lines on the earth’s surface.
above a grounding electrode
is
current intensity of the electrode
even
fatal
The maximum voltage of
a step in the region
a parameter important for applications
may
where
cause exceedingly large values of
a large
Estep
,
with
consequences.
Example 3.15
Hemispherical Grounding Electrode
A hemispherical grounding electrode of radius a = m is buried in
1
the earth of conductivity
— 10 -3 S/m, with its base up, as shown in Fig. 3.26(a). The current of the electrode is
I = 1000 A. Find (a) the grounding resistance of the electrode and (b) the voltage of a
step between points on the earth’s surface that are r i=2m and r2 = r\ + b away from the
electrode center, where b = 0.75 m (average person's step).
a
Solution
(a)
Using the image theory for steady currents, Fig. 3.25, we get the equivalent system
in Fig. 3.26(b), which consists of two hemispherical electrodes pressed onto each
other, forming thus a spherical electrode with current 2/ flowing into a homogeneous
medium of conductivity a. Due to spherical symmetry, the current density vector, J, in
the medium is radial. Consequently, J in the original system is also radial and of the form
given by Eq. (3.46). Applying the continuity equation to a spherical surface S of radius r
shown
Figure 3.26
(a)
Hemispherical
grounding electrode and
(b)
equivalent spherical electrode
in a
homogeneous medium;
Example
3.1 5.
for
(a)
(b)
Section 3.13
165
Grounding Electrodes
positioned around the electrode gives
J(r)
=
~
(a<r<oo).
(3.193)
2tt^
4tt7^
that the same result is obtained by applying the continuity equation to a hemisphersurface of radius r in the original system - in Fig. 3.26(a).
Note
ical
Combining Eqs.
(3.18)
and
(3.193), the electric field intensity in the earth
£ (r =
)
With the use of Eqs.
(3.192)
and
the reference point at infinity and
V* =
(b)
r°°
L
The voltage of a
Eir
step
is
- = Anam
a
(3-194)
2
(3.191), the potential of the electrode with respect to
its
grounding resistance come out to be
_ v*_
—
R gr —
A
/
=^Ta
1
=
159 Q.
grounding resistance of a
hemispherical electrode
as large as
lb
= 21.7 kV
= r2 — r\).
(b
2nar\r2
Two
(3.195)
2Jioa
Edr =
Example 3.16
is
(3.196)
voltage of a step; b
Hemispherical Grounding Electrodes
emf £ = 100 V,
two hemispherical grounding electrodes. The radius of
each electrode is a = 2 m, the conductivity of the earth is a — 10 mS/m, and the distance
between the electrode centers is d = 100 m. Compute the current in the circuit.
The
electric circuit
shown in Fig.
3.27 consists of an ideal voltage generator of
a wire of negligible resistance, and
Solution
Since
d^>
a,
the potentials with respect to the reference point at infinity of the
individual electrodes can be evaluated independently
isolated grounding electrode.
fore 2 7? gr
is
,
The equivalent
where the expression
for /? gr
is
from each other, that
is,
as for a single
resistance seen by the generator equals there-
given in Eqs. (3.195).
As
the resistance of the wire
negligible, the current in the circuit is
1
Example 3.17
=
= naa£ =
6.28
A.
Hemispherical Grounding Electrode
(3.197)
in a
Two-Layer Earth
Conductivity of a layer of earth around a hemispherical grounding electrode shown in
Fig. 3.28 is o\
=
5 x 10
-3
S/m and
that of the rest of the earth
a2
=
10
-3
S/m. The
of the electrode and the boundary surface between the earth layers are a
b
=
2 m, respectively.
The current
is
average person 's step
intensity of the electrode
is
I
=
= lm
radii
and
100 A. Determine (a) the
grounding resistance of the electrode and (b) the power of Joule’s losses
in the earth.
+ £
/
Figure 3.27
cr
Two
hemispherical
grounding electrodes in a dc
circuit; for Example 3.1 6.
an
166
Chapter 3
Steady
Electric
Currents
Figure 3.28 Hemispherical
grounding electrode
in
an earth
v
two concentric
Example 3.1 7.
consisting of
layers; for
Solution
(a)
Because of symmetry, the current flow
both layers of earth
in
is
radial.
Applying the
we obtain the same expression for the current density J in the earth as in Eq. (3.193). The electric field intensity
vector in the earth is Ei = i/a\ for a < r < b and E 2 = i /02 for b < r < 00 By means of
continuity equation to the closed surface
S shown
in Fig. 3.28,
.
Eq. (3.192), the potential of the electrode with respect to the reference point at
is given by
rb
V,
&
Edr =
-F
roo
Ei dr
+
1
(
Using Eq.
(3.32),
—
^
resistance, Eq. (3.191),
2nb
(b)
=
Jb
Ja
from which the grounding
£2 dr
/
and combining
b-a
\ cj\a
j
(3.198)
,o\
is
j_
95.5
(3.199)
fi.
02
with Eq. (3.193), the
it
infinity
power of
Joule’s losses in the
earth turns out to be
Pi
f JEdv
(°°
=
where dv
Fig. 3.28).
power
is
the
—
-=
Jv
2nr2
Ja
volume of a hemispherical
Noting that the
Ej
Edr,
I /
(3.200)
J,
r
shell of radius r
last integral in this
of joule's losses in the
E2nr2 dr =
and thickness dr (portrayed in
Tgr [Eq. (3.198)], we have
equation equals
= Tgr 7 = E gr 7 2
(3.201)
,
earth surrounding a
and Ej
grounding electrode
=
955
Example 3.18
kW
for given numerical data.
Spherical
17
Grounding Electrode Deep
in
the Earth
A spherical grounding electrode of radius a = 30 cm is buried in a homogeneous earth of cona — 10 -2 S/m such that its center is at a depth d = 6 m with respect to the ground
ductivity
surface, as
and
its
maximum
l7
shown
in Fig. 3.29(a).
There
is
a steady current flowing
potential with respect to the reference point at infinity
tangential
component of the
Although derived for a
specific
resistance Rot
and current
/.
Note
electric field intensity vector
grounding electrode, that
of Joule’s losses in the earth in Eq. (3.201)
that this
is l/gr
is
is
in Fig. 3.28,
through the electrode,
=
15 kV. Calculate the
on the earth’s surface.
the expression for the
power
true for an arbitrary grounding electrode, with grounding
actually an expression of Joule’s law, Eq. (3.77).
3
Section 3.1 3
Grounding Electrodes
Figure 3.29
167
(a) Spherical
grounding electrode buried
deep in the earth and (b)
equivalent system - by virtue
of the image theory for steady
currents; for Example 3.1 8.
Solution
Since
d^>
a,
the influence of the boundary with air
meaning
of the electrode can be neglected,
in the infinite
that
R gr
homogeneous medium of conductivity
on the grounding resistance
can be evaluated as for an electrode
a. Invoking the duality relationship in
Eq. (3.102), we can further relate R gr to the capacitance (C) of the same metallic sphere
situated in an infinite dielectric medium of permittivity s. Eq. (2.121) gives this capacitance
for s
= eo, so that
f? gr
£p
—
=26.5
oC 4 noa
Hence, the current of the electrode is I — Vgx /R gx =
(3.202)
£2.
spherical electrode buried
deep
565.5 A.
and the superposition principle,
the resultant current density vector at a point
in Fig. 3.29(b), the position of which is
determined by a radial distance r from the projection of the electrode centers on the plane of
symmetry (point O), can be obtained as
Applying the image theory for steady currents
(Fig. 3.25)
M
J
where J originai and
+ Jimagei
^original
3.2 33
due to the current flow from
Jimage are the current density vectors
Making use of the
vidual electrodes.
—
fact that
d^>
indi-
these current densities can be evaluated
a,
independently from each other, which results in
Jo
'original
(cos a
=
The
r/R and
— J«
=
•'image —
R=
tangential
yjr 2
+ d2
J
^2
^
2/original
COS a
—
(3.204)
2 R3
).
component of the
the original system, Fig. 3.29(a),
electric field intensity vector
on the
earth’s surface in
is
_
1
From
—
^
J
Ir
_
2na(r2
a
the condition
dE
t
(r)
+ J2
)
(3.205)
3 /2
= 0,
(3.206)
dr
we
£ is maximum on
maximum field is
get that the field intensity
centered at the point O. This
t
(Ft)niax
—
Ft(r ma x)
Note that the maximum voltage of a
is
a circle of radius r max
— /V
9nad2
step, taking
=
96 V/m.
= d/sfl =
4.24 m,
(3.207)
0.75
m as an average person’s step,
= 72V,
(3.208)
b
=
approximately
(Fstep) max
^(F
t
grounding resistance of a
)m ax b
in
the earth
168
Chapter
Steady
3
Currents
Electric
which
much
is
less
than the voltage found
because the electrode
in Fig. 3.29(a) is
in
Eq. (3.196), for instance. This
buried deep
Problems'. 3.23-3.29; Conceptual Questions (on
MATLAB
Exercises (on
is
expected,
in the earth.
Companion
Website): 3.27-3.30;
Companion Website).
Problems
3.1.
Charge density
3.5.
ductor. Consider a hollow steel-reinforced alu-
minum wire conductor of length / =
ity
o
coordinates,
=
does not, e
in the
region
density
/ const,
=
(e/<r)Va
of the conductor
is
a
= 5 mm, the thickness
= 1.1 MS/m) rein-
of the stainless-steel (aste ei
W.
•
forcement
is
b
—
a
=
5
mm, and
Various computations for a copper wire with
conductor, whose overall radius
steady current. For a copper wire with a steady
aluminum
Example
density of the
power
power of ohmic
material, (e) the total
the wire,
(f)
wire
room
temperature, and (g) the wire resistance at the
_1
temperature of 100°C (ac u = 0.0039 K ).
A
an
with
capacitor
Parallel-plate
dielectric.
homogeneous imperfect
has
find
(a)
is
the resis-
the ratio of
—
10
5
S/m and
02
=4
x 10 5 S/m. The voltis V = 50 V.
age between the resistor terminals
Find the electric
a
field intensity,
current density,
current intensity, and power of Joule’s losses in
dielectric of permit-
each of the two cuboids if they are connected
as in (a) Fig. 3.30(a) and (b) Fig. 3.30(b).
e
invariant),
is
= 2 cm,
is
Resistor with two cuboidal parts. A resistor is
formed from two rectangular cuboids of the
same size, with sides a = 8 mm, b = 2 mm, and
c — 4 mm. The cuboids are made out from dif-
<j\
and conductivity o. The separation
between plates is d and the plate area is S.
If the voltage between the plates is V (timetivity
c
is
MS/m). What
ferent resistive materials, with conductivities
imperfect
capacitor
parallel-plate
35
ends of the wire?
3.6.
losses in
the resistance of the wire at
=
the rest of the
current intensities lA\/htee\ in th e two materials for a given dc voltage drop between the two
the electric field intensity in the wire, (c) the
of Joule’s losses in the
(ctai
tance of this wire, and what
compute: (a)
the number of electrons that pass through a
cross section of the wire during one hour, (b)
3.1,
voltage between the wire ends, (d) the volume
100 m. The
radius of the cylindrical hole in the central part
const. If the electric potential
given by p
is
whereas permittiv-
V, show that the volume charge
is
current described in
3.3.
Hollow steel-reinforced aluminum wire con-
conductivity of the material varies with spatial
3.2.
terms of the conductivity
in
gradient. In a region with steady currents, the
current distribution
the
in the dielectric, (b) the
current through the
capacitor terminals, (c) the conductance of
the capacitor, (d) the
in
power of
Joule’s losses
the capacitor, (e) the free charge distribu-
tion in the capacitor,
and
(f)
the
bound charge
distribution in the capacitor.
3.4.
Spherical capacitor half
ing liquid.
A
filled
with a conductFigure 3.30
spherical capacitor has a poorly
conducting liquid dielectric occupying a half of
the space between the electrodes.
electrodes are a and b (a
ductivity of the dielectric
conductance of
<
is
The
this capacitor.
rectangular cuboids
parallel; for
Problem
connected
made out from
in (a) series
and
(b)
3.6.
radii of
and the cona. Determine the
b),
Two
different resistive materials
3.7.
Conductivity gradient along the resistor current. Fig. 3.31 shows a right-angled cuboidal
resistor of
dimensions
a, b,
and
c,
made
out
169
Problems
from a continuously inhomogeneous
given by the following function
coordinate: o(x)
where
cto is
=
cr0 /(l
+ 9x/a)
layer near the inner electrode are s x \
-4
4 x 10
and ctj
S/m, and those of
is
=
of the x-
<x <
(0
and conductivity of the
relative permittivity
resistive
The conductivity of the material
material.
The capacitor
a constant. Calculate the resistance
V=
voltage
of this resistor.
12
the
and CT2 = 5x 1CT 6 S/m.
connected to a time-invariant
=7
other layer e r2
a ),
=
is
100 V. (a) Find the current
distri-
bution in the dielectric. Using this result and
integration (field-theory approach),
compute
conductance of the capacitor, (c) the
power of Joule’s losses in each of the layers, and
(d) the free charge on each of the boundaries
of radii a b and c.
(b) the
,
3 11
.
.
,
Solution using equivalent circuit with ideal ele-
ments. Repeat parts (b)-(d) of the previous
problem but using the equivalent circuit with
two ideal capacitors and two ideal resistors in
Figure 3.31 Right-angled cuboidal resistor
made out from
inhomogeneous
Problems
3.8.
3.
Fig. 3.18(b) (circuit-theory
that the conductivity of the
inhomogeneous
imperfect
shows a parallel-plate
capacitor with circular plates of radius a and a
continuously inhomogeneous imperfect dielectric. The permittivity and conductivity of the
function of the
dielectric are the following functions of the
3 . 12
resistive material; for
Assume
material in Fig.
3.31
y-coordinate, a(y)
y <
b,
cto
= const),
a
is
= cto[1 + 9sin(7ry/h)]
and
(0
<
find the resistance of
Continuously
3.9. Integrated-circuit
conductivity
with
resistor
An
profile.
exponential
integrated-circuit
by diffusing a layer of ptype impurity into an rc-type background material. As a consequence of the diffusion process,
the concentration of impurity is not uniform
(IC) resistor
is
built
across the treated region, so that the obtained
3.32
Fig.
and ct(z) =
where cto is a constant
the separation between the plates. The
z -coordinate: e(z)
—
2(1
T 3 z/d), 0 <
z
<
ct0 /
(1
and d
the resistor.
IC
.
dielectric.
7-3. 9.
Conductivity gradient normal to the resistor
current.
approach).
a continuously
is
capacitor
is
+ 3z/d)so
d,
connected to a time-constant
volt-
age V. Find (a) the current distribution in the
dielectric, (b) the
(c)
the
conductance of the capacitor,
power of Joule’s
losses in the capacitor,
(d) the free charge distribution in the capacitor,
and
(e) the
bound charge
distribution in
the capacitor.
can essentially be represented by
resistor
the structure in Fig. 3.31 with the conductivity
varying as cr=a(y). In specific, a(y) exponentially decreases
from
air-resistor surface to 02
=
o\
=
0.1
100 S/m at the
S/m at the inter-
face with the n-type background, which
is
con-
sidered to be nonconducting, and the length,
Figure 3.32
and width of the p-type region are
a = 4 p m, b — 1 pm, and c — 2 p,m, respectively. Determine the resistance of the IC
Parallel-plate capacitor
thickness,
with a continuously
inhomogeneous
resistor.
3.10. Spherical capacitor with
layers.
The
two
= 5 cm
and c
=
15
cm
is
of two concentric imperfect dielec-
where the radius of the boundary
surface between the layers is b — 10 cm. The
tric layers,
Problem
3.1 2.
with
two
lossy dielectric
dielectric of a spherical capacitor
with electrode radii a
composed
lossy
dielectric; for
3.13. Parallel-plate
dielectrics.
capacitor
homogeneous dielectric
The permittivities of the
are sj and £ 2 Both parts
tor with a piece-wise
shown
in Fig. 3.33.
dielectric
lossy
Consider a parallel-plate capaci-
parts
.
1
70
Chapter 3
Steady
Electric
Currents
or
2 The
d and the
are lossy, with conductivities o\ and
separation between the plates
is
cable
-
is /
of the
cross-sectional areas of the dielectric parts are
100 m. The relative permittivities
are
—
=
eri
conductivities
the
and Si The voltage between the electrodes
is V. By employing a field-theory approach
(based on determining the current and field
=
layers
are
=
4 and e T 2
o\
=
10
-12
8,
and
S/m and
12
5 x 10~
distribution in the dielectric), calculate (a) the
S/m. The cable is connected at
an ideal voltage generator of timeinvariant emf £ = 30 V, and the other end of
the cable is terminated in a load of resistance
current through the capacitor terminals, (b) the
R]_
conductance of the capacitor, (c) the power of
Joule’s losses in each of the dielectric parts, and
(d) the free charge distribution in the capacitor.
in the dielectric, (b) the
Si
(72
.
one end
=
to
Compute
1
(a) the current density
conductance per unit
length of the cable, (c) the current intensity
along the conductors, (d) the current density in
each of the conductors, (e) the power of Joule’s
losses in the cable and in the load, (f) the power
of the generator, and (g) the free surface charge
Figure 3.33
density on the boundary between the dielectric
Parallel-plate
layers.
capacitor with
two imperfect
£
dielectric
parts; for
Problem
c
+
d
Rl
b
3.1 3.
1ft.
3.14.
Equivalent circuit with ideal capacitors and
ei,cr\
Repeat the previous problem but
employing a circuit-theory approach, i.e., generating and solving an equivalent circuit with
ideal capacitors and resistors.
e2
resistors.
,
o- 2
/
Figure 3.34 Coaxial cable with two coaxial imperfect
3.15.
Current distribution through circular plates.
Find
the
distribution
of
dielectric layers in a
the upper and lower plate of the parallelplate capacitor with imperfect dielectric
fect dielectric.
conductors
Current distribution through a thin spherical
Consider the spherical capacitor
Example
3.3,
and assume that the inner electrode is hollow,
with a very thin wall. With this assumption,
determine the distribution of current through
the inner electrode. Represent this current
using the surface current density vector.
two lossy dielectric layers.
shows a longitudinal cross section of a
coaxial cable with two coaxial layers of imperFig. 3.34
dielectric.
conducting,
c
=
the
tric
10
The conductors
and
their
mm, and d =
12
radii
are perfectly
are
mm. The
a
=
2
3
mm,
radius
of
boundary surface between the dieleclayers is b = 7 mm and the length of the
A
coaxial cable with perfect
filled
is
with an inhomogeneous
=
£q(1
+
/a 2 )
charges
3.19.
in the dielectric.
Coaxial cable partly
tric.
3.17. Coaxial cable with
fect
3.1 7.
and ad = cro/(l 4- r2 /a 2 ) (a < r < b),
where a and b are the radius of the inner
cable conductor and the inner radius of the
outer conductor, and oq is a constant. The voltage between the cable conductors is V (dc).
Calculate (a) the conductance per unit length
of the cable and (b) the density of volume free
r
vector.
with imperfect dielectric from
Problem
imperfect dielectric of parameters e
be described using the surface current density
electrode.
for
an inhomogeneous imper-
3.18. Coaxial cable with
from
Problem 3.12. Assume that the plates are thin
enough so that the current through them can
3.16.
dc regime;
through
current
is
A
filled
with a lossy dielec-
coaxial cable with perfect conductors
partly filled with an imperfect dielectric of
conductivity
<
7
,
as
shown
ductor radii are a and
in Fig. 3.35.
b, the
The con-
cable length
is /,
and the length of the air-filled part of the cable
is c. The cable is fed by an ideal dc voltage generator of emf £, and the other end of the cable
is
open. Find the expression for the current
intensity through the cable conductors.
171
Problems
the
c
and
strips,
(c)
volume
free charge in the
dielectric.
<j
b
da
l
filled with a condc regime; for Problem 3.19.
Figure 3.35 Coaxial cable partly
ducting dielectric
3.20.
in a
Coaxial cable with a poorly conducting spacer.
Shown
in Fig. 3.36
air-filled
is
line
with a continuously inhomogeneous
imperfect dielectric
in a
dc regime;
a cross section of an
coaxial cable with a spacer
the conductors.
Figure 3.37 Planar
The spacer
is
between
made
3.23.
Grounding electrode
earth.
out of
for
in
Problem 3.22.
an inhomogeneous
Conductivity of the earth around a
hemispherical grounding electrode shown in
an imperfect dielectric of conductivity a and
its cross section is defined by an angle a.
The conductor radii are a and b. What is the
conductance per unit length of this cable?
Fig. 3.38
can be described as the following func-
tion of the radial coordinate
(a
<
r
<
oo),
trode and
where a
cto is
is
a constant.
of the electrode
o(r)
= o^sfajr
The current
Find
is I.
r.
the radius of the elec(a) the
intensity
grounding
resistance of the electrode, (b) the total
of Joule’s losses in the earth,
and
(c)
the
power
power
of Joule’s losses in a layer with thickness a
around the electrode
Figure 3.36 Cross
(a
<
r
<
2d).
section of a coaxial
cable with a poorly
conducting spacer
between conductors;
Problem 3.20.
Figure 3.38
for
Hemispherical
grounding electrode
3.21.
Planar line with imperfect conductors. If the
planar transmission line in Fig. 3.24
in a
are not perfectly conducting, but have a finite
earth; for
conductivity ac find the dc resistance per unit
Problem 3.23.
strips of the
,
length of the
each of the
3.22.
Planar
line
dielectric.
line.
Assume
strips is
with
that the thickness of
t.
3.24.
Grounding electrode in a two-sector earth.
The earth around a hemispherical grounding
electrode of radius a = 1 m can be represented as two sectors with conductivities
10~ 3 S/m and 02 = 8 x 10~ 3 S/m, as
cti = 2 x
shown in Fig. 3.39. The current intensity of the
electrode amounts to 1 = 50 A. Compute (a)
the grounding resistance of the electrode and
(b) the power of Joule’s losses in each of the
two earth sectors.
3.25.
Two
an inhomogeneous lossy
Consider a planar (two-strip) trans-
mission line shown in Fig. 3.37. The strips
are perfectly conducting, very thin,
w
wide,
and d apart from each other. The
dielectric between the strips is imperfect and
continuously inhomogeneous, with parameters
e(x) = (4 + 3x/d)£Q and a(x) = cro/(l + 9 x/d),
0 < x < d, where cro is a constant. The line is
connected to an ideal dc voltage generator of
emf £, and is open-circuited at the other end.
Calculate the distributions of (a) volume cur/
continuously
inhomogeneous
long,
rent in the dielectric, (b) surface current over
short-circuited grounding electrodes.
Two
grounding electrodes
of radii a are galvanically connected to each
other by a wire of negligible resistance, as
identical hemispherical
172
Chapter
3
Steady
Electric
Currents
3.27.
Two deep
spherical grounding electrodes.
Consider two identical spherical grounding
homogeneous
Both the distance
between the sphere centers and their depth
with respect to the ground surface equal d,
where d^> a. What is the resistance between
the two electrodes?
metallic electrodes of radii a in a
earth of conductivity a.
Figure 3.39
Hemispherical
grounding electrode
in
a two-sector earth;
for
shown
earth
is
trode
Fig.
in
Problem 3.24.
The conductivity of
3.40.
the
3.28. Cylindrical
earth.
o and the distance between the elecDetermine the
is d ( d ~S> a).
radius a
centers
grounding resistance of such
a
system
A
10
electrodes.
d
20
cm and
-2
=
S/m such
—
/
=
5
m
is
buried
4
m
200
=
depth
below the ground surface, as depicted
that
in Fig. 3.42. If a
I
length
homogeneous earth of conductivity a
in a
of
=
grounding electrode deep in the
grounding electrode of
cylindrical
its
axis
is
at
a
steady current of intensity
A flows through the electrode, find the
maximum
voltage of a step on the earth’s sur-
—
m as an
face.
Adopt b
step.
Neglect end effects due to the
0.75
average person’s
finite
length
of the cylinder.
Two
Figure 3.40
electrodes; for
3.26
short-circuited hemispherical
Two grounding
The
grounding
Problem 3.25.
electrodes in a layered earth.
electric circuit
shown
in Fig. 3.41 consists
of an ideal current generator of current intensity /
=
100 A, a wire of negligible resistance,
and two
identical
hemispherical
—
electrodes of radii a
2 m.
grounding
The conductivity
Figure 3.42 Cylindrical grounding electrode buried deep
of the earth near the electrodes, within the
radius b
=
4 m,
is
o\
=
=
10~ 2 S/m. The conduc-3
elsewhere is 02
10
S/m. The distance
between the electrode centers is d = 80 m.
What is the voltage between the electrodes?
tivity
Figure 3.41
Two
hemispherical grounding electrodes
layered earth; for Problem 3.26.
in a
in
the earth; for Problem 3.28.
3.29
Two deep
parallel cylindrical electrodes.
Two
grounding electrodes as
the one in Fig. 3.42 are placed parallel to each
other (at the same depth with respect to the
earth’s surface) so that the distance between
their axes is D = 6 m. If they are connected
together in a circuit with an ideal current generator as in Fig. 3.41 (with I — 200 A), compute
(for the numerical data given in the previous
problem) the voltage between the electrodes.
identical cylindrical
4
Magnetostatic Field
in Free Space
[
Introduction:
n our studies of steady electric currents in
preceding chapter, we introduced and discussed physical laws and mathematical techniques
for determining the distribution of currents for
given geometry, material properties, and excitation
I the
from the experimental law of the magbetween two point charges that move
free space (a vacuum or air) - Coulomb’s law
Starting
netic force
in
for the magnetic field,
we
shall derive the Biot-
Savart law, which, in turn, will serve as a starting
of various structures.
We now introduce a series of
new phenomena associated with steady electric cur-
point for the derivation of Ampere’s law. Both
which are essentially the consequence of a
new simple experimental fact - that conductors with
netic field
rents,
currents exert forces on one another. These forces
are called magnetic forces, and the field due to one
current conductor in which the other conductor
is
and which causes the force on it is called
the magnetic field. Any motion of electric charges
and any electric current are followed by the magnetic field. The magnetic field due to steady electric
currents is termed the steady (static) magnetic field
or magnetostatic field. The theory of the magnetosituated
static field, the magnetostatics, restricted to a vac-
uum and nonmagnetic media
chapter.
is
the subject of this
The magnetic materials
in the following chapter.
will
be discussed
laws represent a means for evaluating the mag-
due to given steady-current distributions,
and we shall apply them to many theoretically
and practically important configurations that do not
include magnetic materials. The differential form
of Ampere’s law will introduce a new differential
operator, the curl. We shall derive the law of conservation of magnetic flux (Gauss’ law for the magnetic field), and complete the full set of Maxwell’s
equations for the magnetostatic field in nonmagnetic media.
The magnetic vector
potential will be
introduced as a counterpart of the electric scalar
potential and the magnetic dipole as the equivalent of the electric dipole.
Examples involving
evaluation of magnetic forces and torques will also
be presented.
173
174
Chapter 4
Magnetostatic
Field in Free
Space
MAGNETIC FORCE AND MAGNETIC FLUX DENSITY
VECTOR
4.1
The magnetic
around moving
fundamental property
is that there is a force acting on any electric charge placed in the space, provided that
the charge is moving. That wires carrying electric currents produce magnetic fields
was first discovered by Oersted in 1820. To quantitatively describe this field, we
introduce a vector quantity called the magnetic flux density vector, B. It is defined
through the force on a small probe point charge Q p moving at a velocity v in the
field, which equals the cross product of vectors Q \ and B,
p
field
is
a special physical state existing in a space
electric charges and, thus, electric currents in conductors. Its
1
definition of
B
(unit:
Fm
T)
This force
is
= Qp \
x
called the magnetic force.
B
The
(<3 p
unit for
note that the magnetic flux density vector, which
induction vector,
is
->
is
(4.1)
0).
B
is
tesla (abbreviated T).
We
also referred to as the magnetic
defined analogously to the electric
field intensity vector,
E, in
electrostatics [Eq. (1.23)].
The
basis for determining the magnetic flux density vector due to steady curan experimental law describing the magnetic force between two point
charges that move in a vacuum. This law represents the equivalent of Coulomb’s
law, Eq. (1.1), of electrostatics, and is an underpinning for the entire magnetostatic
theory. With reference to Fig. 4.1, it states that the magnetic force on a point charge
Q 2 that moves at a velocity \2 in the magnetic field due to a point charge Q\ moving
with a velocity vj in a vacuum (or air) is given by
rents
is
magnetic force
Fml2
=
Ho
Q2V2 X (Q1V1 X
the permeability of a
permeability of a
vacuum
vacuum
is
the
is
(4.2)
same
as for
Coulomb’s law
in Fig. 1.1,
and
/xo is
(free space),
/xo
(H
)
R2
47r
where the notation of vectors
R 12
= 47t
x 10
7
H/m
(4.3)
henry, the unit for inductance, which will be studied in a later chapter).
mentioned, is the result of experiments, but it can also be derived
from Coulomb’s law using the special theory of relativity. Note that it is sometimes
referred to as Ampere’s law of force.
Although of the similar form, the magnetic force law in Eq. (4.2) is mathematically more complicated than Coulomb’s law, because two vector cross products are
implied in the equation. We first need to determine the vector resulting from the
Eq.
(4.2), as
cross product Q\\\ x R[2, and then the cross product of (?2 V 2 and that vector [and,
2
of course, multiply the result by hq/(At: R )]. Let us apply this rule to the situation
shown
Figure 4.1 Magnetic force
between two moving point
in a vacuum, given
by Eq. (4.2).
in Fig. 4.2,
mal to the
where vectors Q\v\ and 02 v 2 are parallel to each other and northem. We see that the force between the charges is attractive
line joining
charges
'The cross product (also referred to as the vector product) of vectors a and b. a x b, is a vector whose
is given by |a x b| = |a| |b| sin or, where a is the angle between the two vectors in the product. It
is perpendicular to the plane defined by the vectors a and b, and its direction (orientation) is determined
by the right-hand rule when the first vector (a) is rotated by the shortest route toward the second vector
(b). In this rule, the direction of rotation is defined by the fingers of the right hand when the thumb points
magnitude
in
the direction of the cross product.
175
Magnetic Force and Magnetic Flux Density Vector
Section 4.1
HISTORICAL ASIDE
Hans
Christian Oersted
(1777-1851),
physicist,
a
was a professor
and chemthe
at
istry
University
of Copenhagen.
As
of a classroom
demon-
part
a wire with a current is also deflected in a
magnetic field (by the magnetic force), thus laying the foundation for the construction of the
electric motor. His epoch-making discovery of
the magnetic effects of electric currents was
announced to the French Academy of Sciences
on September 4, 1820. It immediately set off
an explosion of research activity by many bril-
he discovered,
apparently by accident,
that an electric current
flowing through a wire
produces a magnetic field
around the wire. Namely, while preparing an
evening lecture on electricity and magnetism in
April of 1820, in which he wanted to demonstration,
a wire connected
Ampere
(1775-1836), Biot
(1774-1862), and Savart (1791-1841). Oersted’s
experiment was the
between the terminals of
first
demonstration of a con-
a voltaic source (battery), Oersted noticed that
nection between electricity and magnetism, and
the magnetic needle of a compass sitting next
considered the foundation of the modern study
to the wire spun off of the north position every
of electromagnetism. In 1829, Oersted
time the battery was in use.
the
was generally
believed at that time that electricity and magnetism were not related. Therefore, he was
extremely surprised and excited with what he
if
minds, including
liant
strate the heating effects of the electric current
in
continued experimenting during that
which eventually led him to the conclusion that an electric current creates a magnetic
field -and electromagnetism was born. He also
discovered that, not only is a magnetic needle deflected by an electric current, but that
lecture,
physics
of
He
saw.
Danish
the vectors are in the
opposite directions
same
It
direction [Fig. 4.2(a)],
This
[Fig. 4.2(b)].
is
first
is
became
director of the Polytechnic Institute in
Copenhagen, now the Technical University of
Denmark. (Portrait: AIP Emilio Segre Visual Archives,
Brittle
Books
and repulsive
Collection)
they are in
if
formally just opposite to Coulomb’s law,
where like charges repel and unlike attract each other.
Combining Eqs. (4.1) and (4.2), and assuming that the second charge in Fig. 4.1
is a probe charge ( Q 2 = Q ), we can identify the expression for the magnetic flux
p
density vector of a point charge Q moving with a velocity v:
fio
4 rt
Qx x R
R2
where R is the distance from the charge and R
from the source point toward the field point,
perpendicular to
R and
P0
Qx
R 12
ml2
B due
charge
the unit vector along
is
as
shown
in Fig. 4.3.
perpendicular to the plane of the vectors
R
to
a moving point
in free
space
directed
Vector
Q\ and
B
R.
is
Its
P0
,V 2
Vl
(4.4)
f
Qx
<22
_
Q\\x x R12
-
Rli
®
R
1
X
F ml2
Figure 4.2
Two
charges
moving parallel to each other
in the same direction (a) and in
R 12
opposite directions (b)
(a)
(b)
vacuum.
in a
1
76
Chapter 4
Magnetostatic
Field in Free
comes from the
orientation
B
Figure 4.3 Magnetic flux
difference
density vector due to a point
in a
definition of the cross product of vectors,
i.e.,
it
is
vacuum.
The
the direction of the field vector with respect to the source.
E, due to
field intensity vector,
magnetic
flux density vector, B,
other words, the lines of
E
Q
is
radial with respect to
due to
Qy is circular with
Q
electric
whereas the
(Fig. 1.7),
respect to <2v (Fig. 4.3). In
are radials starting at Q, while the lines of
B
are circles
centered on the line containing the vector Q\.
Conceptual Questions (on Companion Website):
4.1
and
4.2.
HISTORICAL ASIDE
Nikola
ant
American
was
inventor,
for-
of
He made
in
(ac)
and had about
Croatia (then in the Habsburg Empire of Austria),
He
area
of polyphase
known
U.S.
alternating
patents
currents
engineer
250 patents issued in the
U.S. and other countries.
Tesla was born in Smiljan
near Gospic, Lika, in
Serbian family.
the
and polyphase systems, including ac generators,
alternating current
practical
in
motors, and transformers, as well as principles
no complete
mal education.
obtained most of his best
brilli-
a
electrical
with
(1856-
Tesla
an
1943),
in a
studied mechanical engi-
power
distribution using alternating currents,
1887 and 1888, and delivered his famous lecture “A New System of Alternate Current Motors
and Transformers” before the American Institute
Engineers (now IEEE) on May
June of 1888, George Westinghouse
(1846-1914), head of the “Westinghouse Electric
of Electrical
16, 1888. In
Company”
in Pittsburgh,
bought the
first
seven
of Tesla’s patents for polyphase systems. This
marked
the beginning of the five-year “war of
Polytechnic School in Graz from
currents” to decide whether Edison’s existing dc
1875 to 1878 and then natural philosophy at the
University of Prague until 1880, but never gradu-
systems or newly proposed Tesla- Westinghouse ac
systems would be the chosen technology for the
ated due to a shortage of funds after his father’s
global electric
death. While working as electrical engineer for
in
he came up with
an idea of the principle of the rotating magnetic
field and polyphase alternating currents producing
it in 1882 and constructed the world’s first induc-
nant transformer
neering
at the
companies
tion (ac)
hours.
Budapest and
Paris,
motor
He
four cents
for
in
in 1883, both during after-work
emigrated to America in 1884, with
in
his pocket,
Thomas Edison
and worked
(1847-1931).
He
get Edison interested in the induction
for a year
failed
Company”
in
New
York
to
motor and
alternating current, and later established his
“Tesla Electric
is
determined by the right-hand rule in rotating Q\ by the shortest route to R.
Note that the magnetic flux density is proportional to the product Q\, which
can be regarded as a measure of sources of the magnetostatic field (the measure
of sources of the electrostatic field is a charge Q ). Comparing Eq. (4.4) to the
expression for the electric field intensity vector due to a point charge, Eq. (1.24),
we observe the same dependence on the amount of sources (£9|v| and Q ) and on
distance R. The constant no corresponds to the constant 1/eo- The only formal
Q
charge moving
Space
own
City. Tesla
power
distribution of the future
America. Tesla invented
transformer)
erating
in
known
1891, which
alternating
an
air-core
reso-
as a Tesla coil (or Tesla
currents
was capable of genof extremely
high
frequencies for that time (up to a hundred of thou-
sands of cycles per second or 100
notation).
The
coil
kHz
in
today’s
could also build up tremen-
dously high voltages and cause spectacular spark
discharges. Tesla’s
new
electric
lamps based on
a
high-frequency power supply were the forerunners
of our fluorescent tubes and neon signs. In a series
of spectacular lectures-demonstrations
in
America
Section 4.2
and Europe from 1891 to 1893 aimed at promoting
alternating current and high-frequency technology,
he was able to power electric lamps without wires,
either through air or by allowing high-frequency
electricity to flow through his body, produce artificial lightning flashes of different shapes, and even
shoot large lightning bolts from his coils to the
audience without harm. In early 1893, he proposed a complete radio system with a transmitter
and receiver in the form of Tesla coils tuned to
resonate at the same frequency as a basis for wireless communications. The “war of currents” ended
on the evening of May 1, 1893, with the opening of the Chicago World Exposition, spectacularly
illuminated by a hundred thousand lamps with
alternating current from Tesla’s generators. Only
couple of months later, Westinghouse was awarded
Tesla’s
is,
less telegraphy. In 1898, Tesla
world’s
first
it
demonstrated the
wireless remote control (of a boat
model) at Madison Square Garden. He continued his investigations of the wireless transmission
of power in 1899 and 1900 in Colorado Springs,
where he built a huge, 200-kW radio transmitter
consisting of a 142-foot metal mast with a spherical top and the world’s largest Tesla coil 51 feet
in diameter. Tesla also patented a bladeless disk
turbine, so-called Tesla turbine, in 1913, as well as
many other devices and systems in various
netic flux density,
in 1960. (Portrait:
©
16,
first
science and engineering.
Nikola Tesla
Museum, Belgrade,
Serbia)
1896. Tesla filed his basic radio
LAW
we have many point charges moving
is,
(/L
is
£v)
= Nin dv Qv d = Nv dv Q\ d = J dv,
in
charges in dv
is
computed
(4.5)
dv
We
the current density vector, defined by Eq. (3.3).
permeability everywhere 2
is /xq.
Hence, the magnetic
assume that the
due to the
flux density vector
as
dB =
/xq (J
An
dv) x
R
(4.6)
R2
which is known as the Biot-Savart law. We see that the product J dv represents a
macroscopic volume elemental source of the magnetic field, and we call it accordingly the
volume current element. By integrating Eq.
(4.6),
we
obtain the expression
2
We shall see in the next chapter that most materials we encounter in science and engineering are
nonmagnetic materials, that is, their permeability is practically that of a vacuum. Examples are commonly used metallic conductors such as copper, aluminum, silver, gold, etc., most frequently encountered
natural environments (air, water, and ground), biological tissues, and practically all insulators and semiconductors.
areas of
ac hydro-
BIOT-SAVART
where J
in fact,
was Guglielmo
Marconi (1874-1937) who put it into practical and
commercial use, and was awarded Nobel Prize
in Physics in 1909 for the development of wirethe invention of radio, although
in free space, the total magnetic flux density
by the principle of superposition, a vector sum of the magnetic flux density
vectors due to individual charges. In a current, a vast number of elementary free
charges moves through a conductor in an organized manner with the macroscopic
average velocity v d (drift velocity). The sum of products Q\ for all charges in an
elementary volume dv is given by
If
vector
177
power plant on Niagara Falls, based on
design, which was put into operation on
November
4.2
patent applications in 1897, and this
Law
The SI unit for the magthe tesla, was named in his honor
the contract to build the world’s
electric
Biot-Savart
Some materials, such as
much greater than /xq.
permeability
iron, steel, cobalt, nickel, ferrites, etc.,
on the other hand, have
Biot-Savart law
178
Chapter 4
Magnetostatic Field
in
Free Space
Figure 4.4 Magnetic flux density vector due to three characteristic current distributions - volume current
(b),
and
line
current
for the resultant
entire
volume
surface current
magnetic flux density vector due to a current distribution
an
in
v [Fig. 4.4(a)]:
Ho
Biot-Savart law for
(a),
(c).
C (J dv) x
volume
R
4 tv Jv
current
R
(4.7)
2
HISTORICAL ASIDE
experimental physics and acoustics, and a
Jean Baptiste Biot (1774-1862), a French physicist, was a student of Lagrange (1736-1813) at
the Ecole Polytechnique and a junior professor of mechanics under sponsorship of Laplace
(1749-1827) at the College de France. Biot’s
most important contributions to science are in
the theory of polarized light and effects
on
member
of the Paris Academy. His most important
work was
in vibration
and
sci-
and
particularly in the physics of the violin. Biot and
Savart showed experimentally in October of 1820,
soon after Oersted’s discovery of magnetic effects
entific
acoustics,
of electric currents, that the magnetic field pro-
it
duced by a current
by organic substances. Felix Savart (1791-1841),
another French physicist, was also a professor at the College de France, where he taught
in
a long, straight wire
is
inversely proportional to the distance from the
wire.
S [see Eqs. (3.12) and
dS
instead
of
/V
dv
in
Eq.
yielding
J s dS as a surface
(3.13)],
(4.5),
v
s
current element, where J s is the surface current density vector. The magnetic flux
In the case of a surface current flowing over a surface
we have
N
density vector due to a surface current distribution
A(o
Biot-Savart law for surface
f (J s dS) x
current
Finally, for a line current
along a
line
e.g.,
/,
thus [Fig. 4.4(b)]
R
(4.8)
R2
4 n Js
is
a current of intensity / flowing
through a (generally curvilinear) thin wire of length / and cross-sectional area 5,
dv = S d/, J = I/S [Eq. (3.5)], and ./ dv = JS d/ = / d /, where d / is an elemental segment along /. We conclude that the line current element equals / dl, where dl is
oriented in the reference direction of the current flow. For curved lines, dl is tangential to the line. The magnetic flux density vector due to a line current is hence
[Fig. 4.4(c)]
Wi
B_
Biot-Savart law for line
f / dl x
47T Ji
current
In
summary, we note
that Eqs. (4.7)-(4.9)
form: integral of /^(current element) x
R
R2
(4.9)
have the same generic mathematical
R/(4nR 2 ),
with the following three current
Section 4.3
179
Magnetic Flux Density Vector due to Given Current Distributions
Figure 4.5 Part of a planar
current loop
P in the
C and
same
a field point
plane.
elements as different source functions:
Jdv
—
«
J s dS
>
—
<
>
/dl,
(4.10)
volume, surface, and
line
A
m)
point
in
current elements (unit:
and they
all
represent one physical law - the Biot-Savart law.
In addition, the general form of the Biot-Savart law for line currents, Eq. (4.9),
can considerably be simplified in the case of a planar current contour (contour lying
in one plane) and a field point P (at which the magnetic flux density vector is calculated) in the same plane, shown in Fig. 4.5. Here, the vector B is normal to the plane
of the contour (the plane of drawing), and its magnitude is given by
S=
Because
(R dO
is
nt—*~
x R|
=
=
d/sina
d/cos/3
= MN = Rdd
(4.12)
the length of an arc representing a differentially small portion of a circle of
R
centered at the point P),
we have
B=
where 0
(4.11)
(Fig. 4.5)
|dl
radius
x R|
/|dl
X
Mo
4
is
the angle between
contour plane. The reference
the reference direction of
B
R
™
4n Jc
and an
B
R
arbitrarily
corresponds to the reference flow of
- field; thumb - current).
Note that the integral expression in Eq. (4.13) implies only scalar integration
and the function that needs to be integrated is much simpler than that in Eq. (4.9).
We shall, therefore, use the simplified form of the Biot-Savart law whenever possible, i.e., when dealing with planar current contours and calculating the magnetic
flux density vector in the same plane. Eq. (4.13) can also be used in cases when a
planar part of a nonplanar contour and a field point are in one plane.
(fingers
Conceptual Questions (on Companion Website):
4.3
for
a loop and
one plane
adopted reference axis in the
I, and
related to the direction of I by the right-hand rule
rise of 0
is
that
4.3.
MAGNETIC FLUX DENSITY VECTOR DUE TO GIVEN
CURRENT DISTRIBUTIONS
Eqs. (4.7)-(4.9), with Eq. (4.13) added, are general means for evaluating (by
superposition and integration) the magnetic flux density vector, B, due to
field
z
.
180
Chapter 4
Magnetostatic Field
in
j
Free Space
given current distributions
Eqs.
( 1
.37)— (1 .39) are the
space (or any nonmagnetic medium), just as
in free
means
for evaluating the electric field intensity vector, E,
due to given charge distributions in free space. In this section, we shall consider various characteristic examples of the application of the Biot-Savart law. The examples
include current distributions that are theoretically and practically important on one
hand and for which the integrals can be evaluated analytically on the other. Most of
the magnetostatic structures we shall analyze have their electrostatic counterparts
in Section 1.5, so that many solution strategies introduced and developed in that
section apply also here.
Example
Magnetic
4.1
Current Loop
Field of a Circular
Consider a circular loop of radius a carrying a steady current of intensity / in
the magnetic flux density vector along the axis of the loop normal to its plane.
We
Solution
field
computation
Example
in
1.6.
contribution to the magnetic flux density vector at a point
current element
/
dl at a point P'
on the contour
yj
2
[X()l
=
dB
R=
With reference
P on
dl
x
R
(4.14)
47tR 2
+ a 2 (z is the coordinate of the point P). As
= 1 (unit vector), |dl x R| = dl. Hence, the
plane; for
Example
its
and R are mutually permagnitude of the vector dB
dl
equals
dB
=
along the axis of a circular
current loop normal to
to Fig. 4.6, the
the loop axis (e-axis) of a
is
pendicular and |R|
Figure 4.6 Evaluation of the
magnetic flux density vector
Calculate
subdivide the loop (contour C) into elemental segments and apply Eq. (4.9),
analogously to the electric
where
air.
The
total
magnetic
Mo 1 d/
47tR-
(4.15)
flux density vector at the point
P
is
obtained as
4.1
B =
dB
(4.16)
Because of symmetry, the radial (horizontal) components of the vector
cel out in the integral
(vertical)
(much
1.11, in
like in Fig.
dB
in Fig. 4.6
can-
the electrical case), and only the axial
components
d Bz
=
dflcos(90° -cr)
=
dfisina
= dB^ =
contribute to the final result. Consequently,
B=
which yields [see Eq.
B due
to
(
1
<j>
d Bz i
M() la
B =
2(z
loop
Magnetic
/ in
2
2
+ a2 )3/2
z
Field of a Finite Straight
Consider a straight conductor of length
current of intensity
Jc
-43)
a circular current
Example 4.2
(4.18)
dl.
4n R}
/
(4.19)
-
Wire Conductor
representing a part of a wire contour with a steady
free space. Find the expression for the
B
field at
an arbitrary point
in
space due to this current conductor.
Solution The conductor and an arbitrary field point (P) always determine one plane, which
that we can use the simplified form of the Biot-Savart law in Eq. (4.13). From Fig. 4.7,
means
/
cos 0
The
=
d/R, so that the magnetic flux density
solution to this integral
Mo
f
4n
J,
7d0
is
given by
MO
R
181
Magnetic Flux Density Vector due to Given Current Distributions
Section 4.3
r
e2
cos 9 dO.
(4.20)
4nd Je=ei
is
B=
And
—
(sin 02
(4.21)
sin0i),
B
due
to
a straight wire
conductor of
length
finite
9\ and 02 are the angles defining the starting and ending point of the conductor,
respectively, and d is the perpendicular distance from the conductor to the point P.
The expression in Eq. (4.21) can be combined for computing the magnetic field due to
any structure assembled from straight line segments with steady current. In addition, it can
be used for the evaluation of the contribution to the total field of straight segments contained
where
in structures that also include curvilinear segments.
By
taking 9\
= —n/2
and
02
= n/2
and
in Fig. 4.7
in
Eq. (4.21),
we
obtain the expres-
sion for the magnetic flux density due to an infinitely long straight wire conductor carrying
a current 1 in free space. With a notation
cylindrical coordinate
d
=
r,
standing for the radial coordinate in the
r
system whose z-axis coincides with the axis of the wire conductor,
this
expression reads
MQ/
B
(4.22)
2 nr'
Example 4.3
Magnetic
Field of a
B
due
to
an
infinite wire
conductor
Square Current Loop
A square loop of side length a in free space carries a steady current of intensity
Obtain the
I.
expression for the magnetic flux density vector at the loop center.
Solution
square
The magnetic
sides,
in Fig. 4.8.
flux density at the loop center
B i, is given by Eq.
By means
(4.21) with
d
=
a/2, 6\
due
to the current along
= —n/ 4, and 02 =
7t/4, as
one of the
can be seen
of the superposition principle, the magnitude of the total magnetic flux
density vector amounts to
(4.23)
na
with respect to the reference direction of
B indicated in Fig. 4.8.
Figure 4.7 Evaluation
Example 4.4
Magnetic
Field of a
of the magnetic field of a
Loop with a Semicircular Part
finite straight
shows a wire contour consisting of a semicircle (of radius a) and a straight line (of
length 2a). The contour is situated in air and lies in the xy-plane of the Cartesian coordinate
system. If the contour carries a steady current of intensity I, calculate the magnetic flux
Fig. 4.9(a)
wire
conductor; for Example 4.2.
® _L
density vector at an arbitrary point along the z-axis.
i
at a point P on the z-axis due
element / dl that belongs to the semicircular part of the contour [Fig. 4.9(a)].
This vector is given by Eq. (4.14), and we need to break it up into components suitable for
integration, which are x-, y-, and z-components in this case, and this is very similar to the
decomposition of the vector dE in Fig. 1.12 and Eqs. (1.46) and (1.47). So, we first decompose
Solution
Let dB' denote the magnetic flux density vector
to a current
dB'
dB'
in the
=
plane of the triangle
dB],
+
dB'z
z,
d B'h
—
APOP'
[Fig. 4.9(b)]:
dfi'cosa,
cos a
=
R
d B'z
=
dB' sin a,
sin
a
=
R
(4.24)
where dB' is given
y-components [Fig.
in
dB'h
=
Eq.
(4.15).
We
then represent the horizontal vector dBj, by
its
x-
and
Figure 4.8 Evaluation of the
magnetic
Example
dB'x x
field at
the center
of a square current loop; for
4.9(c)]:
+
dB'
y y,
d B'x
=
dB'h
coscj),
d B'v
=
d5],sin^.
(4.25)
4.3.
-
182
Chapter 4
Magnetostatic
Space
Field in Free
Figure 4.9 Evaluation of the
magnetic
field
contour with
due to
a current
a semicircular
a linear part; for
Example
and
4.4.
Finally,
we
integrate, as in Eqs. (1.48)-(1.50) in the electrical case, the three Cartesian
com-
ponents of dB' along the semicircle, with respect to the azimuthal angle 0 {-n/2 < 0 < n/2)
as the integration variable [note that d/ = ad0 in Eq. (4.15)]:
Iaz
tiQlaz
^°
B- = /hr'
dB- =
'
/
4
fx 0 Iaz
An R?
t*
f
12
...
Tiei-n/2
—R
Mo laz
T
5
27T
=
cos 0 d(p
B =
n/2
f
.
La****
Hence, the resultant magnetic
comes out to be
=
B‘ =
f
,
°-
MO la
B =
due
n/2
A
f
d0
=
MO la
2
(4.26)
to the semicircular part of the
contour
t’I+ 2
2 R?
y
'
I,
flux density vector
’>
I,
noIa
,
dS <
dB
’>
,
= 4^5 Ln
4/? 3
(4.27)
Z
)
The magnetic
the other hand,
is
flux density vector at the point P due to the linear part of the contour, on
determined from Eq. (4.21) with d = z, 0\ = —a, and 9j = a:
MO^™
B =
Zsinaf— x) =
•
M()/a
,
The
total
B
field
B =
along the z-axis
B'
+B =
Mo la
Example 4.5
Magnetic
2
— x+ n
a
•
3
1
„
.
z
R—
y/ z
2
+ a2
.
(4.29)
2
7TZ
Field of a Finite Solenoid
shows a solenoid
Fig. 4.10(a)
(4.28)
is
'
2 7?
-
x.
2nzR
47tz
(cylindrical coil) consisting of
wound uniformly and densely
N
turns of an insulated thin
nonmagnetic support with a
circular cross section of radius a. The length of the solenoid is / and the current through the
wire is /. The medium is air. Find the expression for the magnetic flux density vector along
wire
the solenoid
Solution
in
one layer on a
cylindrical
wound
in a spiral
axis.
Because of the
coil
being closely
and the wire being
thin, the
current flowing through the turns of the solenoid can be regarded as a thin cylindrical current
sheet with surface current density [Eq. (3.13)]
Js
The current over an elemental
=
NI
(4.30)
~T‘
length dz along the solenoid,
d/
=7
S
dz
=
shown
in Fig. 4.10(b),
Nldz
(4.31)
I
2
Figure 4.10
(a)
Uniformly and densely
wound
183
Magnetic Flux Density Vector due to Given Current Distributions
Section 4.3
solenoidal coil with a steady current and (b) evaluation of
the magnetic flux density vector along the solenoid axis; for Example 4.5.
can be viewed as the current of an equivalent circular current loop of radius a and wire
diameter dz. From Eq. (4.19), the magnetic flux density vector of this loop at an arbitrary
P
point
at the solenoid axis [Fig. 4.10(b)] is
HO d/a
dB =
To
find the total field
B
at the point P,
we
2
„
(4.32)
integrate
dB
to
sum
the contributions of
all
equivalent loops along the solenoid (superposition principle):
B
f
MoATa 2 [ 12 dz
Zl
In order to solve this integral, however,
we note
„
„
r^.
~
d
from the point P)
that the relationship in Eq. (1.55), with
substituted here by a, exists between the length coordinate z (measured
and the angular coordinate 0 of the position of the equivalent loop along the solenoid
Multiplying it by a/R = cos 0 yields
axis.
2
a dz
~w = cos#
With
this,
the integral in Eq. (4.33)
B
is
(4.34)
reduced to a simple form:
rv
r<h
HqNI
cos 6 dO
/
21
which
d6.
J9 =9
(4.35)
z,
\
results in
NI
B = Ho
(sin #2
—
sin#i)
(4.36)
z.
21
This expression
is
B
along the axis of a
finite
solenoid
valid for an arbitrary field point along the solenoid axis, both inside
and
outside the solenoid, with the position of the point being defined by angles 0\ and 6j. Note,
however, that the position of the point P can also be defined by coordinates z\ and
z\
=
a tan 0\ and zi = a tan 02In the case of an infinitely long solenoid, 0\
in Fig. 4.10(b)
and
in
Eq.
1
-* -oo) and 62
=
n/2 ( Z2
where
00)
(4.36), so that
B=
where N
= —n/2 (z\
zj,
ixqN'I,
(4.37)
= N // is the number of wire turns per unit length of the solenoid. The magnetic field
uniform, with flux density given in Eq. (4.37), within an infinite solenoid even at points off
of its axis, as we shall show in a later example.
is
B
inside
an
infinite
solenoid
Chapter 4
Magnetostatic
Space
Field in Free
Magnetic
Example 4.6
An
Field of
an
Infinitely
Long
Conductor
Strip
long conductor in the form of a thin strip of width a carries a steady current of
The permeability everywhere is mo- Determine the expression for the magnetic
infinitely
intensity
/.
flux density vector at
Solution
shown
This
is
an arbitrary point
in space.
a two-dimensional problem,
and we solve
in the cross-sectional
it
plane
Eq. (3.82) tells us that the current I must be uniformly distributed in the
cross section of the conductor. Because of the conductor being thin, however, its current can
in Fig. 4.11.
be regarded as a surface current.
with density
Fig. 3.3,
=
Js
(4.38)
a
By virtue
of width d /
=
we subdivide
of the superposition principle,
and each such
dy,
strip
the conductor into elemental strips
can be considered as an
infinitely
long line current of
intensity
=J
d/
s
dl
=
I
dy
(4.39)
a
The magnetic flux density vector due to individual line currents is circular with respect
line, and, from Eq. (4.22), its magnitude at an arbitrary point P in space (Fig. 4.11) is
dB
=
Mo d /
R=
'
2 jiR
where d
dB
into
the perpendicular distance of
is
its
x-
+ d2
P from the plane of
(4.40)
.
the strip.
We decompose
vector
and y-components,
d Bx
=
dBy = dBcos#,
dBsin#,
and integrate them over the width of the
elemental
2
y)y
to the
strip
(4.41)
conductor to sum the contributions of
all
strips,
Mof
2 na
f
yz
Jy=yi
sin#dy
R
mo /
f y2
2na Jy]
'
cos9dy
(4.42)
R
We now use the relationship in Eq. (1.55), where z isy here, to change the integration variable
from y to 0. Multiplying its left-hand side by RcosO and right-hand side by d which
by the fact that cos# = d/R in Fig. 4.11, we obtain
,
is
justified
cos#dv
~R- = A6
Figure 4.11 Evaluation
of the
an
magnetic
infinitely
field
long
due to
strip
conductor (cross-sectional
view); for Example 4.6.
-
(4.43)
I
Section 4.4
Formulation of Ampere's Law
185
which, multiplied by sin0/cos#(= tan 0) gives, in turn,
—R~ =
sin 6
Based on these two
By
=
r
Ho I
bl
2na
C
Jg
e2
dy
sin 6
,
t d e.
relations, the integrals in Eqs. (4.42) are simple to solve:
sin#
n-nT
cos 0
2na
ms
u.
B
cos 02
i
nJ
"i
fit
2na
'
where the use is made of the substitution u = cos 0 ( dn
Hence, the total magnetic flux density vector due to the
B
with R\ and
R2
(4.44)
cos 9
2na
xln^ +
R\
r
r ez
Jg
= «2 (e2 _ fll)
2na
,
= - sin 6 d 6)
( 4.45)
in the first integration.
conductor turns out to be
strip
-<9 i) y
(6>2
P from
being the distances of the point
d(,
l
(4.46)
B
due
to
a
B
due
to
an
strip
conductor
the starting and ending point,
respectively, of the line representing the cross section of the strip conductor, in Fig. 4.11.
Note
R\
= R2
that
if
we
let
a
00 while keeping I /a
=
const,
we
get an infinitely wide and
long planar current sheet with a uniform surface current density Js
infinitely
= — 7r/2,
(infinite), 6\
and O2
=
= I /a.
Then,
n/2, with which Eq. (4.46) becomes
(4.47)
infinite current
sheet
We
conclude that the magnetic field due to an infinite planar sheet of current
each side of the sheet, with field lines parallel to the sheet.
4. 1-4. 9;
Problems'.
is
uniform
at
Conceptual Questions (on Companion Website): 4.4-4. 6;
MATLAB Exercises (on Companion Website).
4.4
We
FORMULATION OF AMPERE
saw
in
Chapter
1
that the electric field
S
LAW
due
tributions in free space can be determined
to highly symmetrical charge dis-
much more
easily using Gauss’ law,
Eq. (1.133), than by direct application of Coulomb’s law and the superposition
principle,
i.e.,
Eqs. (1.37)-(1.39). In magnetostatics, Eqs. (4.7)-(4.9) (the Biot-Savart
law) provide general solution procedures analogous to those in Eqs. (1.37)-(1.39),
whereas the law that helps us evaluate the magnetic
field
due
to highly symmetrical
more easily is known as Ampere’s law (also called
Ampere’s circuital law or Ampere’s work law). It states that the line integral (circulation) of the magnetic flux density vector around any contour (C) in a vacuum
current distributions in free space
(free space)
we mark
is
equal to ho times the total current enclosed by that contour, which
Figure 4.12 Arbitrary
contour in a magnetostatic
field - for the formulation
of
Ampere's
law.
as Ip,
£
The reference
B
dl
= ho
direction of the current flow
is
c
(4.48)
related to the reference direction of
means of the right-hand rule: the current is in the direction defined
by the thumb of the right hand when the other fingers point in the direction of the
contour, as shown in Fig. 4.12. This law may be derived from the Biot-Savart law
(we recall that Gauss’ law is derived from Coulomb’s law), and the derivation is
carried out in Section 4.10. For the present, we accept Ampere’s law temporarily as
the contour by
another law capable of experimental proof. Eq. (4.48) represents Maxwell’s second
Ampere's law
186
Chapter 4
Magnetostatic
Field in Free
Space
HISTORICAL ASIDE
Ampere
Andre-Marie
French
mathematician and phys(1775-1836),
icist,
a
was a professor of
mathematics,
physics,
to the French
Academy
of Sciences his discov-
ery of magnetic forces between wires carrying
He
electric currents.
two
that
same
discovered experimentally
parallel wires carrying currents in the
direction attracted each other, whereas the
and chemistry in Bourg
and Paris, and inspector
wires with currents flowing in opposite directions
general of the
science of the magnetic field due to electric cur-
national
under
Napoleon. Although with
university system
little
formal
Ampere
education,
acquired
the
knowledge of mathematics and sciences for that time by reading many books and
articles, with a great support and guidance from
his father. As a teenager, he was already deriving his own mathematical theories and writing
papers on some geometrical problems. However,
his life was soon to be shattered when his father
was sent to the guillotine by the Jacobins in
Lyon in 1793 during the French Revolution. The
effect on Ampere of his father’s death was devastating, and he did not return to his studies of
mathematics for almost two years. After several
years of tutoring mathematics in Lyon, he was
appointed professor of physics and chemistry at
Bourg Ecole Centrale in 1802, and then professor
of mathematics at the Ecole Polytechnique in Paris
in 1809. His mathematics research included a wide
best possible
variety of topics in probability, calculus, analytic
geometry, and partial differential equations.
He
chemelectricity and
contributed also to the theory of
light,
and, most importantly, to
magnetism. Only a few weeks after hearing of
Oersted’s experimental results, Ampere was ready,
before the end of September of 1820, to report
istry,
equation for
repelled each other. His experiments founded the
rents.
He
around
described the magnetic force circling
a current-carrying wire in a
way which
is
now known as the right-hand rule. On November
6, 1820, Ampere gave a talk on his circuital law of
addition of magnetic forces - a basis of a general
equation that
He was
the
we
first
now Ampere’s (circuital)
call
law.
to describe current as the flow of
along a wire, analogous to the surge of
water through a pipe. He predicted theoretically
that a wire helix with current would behave as it
were a bar permanent magnet, and called such
electricity
a helix a solenoid.
manent magnets
Ampere
also explained per-
as a collection of tiny electric
currents circling eternally within them, and in
this
he was three-quarters of a century ahead of
A
comprehensive presentation of his
and magnetism, including
both description of experiments and mathematical
derivations, appeared in his most important publication “Memoir on the Mathematical Theory of
Electrodynamic Phenomena, Uniquely Deduced
from Experience” in 1826. The same year. Ampere
was appointed to a chair at the Universite de
France, which he held until his death. In his
honor, the intensity of electric current is measured
time.
his
findings in electricity
in
amperes.
(Portrait:
Edgar Fahs Smith
Collection,
University of Pennsylvania Libraries)
static fields in free
space (as
we know, there is a total of four Maxwell’s
field). By expressing the current in terms
equations for the general electromagnetic
of the volume current density,
J,
Ampere’s law becomes
B
Ampere's law for volume
(f)
Jc
current
•
dl
=
yU()
1 J
dS,
(4.49)
Is
where S is a surface of arbitrary shape spanned over (bounded by) the contour C
and oriented in accordance to the right-hand rule with respect to the orientation of
C (Fig.
4.12).
Section 4.5
Figure 4.13 Closed path
seven
What
is
187
C and
4.7.
Algebraic Total Enclosed Current
the circulation of the magnetic flux density vector along the contour
The medium
Law
line currents in air; for
Example
Example 4.7
Applications of Ampere's
C in
Fig. 4.13?
is air.
According to the right-hand rule given with respect to the reference direction
C indicated in Fig. 4.13, the reference orientation of the surface S spanned
over C, i.e., the reference direction of the vector dS in Eq. (4.49), is from S upward. From
Ampere’s law, the line integral of the magnetic flux density vector along C equals mo times
the algebraic sum of all current intensities passing through S. We note that current I\ pierces
S once in the positive direction (direction in agreement with the orientation of S ), I2 passes it
four times, I4 and I-j once each but in the negative direction, I3 once in the positive and once
in the negative direction, Is twice in the positive and once in the negative direction, while h
does not traverse S at all. Hence,
Solution
of the contour
£
What
is
•
dl
=
+ 4/2 -U+I5- h)
mo (h
(4.50)
very important, the same result for the algebraic total current, and thus for the circu-
lation of B,
and
B
is
obtained for any other surface
we imagine with
the contour
C as the perimeter
totally arbitrary shape.
Example 4.8
Fig. 4.14
Contour inside
a
Conductor
shows the cross section of a very long
cylindrical
copper conductor that carries a
steady current of intensity 7 (7 > 0). Is the circulation of the magnetic flux density vector
along the contour C less than, equal to, or greater than mo 7?
Solution From Ampere’s law,
<
)
j
since the total current appearing
B
•
dl
<
(4.51)
mo?>
on the right-hand
Figure 4.14 Contour inside
side of Eq. (4.49) equals exactly that
portion of the total current of the conductor (7) enclosed by the contour.
Conceptual Questions (on Companion Website): 4.7-4.10.
4.5
APPLICATIONS OF AMPERE'S
devoted to the application of Ampere’s law in evaluating the
due to given steady current distributions in nonmagnetic media.
the case with Gauss’ law, the use of Ampere’s law will also require
This section
magnetic
As
is
LAW
is
field
a
conductor with a steady
Example 4.8.
current; for
188
Chapter 4
Magnetostatic
Field in Free
Space
symmetry of the problem to determine which field
components are present in the structure and which spatial variables the present
components depend on. Eq. (4.48), although always true, enables us to analytically solve for the field only due to highly symmetrical current distributions.
careful consideration of the
Namely, as the unknown quantity to be determined (B) appears inside the intewe can use Ampere’s law to obtain a solution only in cases
in which we are able to bring the magnetic flux density, B outside the integral sign, and solve for it. These cases involve current distributions for which we
are able to adopt a closed path C, called the Amperian contour, that satisfies
two requirements: (1) B is everywhere either tangential or normal to C and (2)
B — const along sections of C where B is tangential. Along the portion of the
path where B is normal to it (if such portion exists), the dot product B dl in
Eq. (4.48) becomes zero. Along the remaining part of the contour, B dl becomes
B dl, and the second requirement (constancy) then permits us to remove B from
the integral sign in Eq. (4.48). The integration we are left with is usually trivial and consists of finding the length of that portion of the path to which B is
gral in Eq. (4.48),
,
•
•
tangential.
In this discussion,
Gauss’ law, discussed
we
in
notice and exploit the parallelism with the application of
Section 1.13. All similarities and differences in the mathe-
matical formalism are direct consequences of the mathematical form of the fields
E and B due
Gauss’ law,
we
to the elementary sources
Q
and Q\, Eqs.
(1.24)
and
With
(4.4).
integrate over a closed surface (Gaussian surface) on the left-hand
Ampere’s law implies integration along a closed path
component contributing to the integral is a comthe Gaussian surface and tangential to the Amperian contour,
side of the equation, while
(Amperian contour). The
ponent normal to
respectively.
On
field
the right-hand side of the equation, Gauss’ law involves finding
the total charge enclosed by the surface, whereas the application of Ampere’s law
involves finding the total current enclosed by the contour. For sources expressed
by volume charge and current densities [Eqs. (1.135) and (4.49)], this means volume integration over the volume enclosed by the Gaussian surface and surface
integration over the surface bounded by the Amperian contour, respectively.
Example 4.9
Magnetic
Field of a Cylindrical
Conductor
(a)
dB'+dB"
An
infinitely
sity /.
long cylindrical copper conductor of radius a carries a steady current of inten-
The conductor
is
situated in
air.
Find the magnetic flux density vector inside and
outside the conductor.
Solution
current
is
Fig. 4.15(a)
shows
a cross section of the conductor.
uniformly distributed
in
the cross section, so that
its
According to Eq.
density
(3.82), the
is
(
Figure 4.15 Cross section
of a cylindrical
conductor
with a steady current
/:
(a)
application of Ampere's law
and (b) proof that the
magnetic field lines are
circles; for Example 4.9.
-
Because of symmetry, the lines of the magnetic field due to the conductor current are
centered at the conductor axis. To show this, consider the direction of the magnetic
flux density vector, B. at an arbitrary point P in space, either inside or outside the conductor.
Let the distance of the point from the conductor axis be r, and let dB' and dB" represent
the fields at P due to two symmetrical current elements denoted as J' dv and J" dv and shown
in Fig. 4.15(b). In accordance to the Biot-Savart law, Eq. (4.6), these two elementary field
vectors are such that their sum dB' + dB" is tangential to the circular contour C of radius
r centered at the conductor axis. The same is true for any other pair of symmetrical current
elements, which can also be in a plane that does not contain the point P, and all current
elements constituting the current / in the conductor can be grouped in such symmetrical
circles
(b)
4 52 )
Applications of Ampere's
Section 4.5
pairs.
We
at the point
P is
tangential to the contour C.
const along C,
i.e.,
the magnitude of
B
conclude that the resultant vector
In addition,
symmetry
only on the radial coordinate
with the conductor
B=
also implies that
axis.
r
B
189
Law
depends
of the cylindrical coordinate system whose z-axis coincides
Hence, we can write
B=
B(r)
(4.53)
<j>,
where
is the circular unit vector in the system.
Based on the preceding discussion, it is now obvious that the contour
<j>
C
in Fig. 4.15
satisfies both requirements for the Amperian contour for our problem. Along C, dl
so that B dl = B dl [Fig. 4.15(a)]. The circulation of B along C thus turns out to be
=
d/
<j>,
•
B
dl
=
(b
B(r) dl
=
B(r)
dl
(b
=
B(r )
l
= B(r) 2 nr,
whereas the current enclosed by
<
r
<
oo,
(4.54)
C amounts to
Ic
Jnr2
=
for r
for r
Eqs. (4.54) and (4.55)
0
Jc
Jc
c
combined
in
B=
Ampere’s
<
>
a
(4.55)
a
law, Eq. (4.48), finally give
noIr/(2na
2
)
for r
<a
for r
>
J
HQl/(2nr)
B due
a
Note
that the magnetic field outside the conductor (for r
>
(e.g.,
a thin wire with current I)
and the
a thick cylindrical
current
a)
is
identical to that of a line
current of intensity I placed along the conductor axis, Eq. (4.22). This
current
to
conductor with a steady
|
means
that the line
(thick) conductor of Fig. 4.15 are equivalent
sources with respect to the region outside the conductor.
Cylindrical
Conductor with an Excentric Cavity
A very long cylindrical nonmagnetic conductor of radius b has an excentric cylindrical cavity
of radius a along
its
entire length, as
shown in Fig. 4.16. The axis of the cavity is offset from
+ d < b). The medium in the cavity and outside the
the axis of the conductor by a vector d {a
conductor
is air.
Compute
the magnetic flux density vector in the cavity, assuming a steady
current of density J flowing through the conductor.
Solution The current density in the cavity is zero, and it can be considered to result from
two currents of the same density (J ) flowing in opposite directions. Accordingly, we can represent the current distribution in the hollow conductor, including the cavity, as a
sum
of current
distributions of a cylindrical conductor of radius b carrying a current of density
J without
and another one of radius a with a current of the same density but flowing in the
opposite direction, as depicted in Fig. 4.16. By the superposition principle, the magnetic flux
density vector of the original (resultant) current distribution, B, can be obtained as
a cavity
B=
Bi+B 2
,
(4.57)
Figure 4.16 Cylindrical
conductor with an excentric
cylindrical cavity, viewed as
a superposition of
two
solid
conductors with currents of
the
same
density but opposite
directions; for
Example
4.1 0.
1
.
190
Chapter 4
Magnetostatic
Field in Free
Space
where Bi and B 2 are the magnetic flux density vectors due to the partial current distributions
and J 2 in Fig. 4.16. These vectors, according to Eq. (4.56), can be written in the form 3
Jt
Ft
=
Bi
qJ x
ri
where
and
rj
Mo(— J) X
B2 =
and
2
(4.58)
are the position vectors of the field point (P) with respect to the
r2
conductor axis (Oi) and cavity axis (0 2 ), respectively.
equations,
we
r2
2
combining the preceding
Finally,
get
- r2
x (n
MqJ
B=
MoJ x d
)
(4.59)
2
2
and B = noJd/2. We conclude that the magnetic field inside the cavity is uniform, as the
above expression for the resultant magnetic flux density vector does not depend on the position of the point P. The field lines are parallel to the plane of the conductor cross section and
at right angles to the vector d.
Example
Magnetic
4.1
Cable
Field of a Coaxial
Conductors of a coaxial cable are made from copper, and its dielectric is air. The radius of
the inner conductor is a, whereas the inner and outer radii of the outer conductor are b and
c,
respectively (a
<
<
b
c).
A
steady current of intensity /
is
established in the cable. Find
the magnetic flux density vector everywhere.
Solution
Referring to Fig. 4.17, current densities in the cable conductors are
and
Figure 4.17 Evaluation
of the
magnetic
B
in
4.1
I
n(c 2
Due
to symmetry, the magnetic flux density vector, B,
axis.
Following a similar procedure as
C of radius
contour
for
all
values of
r,
<
r (0
r
Example
in
< 00 ). The
4.9,
circulation of
—
(4.60)
b2 )
circular with respect to the cable
is
we apply Ampere’s law to the circular
B along C is the same as in Eq. (4.54),
while the expression for the total current enclosed by
field of a
coaxial cable; for
Example
h=
four different characteristic positions of the contour.
between the conductors
is
identical to that
found
in
C
is
different for
The field inside the inner conductor and
Example 4.9 for a single conductor.
1
_
the dielectric of a coaxial
HoJ\ r
_
MO Ir
(0
2 na 2
2
cable
<
r
<
MO-f
B=
a).
{a
<
r
<
b).
(4.61)
2nr
For the contour positioned inside the outer conductor, the enclosed net current equals the
entire current of the inner conductor minus that portion of the current of the outer conductor
which
is
enclosed by the contour, so that
B—
Mo
U ~ h^fr2 ~
b
]
2
)]
= ^"^2
(b
^ )r
<
r
<
c).
lc
Finally,
if
the radius r
is
larger than the outer radius of the outer conductor,
B—
-
/)
=
So, the external magnetic field (for r
>
c)
2nr
mo(^
0
is
(c
<
r
< 00 ).
zero. This,
we
(4.63)
see, results
from currents
of equal intensities and opposite directions in cable conductors (making zero total enclosed
current in Ampere’s law), much as the external electric field of a charged coaxial cable is
zero because of charges of equal magnitudes and opposite polarities per unit length of cable
3
Note
B=
that the magnetic flux density inside the solid cylindrical conductor in Fig. 4.15 can be written as
n<)Jr/2, and, in vector form,
respect to the conductor axis.
B =
/xoJ
x r/2, where
r is the position
vector of the field point with
r
Law
Applications of Ampere's
Section 4.5
191
conductors (making zero total enclosed charge in Gauss’ law). [Such (equal positive and
negative) charges and currents are standard in operations of all two-conductor transmission
a very important property of a coaxial cable; the electrostatic and magnetostatic
lines.]
This
fields
of the cable are concentrated exclusively inside the cable, and
is
decoupled (shielded) with respect to the
Magnetic
Example 4.12
Shown
The
in Fig. 4.18(a)
is
interior,
and vice
exterior
its
is
perfectly
versa.
Field of a Toroidal Coil
a toroidal (“doughnut”-shaped) coil with a rectangular cross section.
N turns of wire that are uniformly and densely wound along the length of
coil consists of
7. The inner and outer radii of the toroid are
and its height is h. The medium inside and outside the coil is nonmagnetic.
Calculate the magnetic flux density vector inside and outside the coil.
the toroid and carry a steady current of intensity
a and b (a
<
b),
Solution Fig. 4.18(b) portrays a horizontal cut of the toroid in Fig. 4.18(a). Because of
symmetry, the lines of the magnetic field density vector, B, due to the current in the coil are
circles centered at the toroid axis [z-axis in Fig. 4.18(b)], and the magnitude of B depends
only on the distance r from the
Eq.
(4.53).
axis.
means
This
B
that the vector
Applying Ampere’s law to a circular contour of radius
of the form given in
is
<
r (0
r
<
oo),
we
obtain
the following equations for three characteristic locations of the contour:
=
B(r) 2nr
0
B(r)
B=
Hence,
(0
2n
<
<
r
=
fio(NI — NI) =
B(r)
a),
2nr
= hqNI
(b
0
<
r
<
<
(a
r
<
b ),
(b)
(4.64)
oo).
of the
0 outside the toroid, while inside
B(r)
it
that this result
(inside a thick toroidal coil).
(4.65)
only rectangular).
If
b ) /2
to
the toroid
is its
is
thin,
mean radius.
i.e.,
b
-
contours (0
we can assume
a <^a,
_
27rc
is
c«c
inside
it,
where
c
=
(a
+
fipNI
2jtc
=
that
<
r
<
oo); for
Example 4.12.
flux density
H-qNI
/
showing the winding
and (b) cross-sectional
view showing Amperian
This implies that the magnetic field inside the toroid can be considered
be uniform, with the
where
field of
three-dimensional view
=
valid for an arbitrary shape of the toroid (vertical) cross section (not
is
magnetic
a toroidal coil: (a)
2jxr
Note
Figure 4.18 Evaluation
MO N'l,
(4.66)
B
inside a thin toroidal coil
l
the length of the toroid and N'
= N // is
the
number
of wire turns per unit
length.
We realize that the final expression for B in Eq. (4.66) is the same as that in Eq. (4.37),
found for an infinite solenoid. Namely, we can visualize an infinitely long solenoid as a toroid
with an infinite radius.
Magnetic
Example 4.13
An
infinitely
long
air-filled
Field inside
an
Infinite
Solenoid from Ampere's Law
solenoid has N' turns of wire per unit of
its
length.
A
steady
current of intensity / flows through the winding. Using Ampere’s law, prove that the
magnetic flux density inside the solenoid
Solution
Knowing
that the vector
solenoid axis) and that
magnitude of
B
it is
B
is
given by the expression in Eq. (4.37).
inside an infinite solenoid
zero outside the solenoid,
in the solenoid
from Ampere’s
law.
it is
now
is
axial (parallel to the
very simple to evaluate the
The Amperian contour is a rectangle
The line integral of B along the
Figure 4.19 Evaluation of
positioned partly inside the solenoid as shown in Fig. 4.19.
the magnetic
edge of the rectangle that is inside the solenoid equals Bl, because B is parallel to that edge and does not vary along it (the structure is infinitely long). The line integral
of B along the remaining three edges of the rectangle is zero, because B is perpendicular to
an
vertical (axial)
infinite
field inside
solenoid using
Ampere's law; for
Example 4.1 3.
192
Chapter 4
Magnetostatic
Space
Field in Free
portions of horizontal (perpendicular to the axis) edges that are inside the solenoid and
zero outside
it.
The
net current enclosed by the contour, on the other hand, equals the
ber of turns over the length
which
/,
is
num-
N'l times / (the current through each turn). Hence,
is
,
Ampere's law gives
=
Bl
hqN'II.
(4.67)
the same as in Eq. (4.37). Note that we did not in any way restrict the location of the left
edge of the contour C in Fig. 4.19 to be on the solenoid axis or at a specific distance from it,
which means that Eq. (4.37) is valid across the entire cross section of the solenoid, and not
only on the solenoid axis (in Example 4.5, the magnetic field is evaluated along the solenoid
axis only). In other words, the magnetic field is uniform (the same) throughout the entire
volume enclosed by the solenoid winding (while zero outside it). We also note that the cross
section of the solenoid in this discussion and calculation is not in any way restricted to be of
circular shape, which implies that the magnetic field is the same for arbitrary shape of the
i.e.,
solenoid cross section (provided that the solenoid
Magnetic
Example 4.14
Charged Cylinder
Field of a Rotating
A
very long cylinder of radius a
p.
The cylinder uniformly
is
very long).
is
its volume by a charge of density
an angular velocity w. The permeability
uniformly charged over
rotates about
axis with
its
mo everywhere. Find the magnetic flux density vector inside and outside the cylinder,
assuming that the charge distribution of the cylinder remains the same during the rotation.
is
Solution
As
the charges of the cylinder rotate with
it,
moves
and the current density vector
in the cylinder are circles
=w
x
at that point
centered
We
r
= wn
=
pv(r)
note that this current density, which
Because of the currents
series of
can be found from Eq. (3.28). The current lines
many very
in
= pwr
is
(0
zero,
whereas
contour
B
is
C shown
r
<
and the current density
a).
(4.69)
electric field,
the cylinder being circular, the cylinder can be visualized as a
axial inside the cylinder.
in Fig.
can be regarded
3.17(a)],
long coaxial solenoids of radii varying from r
4.20(b) results in
(i.e.,
Ampere’s
=
outside
0 to
all
r
—
a.
That
is
why
the
equivalent solenoids)
is
Eq. (4.49), applied to a rectangular
[left-hand side of the equation is the same as in
law,
Eq. (4.67)]
rci
r(l
B(r)l
rotating charged cylinder:
=
equivalent impressed
current density and (b)
Amperian contour; for
Example 4.1 4.
<
independent of the
Jj [Fig.
magnetic flux density vector, B, outside the cylinder
(a)
(4.68)
j>,
at the cylinder axis [Fig. 4.20(a)],
as an impressed electric current density.
Figure 4.20 Evaluation
of the magnetic field of a
the cylinder axis
a function of r given by
J(r)
(b)
away from
r
is
is
v
is
they form a volume electric current.
M that
Referring to Fig. 4.20(a), the velocity at which a point
where dS
is
Mo
•/(/)
/
V=r
/
dr'
= mo pwl
magnitude of B
(4.70)
dS
the surface area of a thin strip of length
variable). Hence, the
r dr,
/
Jr
"TT"
1
/
and width d / ( r is the integration
away from the cylinder axis comes
at points that are r
out to be
B(r)
Example 4.15
Using Ampere’s
Magnetic
law,
=
m)pw(a 2 - r2 )/!
for r
0
for r
Field of
an
Infinite
prove that the magnetic
Eq. (4.47).
a
(4.71)
a
Current Sheet from Ampere's Law
flux density vector
current sheet with a uniform surface current density 7 S
in
<
>
in free
space
due to an
is
infinite
planar
given by the expression
Section 4.6
Solution Because of symmetry, the magnetic
not vary in directions parallel to the sheet.
C
We
from which the same
result as in Eq. (4.47)
=
is
B does
193
Ampere's Law
of
B
BO
along each of the two edges of the contour
Bl and the enclosed current
2BI
the sheet and
Form
apply Ampere’s law to a rectangular contour
B
portrayed in Fig. 4.21. The line integral of
that are parallel to the sheet equals
field lines are parallel to
Differential
jU 0
is
Js l [Eq.
7s /,
Pi Js
(3.13)], so that
(4.72)
c
obtained.
l
Figure 4.21 Evaluation
Problems 4.10-4.19; Conceptual Questions (on Companion Website): 4.11-4.15;
of the
MATLAB Exercises (on Companion Website).
an
:
magnetic
infinite
field of
current sheet
using Ampere's law;
for
DIFFERENTIAL
4.6
In electrostatics,
we
(1.166)] starting
from
FORM OF AMPERE
S
Example
4.1 5.
LAW
derived the differential form of Gauss’ law [Eq. (1.163) or
its integral form [Eq. (1.135)]. Again, an analogous concept
and transformation exist in magnetostatics. In this section, we shall utilize the inteform of Ampere’s law, Eq. (4.49), to derive its differential equivalent. What
we expect to obtain is a spatial differential relationship between the magnetic flux
density vector, B (field), and the current density vector, J (source), at a point in
gral
space.
We consider first the one-dimensional case, and assume that J has only a
z-component in the Cartesian coordinate system which is a function of the coordinate x only, that is, J = Jz (x) z (1-D current distribution). Then, by symmetry, the
only present component of B is By (like in Example 4.15), i.e., B = B y {x) y. We
apply Ampere’s law [Eq. (4.49)] to a narrow rectangular contour C lying in the
xy-plane, with edges parallel to the x- and y-axes, as shown in Fig. 4.22. The dimension of the contour in the x-direction is dx and the length of the edges parallel to the
y-axis is /. The magnetic field is constant along both edges with length / ( By does not
vary with y), so that essentially no integration is needed on the left-hand side of
Eq. (4.49), i.e., the integral along each edge reduces to B I (1 has the same direction
as dl). No integration is needed on the right-hand side of Eq. (4.49) either, because
dx is differentially small and we can take Jz (x) as constant over the surface spanned
over C. Finally, as B is tangential to both edges with length l, B and dl are directed
in opposite directions along the left edge, and in the same direction along the right
edge,
we can
B v (x)
,B ,(x+dx)
}
T(6
BO
©
C
26Figure 4.22 For the
derivation of the one-
dimensional Ampere's law
in differential
form.
write
—B
y
{x)
l
+ By (x +
dx)
/
=
iiqJz {x) /dx.
(4.73)
Noting that
d By
= By (x +
dx)
- By (x),
(4.74)
Eq. (4.73) becomes a differential equation,
(4.75)
is the one-dimensional Ampere’s law in differential form. We observe the
analogy with the 1-D Gauss’ law in Eq. (1.158).
We now generalize Eq. (4.75) to an arbitrary three-dimensional current distri-
This
The current density vector has now all three Cartesian components and
them are functions of all three coordinates,
bution.
of
J
= Jx (x, y, z) X + Jy(x, y, z) y + Jz (x, y, z) z.
all
(4.76)
1-D
differential
Ampere's law
194
Chapter 4
Magnetostatic
Field in Free
Space
C in Eq. (4.49) must, therefore, be differentially small in
both dimensions. Additionally, one contour is not enough; we need three small con-
The Amperian contour
each of the three current density components,
tours, oriented perpendicularly to
respectively. All edges of the contours being differentially small, the line integral
along each of them can be approximated by taking a constant value of the
component
tangential to the edge and multiplying
field
by plus or minus the edge
length. This gives us the result for each pair of integrals along parallel edges within
each contour that has exactly the same form as in the 1-D case [Eqs. (4.73)— (4.75)].
For the contour that lies entirely in the plane normal to the jc-component of J,
shown in Fig. 4.23, we have
<j)
B
dl
it
= -B z (x, y, z) d z + B y (x, y, z) dy + B z (x, y +
-By (x,y, z +
dz)dy
dy, z) d z
= noJx (x,y,z)dydz,
(4 .77)
which, divided by dy dz, results in
§c B dl
dy dz
•
_ B z(x, y +
dy, Z)
- B z (x, y,
z)
dy
B y (x, y,
z
+
dz)
- By (x,y,
z)
dz
dB z
8B,
dy
dz
=
Mo Jx
-
(
478
)
For contours that are oriented perpendicularly to each of the remaining two
coordinate axes, y- and z-axes, analogous procedures lead to equations
dBx
dBz
dx
dz
=
3
Bv
8 By
and
ixoJy
dx
—
MO Jz>
(4.79)
8y
by the unit vector x, and Eqs. (4.79) by y and
and summing the three equations, we obtain the following equation
with the vector J [Eq. (4.76)], multiplied by /xq, on the right-hand side:
respectively. Multiplying Eq. (4.78)
z,
respectively,
l
Ampere's law
(dB z
dBy\
V 9y
dz )
in differential
form
fdB x
„
x+
("97
This partial differential equation
form
B
for
field
We
dBr
dBx
dx
dy
z
(PDE)
in spatial
Figure 4.23 For the derivation
Ampere's law in differential
form for an arbitrary current
distribution.
(4.80)
It
relates the rate of
change of
coordinates to the local current density vector,
see that only those variations of individual
of
MoJ-
represents Ampere’s law in differential
an arbitrary steady current distribution.
components
—
components
J.
that are in directions
Section 4.7
Bx
perpendicular to the direction of the component (change of
and not along x, etc.) contribute in this relationship, which
dependences in the differential Gauss’ law [Eq. (1.163)].
4.7
is
along y and
195
Curl
z,
just opposite to the
CURL
The expression on the
left-hand side of Eq. (4.80)
the so-called curl of a vector
is
4
analogous to the divergence of a vector field (E)
as used to express the differential form of Gauss’ law [Eq. (1.166)]. In addition,
we notice that applying formally the formula for the cross product of two vectors
function (B), written as curl B,
system 5 to V x B, where the del operator
obtain exactly curlB. Hence,
in the Cartesian coordinate
Eq. (1.100),
we
curl
VxB=|
B=
1 dB v \
'3/?,
dy
.
(d bx
+
dz )
dz
y
—
- 3 B z^ \
dx )
is
given by
A
y
dBx'
*(Swhich can also be written
in the
curl
curI in Cartesian coordinates
Y-
dy
form of a determinant,
B=V
B=
x
AAA
dx
3y
(4.82)
3z
Bx By B
form is a convenient way for memorizing the expression for
x B in the Cartesian coordinate system). Note that the curl is an operation that
is performed on a vector, and the result is also a vector. Finally, the differential
Ampere’s law can now be written in a short form as
(using the determinant
V
curl
B=
VxB = /xqJ.
(4.83)
Ampere's law using
curl
notation
In nonrectangular coordinate systems, the differentially small
of
its
edges for
instance,
from
becomes
Amperian con-
and with different expressions for lengths
different current density components in each of the systems. For
tour in Fig. 4.23
curvilinear
Fig. 1.10, the
contour perpendicular to the radial vector component
Jr r in spherical coordinates is a curvilinear quadrilateral of edge lengths r d 0 and
rsin# d <p. Carrying out similar derivations as in Eqs. (4.77)-(4.80) for this contour
and contours oriented perpendicularly to other unit vectors
the expression for the curl in cylindrical coordinates
is
in Figs. 1.25
and
1.26,
obtained to be
(4.84)
4
Note that
5
For vectors given by their Cartesian components,
rot
B is
also used for curl B, “rot” being a short for “rotation” (or “rotational”).
a x b
=
(ax
=
(ay b z
x
+ ay y + a z z)
-
a z by ) x
+
x (bx x
(
a z bx
-
+
by y
+ b z z)
ax b z ) y
+
(ax by
-
ay bx )
z.
curl in cylindrical coordinates
,
196
Chapter 4
Magnetostatic
Field in Free
Space
and that for the spherical coordinate system
curl
—
(sm an
dO
1
B= V x B=
3
1
1
-
+
i
_
dr
30
r [_sin#
dBe
\
•
OBtt,)
07
v
rsin0
curl in spherical coordinates
f
30
3
—
3r
l
+
rR 4> ;
e
1
’
,
.
„
(
3
rBf)
Br
'
)
dO
*
(4.85)
same way as with V E in Section 1.15, we use the notation V x B in cylindriand spherical coordinate systems not to refer to it as the actual cross product of
vectors V and B, which is valid only in rectangular coordinates [Eq. (4.81)], but to
merely symbolize the curl operation. With this, we also emphasize the fact that relations derived employing such vector formalism in the Cartesian coordinate system
In the
•
cal
curl
B
hold true (are identities) generally,
cal quantities
and
relations
coordinate systems (properties of physi-
in all
between them are the
facts that are
independent of the
choice of coordinate system).
Figure 4.24 For the
definition of the curl of a
vector field
in Eq.
(4.87).
(4.81) and (4.78), we find that the jc-component of curlB, that
can be expressed as the net circulation (line integral) of B along the
incremental contour C of Fig. 4.23 divided by the area of the surface spanned over
Combining Eqs.
is,
x
•
curl B,
the contour,
—r —B ——
•
x
•
curl
B=
<f
dl
ay az
We
can
now
(4.86)
.
formally proclaim the Cartesian x-axis to be an arbitrary linear axis
and the contour
(direction) in space
C
small contour bounding a surface AS, as
shaped
and write
to be an arbitrarily
shown
in Fig. 4.24,
Sr B dl
—
—-—
differentially
•
n-curlB=
alternative definition of the
lim
AS—>0
curl
where n
is
(4.87)
AS
1
AS (AS = ASn) and determined
an equivalent mathematical definition of the
the unit vector normal to the surface
using the right-hand rule. Eq. (4.87)
curl of a vector.
is
enables us to obtain a component of the curl of
It
direction at a given point by
computing the circulation of
B
B
along a desired
along a contour in the
plane perpendicular to that direction as the contour shrinks to zero about the point.
The
circulation of
B
about C, per unit area, appearing on the right-hand side of
Eq. (4.87) equals n curl B = |curl B| cosa (|n| = 1). Therefore, for the orientation
of AS defined by a = 0, we get the maximum in the circulation (cos a = 1), and
•
Eq. (4.87) becomes
physical
meaning
|curlB|
of the curl
(
lim
§c B
•
dl
AS
\AS->0
(a
=
0).
(4.88)
max
This means that (1) the magnitude of curl B equals the maximum (as the direction
of the surface element AS = ASn is varied) net circulation of B per unit area with
the area of the surface element tending to zero and (2) the direction of curl B is in
the direction that gives the maximum value for the magnitude of the net circulation
per unit area (a
= 0). From
it,
we may regard
the curl of a vector field (not only
the magnetostatic field) physically as a measure of those local sources of the field
field components with respect to that
measure of sources that produce radial field
components with respect to the point [Eq. (1.172)]. Ampere’s law tells us that the
sources producing locally circular field components in the case of the magnetostatic
at a point in
point.
We
space which produce circular
recall that the
divergence
is
a
>
Section 4.7
field in free
Curl
space are elemental currents of intensities J AS. Quantitatively, the curl
represents the surface density of such sources, and in our case this density
concept of the curl of a vector
field is
used in numerous applications, in
is J.
many
The
areas
of science and engineering.
Replacing q,oJ by
Eq. (4.49), leads to
V
x
B
[from Eq. (4.83)] in the integral form of Ampere’s law,
£
Bd,=
/s
(V x B)
•
dS.
(4.89)
Although obtained here specifically for the magnetostatic field in free space, this
equation is an identity, holding for any vector field (for which the appropriate partial derivatives exist). It is widely used in electromagnetics and other areas of science
and engineering, and is known as Stokes’ theorem. In words, it states that the net circulation of a vector field along an arbitrary contour is the same as the net flux of its
curl through any surface bounded by the contour, where the reference orientations
of the contour and the surface are interconnected by the right-hand rule. We notice
the parallelism with the divergence theorem, Eq. (1.173), which applies to a closed
surface and relates the flux of a vector field with a volume integral of its divergence.
To prove Stokes’ theorem (for an arbitrary vector field B), imagine the surface S
subdivided into a large number of differentially small patches AS/ (A Si — 0) which
are bounded by infinitesimal contours Q, as depicted in Fig. 4.25. By applying the
definition of the curl in Eq. (4.87) to one of these patches, we can write
£
B
dl
=
ASi n curl B
=
(V x B)
•
AS/n
=
(V x B)
•
AS/.
(4.90)
A 5, comprising S and sum all the
on the left-hand side of the resulting equation is the line integral of B along the overall contour C bounding the surface S, since the terms arising
from the sides of small contours shared by any two patches cancel out during the
summation, as can be seen in Fig. 4.25, and the only boundaries for which the
Now,
let
results.
us determine this circulation for every
What we
get
cancelation does not occur are those forming the contour C. Hence,
l
B
dl
lim
AS/-».0
V
(V x B)
•
AS/.
Figure 4.25
(4.91)
Open
surface 5
subdivided into differentially
small patches - for the proof of
Stokes' theorem, Eq. (4.89).
Stokes' theorem
197
198
Chapter 4
Magnetostatic Field
Free Space
in
summation on the right-hand side of this equation becomes an
AS, becomes dS, and we have fs (W x B) dS, thus proving the theorem.
By applying Stokes’ theorem to Eq. (1.75) or simply by analogy with the differential form of Ampere's law, Eq. (4.83), we arrive to the differential form of
Maxwell's first equation for the electrostatic field:
In the limit, the
integral,
curl of
E
•
V
in electrostatics
We
see that the electrostatic field
0.
(4.92)
and
a curl-free or irrotational field,
is
property of any conservative vector
E=
x
this is a
with zero circulation along any
field (the field
closed path).
Redo Example
Solution
This
4.9 but
is
Conductor Using
Cylindrical Current
Example 4.16
now
Differential
Ampere's Law
using the differential form of Ampere’s law.
completely analogous to the application of the differential Gauss’ law to
problem with spherical symmetry in Example 1.22. For the interior of
= J z and J = const), combining Eqs. (4.53), (4.84), and
we now have in place of Eqs. (1.174) and (1.175)
solve an electrostatic
the conductor in Fig. 4.15 (where J
(4.83),
7xB =
with C\
=
0,
~
[rfi(r)] z
=
since there exists
Fig. 4.15 [otherwise, this
space (with J
=
—
—
ixqJ i
no
>
rB(r)
»
B(r)
=
hqJ
=
J
rdr
y
+
=
(0
<
+
r
<
a)
C\
(4.93)
,
along the z-axis (for
line current (of intensity Iq)
constant would be C\
+ Ci =
r
= 0)
in
surrounding
/xq/o/( 2^)]. Similarly, in the
0),
9
[rB{r)\ = 0
—
dr
—
>•
rB(r)
—
= C2
B(r)
>
=
Ci
—
<
(a
r
< 00 )
/
.
(4.94)
2
goJa /2, from B(a~) — B(a + ). Both Eqs. (4.93) and (4.94) give the same
corresponding results as in Eq. (4.56).
where
C2 =
Companion Website):
Companion Website).
Problems'. 4.20-4.24; Conceptual Questions (on
4.17;
MATLAB
4.8
LAW
As
vector
B
is
Exercises (on
OF CONSERVATION OF MAGNETIC FLUX
called the magnetic flux density vector,
called, accordingly, the
magnetic
flux (unit:
magnetic
flux.
O
Wb)
and measured
4/
,
is
4.16 and
in
It is
-L
webers (Wb), where
its
flux
through a surface S
is
denoted as O,
B
dS,
(4.95)
Wb = T m 2 We
•
.
recall that the electric flux,
defined as the flux of the electric flux density vector, D, through a designated
surface [Eq. (2.42)], and that the unit for
The generalized Gauss'
is
C.
law, Eq. (2.43), tells us that the net
through a closed surface is equal to the
analogy between electric and magnetic
total
outward
electric flux
charge enclosed by the surface.
fields naturally
Our
imposes a question: what
Section 4.8
Law
of Conservation of
199
Magnetic Flux
HISTORICAL ASIDE
George Gabriel Stokes (1819-1903), a British
mathematician and physicist, was a professor
of mathematics at Cambridge University. Stokes
was the oldest of the trio of Cambridge professors, James Clerk Maxwell (1831-1879) and Lord
Sir
dally contributed to the fame of the Cambridge
was University’s Member of Parliament, and
President of the Royal Society, the three offices
that had only once before been held by one person,
school of mathematical physics in the 19th century.
Sir Isaac
who
Kelvin (1824-1907) being the other two,
espe-
Wilhelm Eduard Weber
(1804-1891), a
at
German
Newton
(1642-1727).
the coils). Furthermore,
was used,
it
many
or similar realizations, by
in the
same
researchers and
was a professor
students over decades both to directly measure
Gottingen University.
the magnetic forces and torques due to given cur-
physicist,
Weber
received his docdissertation
toral
from
the University of Halle
in 1826, with a topic
on
theory
of
the
acoustic
reed organ pipes, and was
appointed
professor
of
and to determine (indirectly measure) the
(unknown) current from the measured torque
(as a sort of ammeter). During his later years
at Gottingen, Weber worked on a theoretical
generalization and unification of laws describing forces between charges at rest (Coulomb’s
law) and in motion (Ampere’s force law), as well
rents
Faraday’s law of electromagnetic induction
physics at Gottingen in 1831. In collaboration
as
with Karl Friedrich Gauss (1777-1855), he built
for time-varying currents (charges in accelerated
the
first
practically useful telegraph (3
km long)
in
motion).
He
also contributed to understanding
1833, to connect his physics laboratory with Gauss’
the connection between light and electromagnetic
astronomical observatory at Gottingen. They also
phenomena and establishing the link between elec-
mag-
tromagnetism and optics, which was crucial for
Maxwell’s development of electromagnetic field
worked together on
investigating terrestrial
netism (earth’s magnetic
is
His main contributions were in the areas of viscous
fluids, sound, light, spectroscopy, fluorescence, and
X rays. During a period of time, he held the post of
the Lucasian Chair of Mathematics at Cambridge,
field).
From 1836
to 1841,
they organized a network of observation stations
theory (that includes
around the world to correlate measurements of terrestrial magnetism at different locations. In 1841,
he developed the electrodynamometer, which
could precisely measure the angular displacement
of a coil with a current caused by another coil positioned perpendicularly to it and carrying the same
current. Weber used this instrument for a final validation of Ampere’s previous conclusions about
magnetic forces due to currents in wire loops (in
with Gauss,
in the early stage of the
this,
we
consider
first
Working again
jointly
a significant impact
development of a new
coherent system of units to also include electromagnetic phenomena, which gradually evolved
into the present International
The SI includes the weber,
named
the magnetic flux density
Brittle
Books
The answer
dB
System of Units
(SI).
unit of magnetic flux,
in his honor. (Portrait: AIP Emilio Segre Visual
Archives,
the net magnetic flux through a closed surface equal to?
To prove
light).
Weber provided
is:
Collection)
zero.
of a single current
element J dv in free space, as depicted in Fig. 4.26. From the Biot-Savart law and
the expression for dB in Eq. (4.6), we know that the lines of this field are circles
centered on the straight line containing the current element (Fig. 4.26). We can
imagine the entire space surrounding the element divided into thin closed tubes
of uniform cross section formed by bunches of field lines, where one such tube is
shown in Fig. 4.26. According to Eq. (4.6), again, the magnitude of the magnetic
I
200
Chapter 4
Magnetostatic Field
in
Free Space
flux density vector
is
constant along each
and thus over the entire volume
field line,
of each tube (because the tubes are thin). Consequently, the magnetic flux through
any tube is the same in magnitude at any cross section of the tube, regardless of
whether that cross section is perpendicular to the tube axis or not.
Imagine now an arbitrary closed surface S in the field, Fig. 4.26. Some of the
elementary tubes pass through S, but always an even number of times. Therefore,
the net outward flux of dB through all small surfaces representing intersections of
thin tubes and the surface S equals zero. Since such intersections cover the entire
surface S, we have
dB dS =
•
i
Our proof continues then by invoking
of which the actual magnetic flux density
Figure 4.26 For the
derivation of the law of
conservation of magnetic
free space can be
decomposed
(4.96)
0.
the superposition principle, by
B
means
of an arbitrary current distribution in
into elementary flux densities
dB due
to individual
current elements making the current distribution, so that
flux.
B
dB,
-L
(4.97)
Vcur
where
v cur
is
the
domain with currents (sources of the magnetic
field).
Applying the
integration over v cur as an operator to Eq. (4.96), and interchanging the order of
integral signs, we obtain
m
dB
-dS
=
(4.98)
0.
Vcur
Finally, substituting
Eq. (4.97),
law of conservation of
magnetic
B dS =
(4.99)
0,
•
i
flux
which completes our proof. This relation is known as the law of conservation of
magnetic flux, and also as Maxwell’s fourth equation. For obvious reasons, it is
sometimes referred to as Gauss’ law for the magnetic field. Together with Ampere’s
law (Maxwell’s second equation), it forms a complete set of Maxwell’s equations
for the magnetostatic field in free space (or any nonmagnetic medium). We notice
that the law of conservation of magnetic flux has the identical form as the continuity
equation for steady currents, Eq. (3.40).
Consider next an arbitrary contour
arbitrary shape that are both
C
bounded by
and two open surfaces, 5] and S2 with
the contour and oriented in the same
,
way - according to the right-hand rule with respect
as shown in Fig. 4.27. The magnetic fluxes through
Oi
Figure 4.27 For the proof
a
contour
is
unique.
[ B dS
S2
(S =
Si
flux
<^)
which
flux
through a contour,
Fig.
4.27
zero,
is
B dS = 0 — 0 2
1
all
that the magnetic flux through any
have a
common
(4.100)
(4.101)
,
from the law of conservation of magnetic
|
that
dS.
through the closed surface S formed by Si and
0 = 02
meaning
•
2
{
The total outward
U S2 ) is
respectively.
the surfaces are
02 = [ B
Js
and
•
Js
that the magnetic flux
through
=
to the orientation of the contour,
flux.
Hence,
(4.102)
,
number
contour bounding them
is
of surfaces of arbitrary shape
the same, provided that
all
the
r
,
Section 4.9
201
Magnetic Vector Potential
same way. This enables us to link the flux to a contour
bounded by the contour (and there is an infinite number
surfaces are oriented in the
rather than to a surface
of such surfaces), and to use the term flux through a contour or flux linked by a
is uniquely determined by the
shape of a contour and by its orientation, and is the same for any surface spanned
over the contour and oriented in accordance to the right-hand rule with respect to
contour. In other words, the flux (through a contour)
the orientation of the contour.
By analogy
with the differential form of the continuity equation for steady cur-
rents, Eq. (3.41), the law of conservation of magnetic flux in differential form
given by
V B=
We
see that the
B
(4.103)
0.
•
is
flux
conservation law
differential
field is
means
in
form
another divergenceless (divergence-free) or solenoidal
be local sources of radial magnetic field
components with respect to a point, i.e., that there exist no positive or negative
“magnetic charges” and no free north or south magnetic poles, which would correspond to electric charges, with density p [Eq. (1.166)]. Equivalently, the magnetic
field lines close upon themselves, as there are no “magnetic charges” for the lines to
begin and terminate on.
We can now summarize the two Maxwell’s differential equations governing the
vector
This
field.
that there cannot
electrostatic field in free space,
VxE = 0
V-E =
and
—
(4.104)
conservative
(4.105)
solenoidal field
field
eo
and those
for the magnetostatic field in free space,
V
x
B=
which, in a condensed form, shows
between the two
•
0,
the similarities and fundamental differences
fields.
Problems 4.25 and
:
all
V B=
and
ixq J
4.26;
Conceptual Questions (on Companion Website):
MATLAB Exercises (on Companion Website).
4.18-4.23;
MAGNETIC VECTOR POTENTIAL
4.9
In electrostatics,
electric fields
we
introduced the electric scalar potential (E) to help us describe
and evaluate the
electric field intensity vector (E).
potential due to a point charge
free space
is
Q
,
as
The
an elementary source of the
electric scalar
electric field, in
[Eq. (1.80)]
V=
1
Q
Atzsq
R
Following the analogy established in connection with Eq.
(4.106)
(4.4),
which
is
the basis of
the Biot-Savart law, the potential due to an elementary source of the magnetostatic
field,
Q\,
is
defined as
A =
A
Mo
Qy
(4.107)
s?i
magnetic vector potential
(unit:
is called the magnetic vector potential, and its unit is T
m. It is a
whose direction is very simple to determine - the same as the direction of Qy,
and whose magnitude is proportional to 1 /R (and not to 1/R 2 which is present in
This quantity
•
vector
,
T m)
•
—
202
Chapter 4
Magnetostatic Field
in
Free Space
B due
the expression for
to Q\).
By
the
same reasoning
as with obtaining the three
versions of the Biot-Savart law for volume, surface, and line currents, Eqs. (4.7)(4.9), the
corresponding integral expressions for the magnetic vector potential
are:
A
due
to
volume current
(
4 108 )
A
due
to surface current
(
4 109 )
(
4 110 )
A due to line current
In general, the solutions for the magnetic vector potential
due
.
.
.
to given current
distributions are substantially simpler than the corresponding solutions for the
magnetic flux density vector.
Let us find the curl of A. By representing the expression in Eq. (4.107) in the
spherical coordinate system shown in Fig. 4.28, in which Q is at the coordinate
origin, so R = r (r being the radial spherical coordinate), and v is z-directed.
A = A z = A cos 9 r — A sin# 0 = A r + AyO, A —
r
MoQv
4nr
(
4 111
(
4 112 )
.
)
where
A = A cos 9 =
HoQvcosd
r
(Afj,
=
0).
Aq
4nr
HoQv sin 9
= —A sin 9 — —
4nr
Using the expression for the curl
in spherical coordinates,
.
Eq. (4.85),
we have
V
x
A=
r
(rAg)
dr
mo Qv
4^r
Noting that
we
- 3A r
4>
~d9
d
—
(— sin
dr
cos 9
d
.
6>)
2
<*>
=
— —
/u 0
Qvsin9
(
4 113 )
(
4 114 )
(
4 115 )
:
;
9
4nr 2
d9
<P
.
(Fig. 4.28)
= Qv
Q\ xr = Qv i
x
?
V
x
A=
|
z x
r|
<|>
x
r
= Qv sin 9
<J>,
.
obtain
Mo
47t
Figure 4.28 Magnetic vector
due to a point charge
moving in free space, and its
components in a spherical
potential
coordinate system.
gv
2
r
.
which
R=
r.
exactly the magnetic flux density vector due to Q\, Eq. (4.4) with
is
R=
203
Magnetic Vector Potential
Section 4.9
r
and
Hence,
B=V
x A,
(4.116)
magnetic
flux density
from
potential
and the same is true for the magnetic vector potential dA due to an arbitrary
ume, surface, or line current element, Eq. (4.10). Integrating the expression V x
over a domain v with volume currents, we have
J
(V x dA)
where the integration and
=V
dA = V
x
)
x A,
vol-
dA
(4.117)
differentiation (del) operators can readily interchange
places because they are completely independent
~ the integration
is
performed with
respect to the coordinates of the source point (point at which the current density
vector
is
J),
while the differentiation
carried out with respect to the coordinates
is
of the field point (point at which the magnetic vector potential
is
A). Consequently,
the magnetic flux density vector (B) and the magnetic vector potential (A) due to
an arbitrary volume current distribution, Eq. (4.108), are related
at
an arbitrary
point in space as in Eq. (4.116), and similar proofs can be carried out for the magnetic potential
due to surface currents, Eq.
(4.109),
and
line currents,
Eq. (4.110).
This relationship, in conjunction with Eqs. (4.108)-(4.110), provides an alternative
produced by steady currents, where A is
found from the potential by differentiation. Potentials, generally, are auxiliary quantities that are used to determine fields
indirectly. Finally, we note that, contrary to V in electrostatics, A does not have any
general
method
evaluated
first
for evaluating the
B
field
by integration, and then
B
is
simple physical interpretation in magnetostatics.
On
the other side, the divergence of the magnetostatic potential
which
A
is
always
consequence of the steady current density J being a
divergenceless vector, i.e., of the continuity equation in Eq. (3.41) or (3.40). Namely,
in analogy to Eq. (4.117), we can apply the divergence operator to the integral
expression in Eq. (4.108) and write
zero,
V
•
is,
essentially, a
A=
(4.118)
is the closed surface bounding v, and the use is made of the divergence
theorem, Eq. (1.173), to convert the volume integral to a surface (flux) one. For
the special case when v is a spherical domain centered at the field point (where the
potential is being computed), 1/R can be brought out of the flux integral, as S in
where S
R and the integral, in turn, becomes zero,
by virtue of the continuity equation, Eq. (3.40). In the general case, the
arbitrary domain v can be subdivided into a stack of concentric spherical layers with
that case
is
§s J dS =
a spherical surface of radius
,
0,
each of them being a part of a full spherical shell of thickness d R centered at the
with the current of density J flowing through that part and no current
field point,
The same conclusion about a zero flux integral
based on the continuity equation can then be derived for the surface of each shell,
and thus
in the rest of the shell, outside v.
V A=
•
0,
(4.119)
divergence of
currents
which holds true for the magnetic potential due to surface and
as well.
line steady currents
A
due
to
steady
V
204
Chapter 4
Magnetostatic
Field in Free
Space
The magnetic
expressed
flux through an arbitrary contour, Eq. (4.95), can
terms of the magnetic vector potential,
in
4>
jT_B-dS
=
now
= y (VxA)-dS,
be
(4.120)
which, by using Stokes’ theorem, Eq. (4.89), becomes
magnetic
flux
O=
from the
£
potential
Adi.
(4.121)
The orientation of the contour C and the surface S are in accordance to the righthand rule, as shown in Fig. 4.29. Eq. (4.121) represents a means for determining the
magnetic flux by evaluating a contour integral (of A) rather than a surface integral
(of B), which is very convenient in some computations and derivations. Note that
this is yet
same
for
another proof that the magnetic flux through a contour
surfaces bounded by C ).
is
unique (the
all
Figure 4.29 For the
evaluation of the magnetic
through
by
integrating the magnetic
flux
a surface
vector potential along
Problems: 4.27; Conceptual Questions (on Companion Website): 4.24 and 4.25;
MATLAB Exercises (on Companion Website).
its
boundary.
PROOF OF AMPERE
4.10
We
now ready
are
On
A.
to prove
S
LAW
Ampere’s law by
utilizing the
magnetic vector potential,
we have, by
the left-hand side of the differential form of the law, Eq. (4.83),
means of Eq.
(4.116),
VxB = Vx(Vx A).
(4.122)
Applying formally (symbolically) the vector identity for expanding the vector
triple
product, 6
a x (b x
to the product
V
c)
=
b(a
•
—
c)
c(a
•
b),
(4.1
23)
x (V x A), we get the following identity for expanding the curl of
the curl of an arbitrary vector field (A):
V
The
div
first
x (V x A)
term
in this
A=V A=
•
= V(V
A) - (V V)A
expansion
•
is
V2A
is
x
this,
in
it
is
24)
zero, because
Eq. (4.122) becomes
B=—
2
A.
called the Laplacian of the vector function
Eq. (1.100), can be written
(4.1
•
grad(div A), and in our case
0 [Eq. (4.119)). With
V
Vector
= V(V A) - V 2 A.
(4.125)
A, which, having
in
mind
Cartesian coordinates as
2
'2
2 \
(
6
A
cross (vector) product of a vector (a) with a cross product of two (other) vectors (b and c)
the vector triple product.
Note
also that the vector triple product identity in Eq. (4.123)
the “bac-cab" or “back-cab" identity, as the
equation.
word “bac-cab" can be read on
is
is
called
known
as
the right-hand side of the
z
Section 4.1 0
d
2
Ax
~dx 2r
'd
+
We
2
A x 3 2A x \ „
X+
+
~d^ l 2 ')
A
7
3x 2
2
3
+
+
d
2
A
7
d
+
2
A
7
\
'd
2
Av
dx
+
2
2
Av
3
+
2
3z
3y
3y
Law
A,
2
„
z.
(4.126)
3 z2
see that Cartesian components of the Laplacian of
components of A, where the
Eq. (2.94), and hence 7
the corresponding
scalar field,
3
Proof of Ampere's
A
equal the Laplacian of
operator
latter
is
the Laplacian of a
V 2 A = V 2 A* x + W 2 A y y + V 2 A Z z.
(4.127)
Laplacian of a vector in
Cartesian coordinates
From Eq.
(4.108),
A = Ax x + Ay y + A
=
z
Jx dv
Mo /*/*/*
„
(Jx
Mo
_
z
4tt
[
JVv
f Jy
dv
x
+ Jy y + Jz z) dv
R
f
„
Jz dv \
(4128)
and we conclude that each Cartesian component of A is actually produced by the
same component of the current density vector, namely, A x is produced by /*, and so
on. Recalling the expression for the electric scalar potential due to a volume charge
in free space, Eq. (1.82), we identify the duality between V due to p and A x due
to/*:
V=
1
f
p dv
Ax =
,
i
/ ^-=r
47T£ 0 Jv
R
Mo / Jx dv
^l-Z—.
(4.129)
R
4 n Jv
By
the same duality principle, in turn, there must be a differential equation for
which has the same form as the differential equation for V, that is, Poisson’s
equation [Eq. (2.93)]. Therefore, by changing the variables, we get
Ax
V2 E = - —
=
-mo/*.
V 2A Z =
~p-qJz
VL4*
£o
(4.130)
Similarly,
V 2 A y = — Mo/y
and substituting
all
and
these back in Eq. (4.127),
V2 A =
This
is
(4.131)
,
-MoJ-
(4.132)
a second-order vector partial differential equation
(PDE)
for the magnetic
vector potential of a volume current in free space (or any nonmagnetic medium),
usually referred to as the vector Poisson’s equation.
in solving (analytically or
We use it
numerically) for
can be used as a starting point
we can
A
V
Note
It
to a given current distribution J.
here for deriving Ampere’s law. Namely, returning to Eq. (4.125),
for the Laplacian of
and obtain
now substitute
7
A due
x
B = -V 2 A = moT
(4.133)
does not have equally simple counterparts in cylindrical and
and 1.26). In these systems, the Laplacian of a vector func2
tion (A), V A, is computed, from Eq. (4.124), as V 2 A = V(V A) — V x (V x A) = gradfdiv A) — curl
(curl A), namely, as the gradient of the divergence of A minus the curl of the curl of A, using
the expressions for the gradient, divergence, and curl in Eqs. (1.105), (1.108), (1.170), (1.171), (4.84),
and (4.85).
that the expansion in Eq. (4.127)
spherical coordinate systems (Figs. 1.25
vector Poisson 's equation
205
206
Chapter 4
Magnetostatic
Field in Free
which
Space
Eq. (4.83), and this completes our proof of the Ampere’s law in differtheorem, Eq. (4.89), which is derived from the mathematical
is
ential form. Stokes’
definition of the curl of a vector, gives
Note
its
integral equivalent, Eq. (4.49).
that the proof carried out in this section
of the magnetic vector potential, which
is
essentially based
on the concept
defined by Eq. (4.107) and related to the
magnetic flux density vector by Eq. (4.116). This latter relation is obtained from
Eq.
(4.4),
we
which
is
is
the rudimentary version of the Biot-Savart law. This
means
that
have, in fact, derived Ampere’s law from the Biot-Savart law.
4.1
MAGNETIC DIPOLE
1
A small
loop with a steady current constitutes the magnetic equivalent of the elec-
dipole of Fig. 1.28, and
is referred to as a magnetic dipole. The reason for
and what we mean by “small” will soon be evident. A magnetic dipole is
characterized by its magnetic moment, defined as
tric
this
m = 7S,
magnetic dipole moment
where
/
is
(4.134)
the current intensity of the loop and S
=
Sn
is
the loop surface area
vector, oriented in accordance to the right-hand rule with respect to the reference
direction of the current.
Note
that
m analogous to the electric dipole moment, p,
m A m
is
defined by Eq. (1.116). The unit for
We would like
is
2
•
.
magnetic vector potential and the
magnetic flux density vector due to a magnetic dipole at large distances compared
with the loop dimensions. To this end, we consider a rectangular loop with sides
a and b, shown in Fig. 4.30. Let the axis of the loop coincide with the z-axis of a
spherical coordinate system and the center of the loop be located at the origin of
the system. Far away from the loop ( r
a, b ), the loop is observed as being small
and the magnetic vector potential of the loop can be evaluated as that due to four
line current elements, given by / a, 7b, /(—a), and /(— b), using Eq. (4.110). Let us
first calculate the potential at a point P due to the pair of parallel elements of length
to find the expressions for the
a. Fig. 4.30,
(4.135)
where r\ and r2 are the distances of the point P from the centers of the first and second element, respectively. Noting that the position vector of the first element with
respect to the second element is d = — b, we can now use the same approximations
z
Figure 4.30 Magnetic dipole.
X
A,•aa
A
A
:
Section 4.11
207
an electric dipole, with which Eq. (4.135) becomes
as in deriving Eq. (1.115) for
i^o/a
d
•
Mo^a(b
r
r)
(4.136)
4 nr2
47T
due to the other pair of elements, those of length b
Similarly, the potential
Magnetic Dipole
,
is
given by
~ ^ oIh(a
~
4 icr2
a
Abb
'
?)
(A"
'
1
J
By superposition and using the formula for the vector triple product in
Eq. (4.123), the total magnetic vector potential of the dipole turns out to be
A = A aa + A bb =
As
the magnetic
moment
[b(a
•
of the loop
r)
-
is
(Fig. 4.30)
a(b
r)]
•
= £j^(a
m = /S = lab z = /a
A can be expressed in terms of m as
_ ixq m x r
(4.1 38)
r.
(4.139)
b,
(4.140)
r2
471
We
x
x b) x
moment and
on the shape of the
see that the magnetic potential depends on the dipole magnetic
the position of the field point with respect to the dipole, and not
mxr = msin0f
be arbitrary (not only rectangular). Finally, since
which directly comes from Eq. (4.114), Eq. (4.140) can be rewritten as
loop, which can
Horn sin 9
A=
*
(4.141)
4nr2
magnetic dipole potential
We
conclude that the magnetic vector potential due to the magnetic dipole has
only a 0 -component in the spherical coordinate system, which is a function of
coordinates r and
The magnetic
9.
flux density vector of the dipole
formula for the curl in spherical coordinates, Eq.
is
now determined
applying the
(4.85), to the expression for
A in
Eq. (4.141),
B=V
x
A=
(sin 6
rsin# 3 9
ix 0
m
47rr 3
Comparing Eqs.
(1.117)
and
2 cos 9 r
(4.142),
1
3
r
dr
(rA<p) 0
#)
+ sin 9 0)•
we observe
(4.142)
that the
E
and
B
field of
the electric and magnetic dipoles, respectively, are identical in form, so that
the corresponding field lines have identical shape.
large distances
from the dipoles
trated in Fig. 4.31.
normalized
We
field lines
However,
this is true
(relative to their dimensions),
which
only at
is
illus-
see that, although identical far from the sources, the
due to the two dipoles are fundamentally different close
on the two charges forming the electric
to them; the electric field lines terminate
dipole,
whereas the magnetic
field lines close
upon themselves through
the current
loop.
We shall see in the next chapter that the concept of a magnetic dipole is fundamental for understanding the behavior of magnetic materials, much like the
electric dipole was used in Chapter 2 in studying the electric field in the presence of
dielectric materials. The field of a magnetic dipole, Eq. (4.142), is also used, again in
magnetic dipole
field
208
Chapter 4
Magnetostatic Field
in
Free Space
Figure 4.31 Normalized
electric field lines of
an
electric
dipole (a) and magnetic field
lines of a
magnetic dipole
(b).
parallel to the electric field of
an electric dipole, as an approximation for the
and quasistatic (low-frequency) magnetic
which is important in EMI considerations.
field
produced by an
Problems 4.28-4.31; Conceptual Questions (on Companion Website):
:
MATLAB
Exercises (on
Companion
Website).
static
electrical device,
4.26;
Q
Q
The Lorentz Force and
Section 4.12
4.12
THE LORENTZ FORCE AND HALL EFFECT
From
the definition of the electric field intensity vector, Eq. (1.23),
electric force
on a point charge
we have
Eq.
that the
Q situated in an electric field of intensity E is
Fe = QE.
Similarly,
209
Hall Effect
(4.1) tells us that the
(4.143)
magnetic force on a point charge
electric force
on a
particle
Q moving at a
velocity v in a magnetic field equals
Fm = Qy X
B,
(4.144)
magnetic force on a particle
where B is the flux density vector of the field. Finally, the force on a moving charge
due to both an electric and a magnetic field is obtained by superposition,
F = Fe + Fm —
This equation
+
tric
is
magnetic
known
=
Q E + Qx
x B.
(4.145)
as the Lorentz force equation or law.
electromagnetic) force on the particle
is
The
Lorentz force
total (elec-
called the Lorentz
force.
An
and important manifestation of the motion of free charges in
is the Hall effect, which we
describe briefly here. We consider a conducting strip of width a situated in a uniform
steady magnetic field of flux density vector B that is perpendicular to the strip, as
interesting
a material under the influence of the Lorentz force
depicted in Fig. 4.32. Let a steady current of density J flow through the
charges constituting the current can be positive or negative
in a semiconductor),
and
shows both
Fig. 4.32
cases.
(e.g.,
Due
strip.
The free
holes and electrons
to the
magnetic force
given by Eq. (4.144), the charges move (deflect) across the strip in the direction
perpendicular to both B and J, which results in a charge separation on the two
sides of the strip. In Fig. 4.32(a), the free charges are positive
velocity)
is
in the
same
direction as J [Eq. (3.11)],
x
v<j
(
>
0),
v
=
VH
(a)
vj (drift
B is directed to the right, and
move to the right. In Fig. 4.32(b), on the other
hand, the free charges are negative (
< 0), Vd is in the direction opposite to J, Vd x
B is directed to the left, and Fm is again directed to the right (because Q < 0); hence,
so
is
Fm
;
thus, the positive charges
edge of the strip. Accumulated charges
E) across the strip. This field, in turn, acts on
the free charges with an electric force, Eq. (4.143), which is in the opposite direction
to the magnetic force. In the equilibrium, the two forces are equal in magnitude, i.e.,
the Lorentz (total) force on the charges is zero,
the negative charges
produce an
end up
at the right
electric field (of intensity
Lh
(b)
F = Q(E +
Vd x B)
=
(4.146)
0,
from which,
The voltage between the
strip
The
is
known
as the Hall voltage,
left in Fig. 4.32(a),
the polarity of the Hall voltage
Note that
is
v&Ba.
this represents a
p-type or n-type.
tells
method
(4.148)
and the
is
free charges
and
(b)
negative free charges.
edges amounts to
direction of the Hall voltage drop
charges [from right to
(4.147)
Vd-B.
Eh = Ea =
effect.
in a
material with (a) positive
E=
This voltage
Figure 4.32 Hall effect
effect itself
is
called the Hall
different for positive
and from
and negative
left to right in Fig. 4.32(b)]. So,
us the sign of free charge carriers in a material.
for determining
whether a given semiconductor
Hall voltage
210
Chapter 4
Magnetostatic Field
Free Space
in
HISTORICAL ASIDE
Hendrik Antoon Lorentz
(1853-1928), a Dutch mathematician and physicist,
a professor of mathematical
physics
the
at
(of space and time coordinates), which describe
time dilation and length contraction for a body
moving
at velocities close to the speed of light
and represent the foundation of Einstein’s (1879-
He
University of Leiden, was
1955) special theory of
the winner of the
the Lorentz force law for a
parti-
cle in
fields.
Nobel Prize
He
further
Maxwell’s
in
1902
Physics.
developed
electromag-
and
proposed the electron
theory according to which oscillating electrons
within atoms constitute miniature equivalent
Hertzian radiators that emit light. In 1904, he
netic theory of light,
Edwin
Herbert
Hall
(1855-1938), an American
was a profesHarvard University.
physicist,
sor at
He
received his doctor-
from Johns
Hopkins University under
Professor Henry Augusate in physics
tus
Rowland (1848-1901),
who
was
one
of
the
world’s most brilliant physicist of the last quarter of the 19th century.
introduced his famous Lorentz transformations
As
a part of his dissertation
work. Hall pursued the question as to whether
the resistance of a current-carrying conductor was
affected by the presence of an external magnetic
field. In experiments with guidance from Professor
Example 4.17
Hall
relativity.
formulated
moving charged
the presence of electric and magnetic
Lorentz
is
also the author of the so-called Lorentz-
Lorenz formula, jointly with Danish physicist
Ludwig Lorenz (1829-1891), who discovered it
independently. The formula provides a mathematical relationship between the index of refraction
(of light) and density of a medium. (Portrait: AIP
Lande
Emilio Segre Visual Archives,
Rowland
in 1879,
Collection)
he used a conductor
mounted on
of a strip of a gold leaf
in the
form
a glass plate
and placed it between the poles of an electromagnet such that the magnetic field lines were
perpendicular to the current flow in the
strip.
What
he observed was the development of a significant
transverse electric field in the conductor and the
associated voltage across the strip as the result of
the applied magnetic
ings in the
Magnet on
famous
field.
article
Hall published his find-
“On
a
Electric Currents” in
New Action of the
American Journal
November of 1879, and this
phenomenon soon came to be known as the “Hall
of Mathematics in
effect.”
He was
Harvard
10,
appointed professor of physics at
Como, Italy -Sept.
in 1895. (Portrait: “Voltiana,”
1927 issue, courtesy AIP Emilio Segre Visual Archives)
Element for Measuring the Magnetic Flux Density
A
Hall element for measuring the magnetic flux density is in the form of a strip with width a
and thickness d, shown in Fig. 4.33. The concentration of free charge carriers in the strip is
/V v and the charge of each carrier is Q. The strip carries a steady current of intensity /, and
the magnetic flux density vector is perpendicular to the strip. A voltmeter shows a voltage
V \2 between the strip edges. What is the algebraic intensity of the magnetic flux density
a
Figure 4.33 Hall element
for
vector (B) with respect to the reference direction indicated in Fig. 4.33?
Solution
From
Eqs. (3.11) and (3.5), the current density
in
the element can be written as
measuring the magnetic
flux density; for
Example
4.1 7.
J
= Nv Qv d =
—
,
(4.1
49)
Section 4.1 3
=
which yields v d
Fig. 4.33,
namely,
Evaluation of Magnetic Forces
211
I/(Nv adQ). The Hall voltage is given in Eq. (4.148), and Eh = V 21 in
opposite to the measured voltage V 12 Hence, the magnetic flux density
it is
.
turns out to be
=
B=
50)
Lorentz Force due to a Rotating Charged Contour
Example 4.18
A
(4.!
vd a
vd a
uniformly charged circular contour of radius a and total charge
Q
rotates in free space
uniform angular velocity w = w z, as shown in Fig. 4.34. A charged
particle q moves with a uniform velocity v = v y along a path that belongs to the plane z — a
and is parallel to the y-axis. Find the Forentz force on the particle at an instant when it is at
about
axis with a
its
the point P(0,
The
Solution
with z
0, a),
above the center of the contour.
2^/2
E=
The time period
for
16nsQa 2
z
in Fig. 4.34
is
given by Eq. (1.44)
^
=
one rotation of the contour
_
T=
(full
P
electric field intensity vector at the point
= a,
Ez.
(4.151)
is
2n
—
w
(4.152)
z
Fe
1
angle divided by the angular velocity). Noting that the total charge of the contour,
Q, passes any given reference point on the contour during the time T, we conclude that
is equivalent to a line current along the contour of intensity
the rotating charged contour
[Eq. (3.4)]
,
_QD _QDu;W
T
From Eq.
(4.19), the
(4.153)
'
magnetic flux density vector due to
B
/r 0
/V2
„
z
(4.145), the
this current at the point
P
(z
noQwV2
—- z„ = B z.„
=—
Forentz force on the charge q comes out to be
qQV2
F = q(E + vyxB) = qEz + qvB x =
2
=
a)
is
(4.154)
I6na
8a
Using Eq.
2n
(eojUQVH'a
x+
(Fig. 4.34)
z).
(4.155)
16^-eofl
Problems'. 4.32.
EVALUATION OF MAGNETIC FORCES
4.13
Eq. (4.144) gives the Lorentz magnetic force (i.e., the magnetic component of the
total Lorentz force) on a single point charge moving in a magnetic field. If we have
many
charges constituting a current in some domain,
principle as in Eq. (4.5)
element J dv
is
B
is
utilize
the superposition
given by
dFm
where
we
and conclude that the magnetic force on a volume current
= (Jdv)xB,
(4.156)
the flux density vector of the external magnetic
Eq. (4.10), the magnetic force on a surface current element
dF m
=
(J s
dS) x B,
field.
Then, from
is
(4.157)
TV
Q
—
Figure 4.34 Lorentz force
on a charged particle
moving above a rotating
charged contour; for
Example 4.1 8.
212
Chapter 4
Magnetostatic Field
in
Free Space
and that on a
line current
element
dF m
= /dlxB.
(4.158)
Integration of Eqs. (4.156)-(4.158) leads to the following integral formulations for
the total magnetic force for volume, surface, and line current distributions:
magnetic force on volume
(4.159)
current
magnetic force on surface
(4.160)
current
magnetic force on
line
current
(4.161)
In the last integral, I can be taken out of the integral sign because
it
is
always
constant along the line (continuity equation for steady currents).
homogeneous conductor of arbitrary cross section with
we have J = const [Eq. (3.82)]
x B = const and Eq. (4.159) becomes
In the case of a straight
a steady current placed in a uniform magnetic field,
and B
=
const, so that J
Fm =
where
v
and
/
x B)
(J
are the
surface area of
its
l
dv
=
(Jv) x
B=
(J57)
xB =
volume and length of the conductor,
cross section, the direction of the vector
direction of the current flow along the conductor, and
intensity of the conductor, /
F m on
a straight conductor
a uniform magnetic
(751)
|1|
=
/.
x B,
(4.162)
5 is the
same as the
respectively,
is
I
the
Introducing the current
= 75, we obtain
Fm
in
=
71
x
B
(4.163)
field
Example 4.19
Two
Force Between
parallel, very long
Two Long
and thin wires
flowing in the same direction.
Parallel
Wires with Current
in air carry currents of intensities
The distance between
the wire axes
d.
is
I\
and
/2
,
both
Find the magnetic
forces on wires per unit of their length.
Solution
shows the cross section of the two wires. From Eq. (4.22), the magnetic
due to the current in the first wire (/i), assuming that it is infinitely long,
of the second wire is
Fig. 4.35
flux density vector
at the axis
B,
doh
2nd
«
(4.164)
y
we can assume
uniform and given
Since the wires are thin (as compared to the distance between their axes),
that the magnetic field across the entire cross section of the second wire
Figure 4.35 Evaluation of the
force
between two long
parallel
current-carrying wires; for
Example 4.19.
is
I
Section 4.1 3
Evaluation of Magnetic Forces
by Eq. (4.164), so that Eq. (4.163) can be used. Hence, the force on the part of the second
conductor that is / long equals
Mo hhl
Fm 2 — 12(H) x B —
2nd
*
Mo hhl
x y
z
1
„
(4.165)
2nd
and the force per unit of its length
dohh
Fm2
Ki = ~T
The per-unit-length force on the
We
same
conductor
first
F'
is
ml
see that the magnetic force between the wires
and
-
(4.166)
2nd
= — F^ 2
.
attractive
is
if
the currents are in the
and repulsive if they are
in opposite directions (/ 1/2 < 0). So, just in contrary to charges Q 1 and Q 2 and Coulomb’s
law, “like” currents (in parallel wires) attract and “unlike” currents repel each other (also see
direction,
i.e., if
both
I\
I2 are either positive or negative,
Fig. 4.2).
Force on a Loop near a Long Wire
Example 4.20
A
long straight wire carries a steady current of intensity
lies in
the
same plane
as the wire, with
sides (of length a) perpendicular.
of the loop
two
I.
A
rectangular conducting loop
sides (of length b ) parallel to the wire
The distance between
The loop carries a steady current of the same intensity, and the
shown in Fig. 4.36. Determine the net magnetic force on the loop.
is c.
currents are
and two
the wire and the closer parallel side
directions of
Solution The magnetic flux density vector due to the current in the long wire at any point in
is normal to the plane, and at a distance x from the wire its magnitude is
the plane of the loop
= ^~.
2nx
B(x)
The
forces
on each
given in Fig. 4.36.
in
magnitude and with opposite
it is
forces
on the
and
sides 1
and
their directions are
directions,
—
Fm4
(4.168)
.
3 are also in opposing directions, but their
from the long wire
different because of their different distances
v
Fm
1 =
(4.163),
obvious that the forces on the sides 2 and 4 are equal
Fm2
The
from Eq.
side of the loop are obtained
From symmetry,
(4.167)
=
tutu \
IbB(c)
do
—
2
(x
=
c
and x
magnitudes are
= c + a),
b
(4.169)
(repulsive),
2nc
2
Fjn 3
=
IbB(c
Mo I b
2n(c + a)
+ a)
(4.170)
(attractive).
y
©
j
(D
—
L
c
/
F ml
B
|fVl
1
a
Fm2
(D
Fm3
b
x
©
Figure 4.36 Evaluation of the
magnetic force on a rectangular
current loop near a long wire
with current; for Example 4.20.
213
.
214
Chapter 4
Magnetostatic
Field in Free
Space
The
total force on the loop, assuming that it is rigid (i.e., the loop maintains
under the influence of the magnetic forces on its sides), is hence given by
its
shape even
Fm — F m T F m2 + Fm3 + F m 4 — F m T F m 3
i
l^pl
i
2
b /I
2n
\c
\
1
_
c
.
_
+ aj
2
mo I ab
2nc(c + a)
.
(4.171)
is pushed away from the long wire. Note that if the polarity of either (but not
both) of the two currents (in the wire and the loop) were reversed, the loop would be pulled
Thus, the loop
toward the long wire.
Force on a Loop in a Uniform Magnetic Field
Example 4.21
Prove that the net magnetic force on a contour of arbitrary shape with a steady current
uniform magnetic field is zero.
Solution
For a uniform magnetic
Fm
Figure 4.37
In
it is
=/®
B=
dl
x
const, so that Eq. (4.161)
B=
/
(b
dl
becomes
x B.
(4.172)
view of the
head-to-tail rule for vector
addition,
field,
in a
For any closed path,
obvious that
dl
=
(4.173)
0,
the integral of dl along a
closed path
for
is
always zero;
which
is
evident from Fig. 4.37, and hence
Fm =
0.
Example 4.21
Torque on a Current Loop
Example 4.22
A
in a
Uniform Magnetic
Field
is situated in a uniform steady magnetic field of flux
There is a steady current of intensity I in the loop. The
angle between the plane of the loop and the plane normal to the vector B is 6. The loop is
mounted such that it is free to rotate about the axis O-O' which is perpendicular to the plane
of the drawing. Find (a) the net force and (b) the net torque on the loop.
rigid
square loop of side length a
density B, as
shown
in Fig. 4.38(a).
Solution
(a)
From Eq.
(4.172), there
is
no net force on the loop,
magnetic forces on the loop sides
(b)
The
is
i.e.,
the vector
sum
of individual
zero.
on the sides of the loop that are normal to the axis O-O' tend to stretch the
do not produce torques on it. The torques (moments) of the forces on the sides
and 2 of the loop (sides parallel to the axis O-O') calculated with respect to the center
forces
loop, but
1
of the loop are
Ti
=
rj
x Fm i
Figure 4.38 Evaluation of the torque
in a
uniform magnetic
field, (b)
field: (a)
and
T
of
T2 =
x
F m2
,
magnetic forces on a current loop
position of the loop relative to the magnetic
magnetic forces on loop sides producing
relationship with the magnetic
r2
moment m
and (c) the
Example 4.22.
a torque,
of the loop; for
(4.174)
Problems
where Fm i and Fm 2 are the magnetic forces on the
respectively,
sides,
and
ri
and
r2
215
are
the position vectors of the centers of sides with respect to the loop center [Fig. 4.38(b)].
Since
Fm i+Fm2 =
the resultant torque on the loop
T=
Ti
+T2 = ri
x Fm i
(4.175)
0,
given by
is
+r2
x Fm2
=
-
(ri
r2 )
x Fm i
= ri 2
x Fm i.
(4.1 76)
r X2 = ri - r 2 joins the point of application of F m2 to that of F m i and is independent of the choice of origin of the two vectors iq and r2 Therefore, the torque is
also independent of the choice of origin, i.e., it is the same when calculated about any
reference point, provided that the total force on the loop is zero.
Using Eq. (4.163), the magnitudes of the forces Fm i and Fm2 are
The vector
.
Fm i=Fm2 =
IaB
(4.177)
,
which, substituted in Eqs. (4.174), gives the magnitudes of the corresponding torque
vectors:
m =
T2 =
T\
(|ri
|
=
a/2).
„
|ri
x Fm i|
=
a „
-F
m \ smd
=
la
2
B sin 6
..
,
(4.1
Hence, the magnitude of the resultant torque vector
78)
T is
T = 2TX = Ia2 Bsm9,
(4.179)
and its direction is shown in Fig. 4.38(b). Of course, the same is obtained from Eq. (4.176).
Noting that the angle between the unit normal n on the loop surface oriented in
accordance to the right-hand rule with respect to the reference direction of the loop
current and the vector
B
is
also 9 [Fig. 4.38(c)],
we conclude
that the resultant torque of
magnetic forces on the loop can be compactly expressed as
T=
m x B,
(4.180)
torque on a current loop
uniform magnetic
m
where
is
the magnetic
moment
in
a
field
of the loop, given by Eq. 4.134, so that
T = |m x
B|
= mBsm.6 =
2
Ia Bsind.
(4.181)
a general expression for the torque on a current loop of arbitrary shape
uniform magnetic field. We see that the torque on the loop always tend
to turn the loop so as to align the vectors
and B. In other words, it tends to align
the magnetic field produced by the loop current (which coincides with the direction of
- see Figs. 4.6 and 4.8) with the applied (external) magnetic field that is causing the
torque. Finally, the magnetic field across a small current loop, i.e., a magnetic dipole
(Fig. 4.30), can always be considered as locally uniform, which means that Eq. (4.180)
gives us the torque on a magnetic dipole (with any shape and the magnetic moment m)
in any (generally nonuniform) magnetic field (with the local flux density B). We observe
the analogy with the torque on an electric dipole, T = p x E, in Eq. (2.3).
Eq. (4.180)
and
is
size in a
m
m
Problems 4.33-4.39; Conceptual Questions (on Companion Website): 4.27-4.30;
:
MATLAB Exercises (on Companion Website).
Problems
4.1.
Rectangular current loop. Consider a rectangular loop with sides a and b in air. If the
loop carries a steady current of intensity /, find
the magnetic flux density vector at an arbitrary
point along the axis of the loop perpendicular
to
its
plane.
216
4.2.
Chapter 4
Magnetostatic
Triangular current loop.
Field in Free
A
loop
Space
form
in the
4.6.
Solenoids with different length-to-diameter
of a triangle representing a half of a square
ratios.
of side a carries a steady current of intensity
core and
/,
as
shown
in Fig. 4.39.
The medium
Calculate the magnetic flux density vector at
P located
a point
of the solenoid
vertex of the
at the fourth
of the
P
(or last) wire turn,
first
<
oo) for the solenoid radius equal to (a)
a
=
25
cm and
= 2 cm, respectively.
(b) a
Helmholtz
coils. Fig. 4.41
Figure 4.39 Triangular
very short
coils,
current loop; for Problem 4.2.
wire, in
Current loop with circular and linear parts.
shows a wire contour composed from
air.
a half of a circle of radius a
and a half of a
is
d
situated in air
carries a steady current of intensity
I.
shows two
each with
N
identical
circular turns of
The distance between the
d,
is
and the wire turn
centers
radii are a.
The coils carry steady currents of equal intensities, /, and are oriented in the same way. When
Fig. 4.40
square of side 2a. The contour
<
z
of the coils
and
and sketch the
function B(z) along the solenoid axis (— oo
4.7.
4.3.
=
is /
field at the solenoid center, at the center of the
250th (or 750th) wire turn, and at the center
square.
a
= 1 A. The length
50 cm. Calculate the B
carrying a steady current I
air.
is
Consider a solenoid with a nonmagnetic
N = 1000 tightly wound turns of wire
— a,
the magnetic field near the center of
(midway between the two
the structure
Find
coils)
approximately uniform, and the structure is
referred to as Helmholtz coils, (a) For an arbiis
the magnetic flux density vector (a) at the central
point
the point
O and (b) at a point la apart from
O along the line perpendicular to the
d (d
trary distance
a), find the
expression for
the magnetic flux density B(z) at an arbitrary
plane of the contour.
point along the axis of the coils (z-axis), assuming that
B due
to each of the coils equals
N
times the flux density of a single turn of wire
(circular current loop, Fig. 4.6) at the respec-
a
and three linear
Problem 4.3.
current distribution. There
is
z
b with a hole of radius a (a
<
=
d/2 (and any
verify that d
d
a
=
2
that
0 at
d, relative to a), (c)
£/dz 2
=
a (note that even
Then
C when
3
d 5/dz 3 = 0 at C in this
0 at the point
case).
surface current over a circular surface of radius
lies in
=
the center of the structure (point C), so for
parts; for
The surface
d5/dz
contour composed from
semicircle
4.4. Circular surface
Show
Figure 4.40 Current
tive coil location, (b)
b) in free space.
the plane z
=
d
0 of a cylin-
drical coordinate system, with the coordinate
origin coinciding with the surface center.
surface current density vector
(a
<
r
<
b ), where 7s o
is
is
Js
The
= J o(a/r)ty
s
Compute
a constant.
the magnetic flux density vector along the
z-axis.
4.5.
Magnetic field of a rotating charged disk. A ciris uniformly charged over
its surface by a charge of density p s The disk
uniformly rotates in air about its axis (perpen-
Figure 4.41
cular disk of radius a
Helmholtz coils;
for Problem 4.7.
.
dicular to the disk) with an angular velocity
4.8.
Spherical
coil.
Consider a
w. Find the magnetic flux density vector at
N
an arbitrary point along the axis of rotation.
formly and densely
Assume
netic sphere of radius a.
that the charge distribution over the
disk remains the
same during the
rotation.
coil consisting of
turns of an insulated thin wire
the wire
is /.
in
wound
uni-
one layer on a nonmag-
The current through
The surrounding medium is air.
217
Problems
Find the magnetic flux density vector
sphere center.
4.9.
at the
calculate
the
everywhere
Two parallel strips with opposite currents. Two
4.12.
(0
magnetic flux density vector
r < 00 ).
<
An
Rotating cylinder with a surface charge.
long conducting nonmagnetic cylin-
parallel, very long identical strip conductors of
infinitely
width 2 a carry currents of the same intensity 7
der of radius a
The distance between
the strips is 2 a as well. The cross section of the
structure is shown in Fig. 4.42. The permeability everywhere is p-o- Compute the B field at
surface with a charge density p s The cylinder rotates in air about its axis with a uniform
and opposite
directions.
the center of the cross section (point O).
is
uniformly charged over
angular velocity w. Find the
cylinder and outside
4.13.
B
field inside the
it.
Rotating nonuniformly charged hollow cylinder.
An infinitely long hollow cylinder of inner
radius a and outer radius b (a
Figure 4.42 Cross section
of
two
its
.
<
b) in air
is
charged with a volume charge density p(r)
=
Por/a (a < r < b), where po is a constant and
r the radial distance from the cylinder axis.
parallel strip
conductors with currents
The
same magnitude
and opposite directions;
for Problem 4.9.
of the
cylinder uniformly rotates about
its
axis
with an angular velocity w. Assuming that
the charge distribution of the cylinder does
not change during the rotation, determine the
4.10.
Magnetic
ductor.
of a hollow cylindrical con-
field
A
through an infinitely long hollow cylindrical
copper conductor of radii a and b (a < b ).
The cross section of the conductor is shown in
Fig. 4.43.
magnetic flux density vector everywhere.
steady current of intensity 7 flows
The medium
the conductor
is
4.14.
Two parallel infinite planar current sheets. Two
parallel infinite planar current sheets in air
have the same uniform surface current density
7S The distance between the sheets is d. Find
the magnetic flux density vector everywhere if
the currents of the sheets run in (a) the same
direction and (b) opposite directions.
.
and outside
nonmagnetic. Find the magin the hole
netic flux density vector everywhere.
4.15.
Magnetic field inside a thin plate with current.
A thin copper plate of length b, width a and
,
thickness
Figure 4.43 Cross section of
intensity
a hollow cylindrical
Magnetic
field
same
« a) carries a steady current of
shown
in Fig. 4.44.
The
Neglecting the end
plate
effects,
is
use
to find the distribution of the
magnetic flux density vector inside the plate
[note that this is mathematically completely
analogous to the application of a similar integral equation in Eq. (6.146) and Fig. 6.25(b)],
of a triaxial cable. The cross
cylindrical conductors, looks the
as
air.
Ampere’s law
section of a triaxial cable, having three coaxial
7,
situated in
conductor with a steady
current; for Problem 4.1 0.
4.11.
d (d
as
the cross cut of the system of three concentric
spherical conductors in Fig. 1.56.
The radius
of the inner conductor of the cable
1
mm,
is
conductor are b
= 2 mm
and
c
= 2.5 mm,
= 5 mm
those of the outer conductor d
e
=
5.5
tric,
all
mm. The
=
and
and
cable conductors and dielec-
as well as the surrounding
Figure 4.44 Thin
conducting plate with
medium, are
nonmagnetic. Assuming that steady cur-
rents of intensities I\
/3
a
the inner and outer radii of the middle
= —1 A
—
2 A, I 2
= —1
a
A, and
steady current; for
Problem 4.15.
flow through the inner, middle,
and outer conductor, respectively,
with respect to the same reference
all
given
direction,
4.16.
Magnetic
tor.
field in
a leaky parallel-plate capaci-
Calculate the magnetic flux density vector
218
Chapter 4
in the
4.17.
Magnetostatic
Field in Free
Space
imperfect dielectric of the parallel-plate
10
cm and
Magnetic
vector
field in a
from Example
leaky spherical capacitor.
der
from Problem 3.17
Magnetic
field
4.27.
Magnetic
(Fig. 3.34).
around a grounding electrode.
Find the magnetic flux density vector on the
surface of the ground for the grounding electrode from
Example
3.15 [Fig. 3.26(a)].
B makes
Law
Magnetic field in a leaky coaxial cable.
Determine the magnetic flux density vector in
height h
Current distribution from field distribution.
Using the differential Ampere’s law, show that
the magnetic field given by Eqs. (4.53) and
from vector
flux
(a)
4.22.
4.23.
is
A = 2 r2 T m (r in m).
<\>
this
region, (b) Obtain the magnetic flux through a
circular contour
plane z
=
origin, (c)
circulation of
4.11 but
employing Ampere’s
in differential form.
charged
Rotating
4.28. Potential
cylinder
by
A
differential
m
is
in radius that lies in the
centered at the coordinate
Check the
by evaluating the
results
A along the contour.
and
magnetic
Am
1
0 and
Redo Example
field
due to a magnetic dipole.
dipole
with
moment
a
m=
Ampere’s law. Redo Example 4.14 but with the
use of the differential form of Ampere’s law.
400
/x
ical
coordinate system. Calculate the magnetic
Thin plate with current, differential Ampere’s
law. Redo Problem 4.15 but applying the
differential Ampere’s law [adopt the vectorcomponent notation as in Fig. 4.22 and
Eq. (4.75), and use the analogy with the
application of the differential Gauss’ law in
potential
Ampere’s law
B=
by
in
in differential
and
is
Magnetic
through a cylindrical surface.
Calculate the outward magnetic flux through
the
flux
lateral
m,
surface of a cylinder of radius
at the follow-
0, 0), (b) (1
(1
magnetic dipole
4.30.
magnetic dipole.
far
away from
it.
Rectangular current loop as a magnetic dipole.
Check the
result for the
B
field
due
to a
rectangular loop obtained in Problem 4.1 by
sides, of length 2
in (b).
B
flux density
giving the magnetic flux density vector of a
m, parallel to the
x- and y-axes. (c) Confirm Ampere’s law in
integral form and Stokes’ theorem by evaluating the net circulation of B along the contour
defined
and
»
given
3
the xy-plane, with the center at the coordinate
and
located at the origin of a spher-
Refer to the circular current loop in Fig. 4.6,
a, Eqs. (4.19) and
and show that for |z|
(4.142) become the same, the latter equation
integral form.
current enclosed by a square contour lying in
origin
A
is
m, ;r/2, n/2), (c) (1 m, n, 0),
m, 7r/4, 0), (e) (10 m, jr/4, 0), and (f)
(100 m, 7r/4, 0). The dipole dimensions are
much smaller than 1 m.
(1
(d)
m). The
density,
z
ing points defined by spherical coordinates: (a)
— l) x + 2x y + xy z] mT {x, y, z
medium is air. (a) Find the current
(b) From the result in (a), find the total
2
[4 (z
2
4.29. Circular current loop as a
1.23].
In a certain region, the magnetic field
4.25.
potential. In a cer-
magnetic vector potential
is
Example
4.24.
cylin-
Find the magnetic flux density vector in
Coaxial cable using differential Ampere’s law.
law
uniform
of conservation of magnetic flux. For the
magnetic field defined in Problem 4.24, confirm the law of conservation of magnetic flux
in differential and integral forms by evaluating
(a) the divergence of B and (b) the outward
flux of B through the surface of a cube, respectively. Let the cube be centered at the coordinate origin, with edges parallel to coordinate
axes and 2 m long.
cal coordinate system:
infinitely long cylinder of radius a in free space.
4.21.
in a
S = 1T. The
given as the following function in a cylindri-
produced by a uniform volume current of density given by Eq. (4.52) along an
(4.56)
cm
20
an angle of 60° with the
tain region, the
4.20.
=
field of flux density
axis.
4.26.
3.3 (Fig. 3.6).
the imperfect dielectric of the coaxial cable
4.19.
—
magnetic
Find the magnetic flux density vector in the
imperfect dielectric of the spherical capacitor
4.18.
a
capacitor from Problem 3.12 (Fig. 3.32).
comparing
it
with the corresponding dipole-
field expression,
away from
4.31.
from Eq.
(4.142), at points far
the loop.
Dipole equivalent to a surface current distribuConsider the circular nonuniform surface
current distribution from Problem 4.4, and
a),
show that far away along the z-axis (|z|
tion.
»
219
Problems
this
4.32.
current distribution can be replaced by
4.35. Wire-strip transmission line. Fig. 4.46
shows
a
an equivalent magnetic dipole located at the
cross section of a two-conductor transmission
coordinate origin. Find the moment, m, of the
line consisting of a
equivalent dipole.
of the strip
Lorentz force due to a rotating charged disk.
Refer to the rotating charged disk from
Problem 4.5, and assume that a charged particle Q moves with a uniform velocity v along
a path parallel to the plane of the disk. Find
the Lorentz force on the particle at an instant
when it is at the point that belongs to the axis
the conductors
of disk rotation and
is
at a distance a
is
wire and a
strip.
The width
2 a and the separation between
is
The
a.
dielectric
air.
is
If
the current I runs along the line, calculate the
magnetic force on the
of
its
strip
conductor per unit
a
section of a
length.
Figure 4.46 Cross
from the
transmission line
disk center.
consisting of a wire
4.33.
Forces between three parallel wires with current. Three parallel very long and thin wires
—
in air carry currents of intensities I\
I2
to
1
A,
— —1 A, and I3 = 2 A, all given with respect
the same reference direction. The distance
between any two
of the wires
is
and a strip; for
Problem 4.35.
4.36.
A
d=lm,
which is made from two identical strips
of width a carry steady currents of intensities I and —I given with respect to the same
tor,
so their cross section constitutes an equilateral triangle (with sides
d ).
(a)
,
Determine the
The
direction and magnitude of the per-unit-length
magnetic force on the wire with current I3 (b)
Redo the problem if I2 = 1 A.
reference direction.
Force on a wire due to a semicylindrical conductor.
very long aluminum conductor in
portrayed in Fig. 4.47. The materials and the
the form of a half of a thin cylindrical shell of
per-unit-length magnetic force
A
radius b
40
situated in
mm
air.
and thickness d
—
0.5
mm
is a.
cross section of such a transmission line
is
ambient medium are nonmagnetic. Find the
A
=
distance of the wire
from both ends of the corner conductor
.
4.34.
Line composed of a wire and a corner strip
wire and a 90° corner conducconductor.
on the
wire.
is
Another very long aluminum
conductor, in the form of a very long wire
= 1 mm,
of radius a
is
positioned along the
The two conductors
carry steady currents of the same magnitude
I = 100 A and opposite directions, as shown in
Fig. 4.45. Find the magnetic force on the wire
Figure 4.47 Cross
axis of the semicylinder.
section of a transmission
line consisting of a
and
wire
a 90° corner strip
conductor; for Problem
conductor per unit length.
4.36.
4.37.
1
Coaxial cable with off-centered cavity.
in Fig. 4.48
is
Shown
a cross section of a coaxial
cable in which the cylindrical cavity of radius
b representing the inner surface of the outer
conductor is off-centered by a vector d with
respect to the
common
axis of the inner con-
ductor and the outer surface of the outer
conductor. The other two radii are a and
Figure 4.45 System
composed
of a wire
conductor along the
axis of
c,
and the relationship b + d < c is satisfied. The
permeability everywhere is hq. If a steady curestablished in the cable,
a thin semicylindrical
rent of intensity I
conductor; for Problem
determine the magnetic force on the inner
conductor per unit of its length.
4.34.
is
i
i
i
220
Chapter 4
Magnetostatic
Field in Free
Space
Figure 4.48 Coaxial
cable with
off-centered inner
Figure 4.49
surface of the outer
Triangular current
conductor; for
loop
in a uniform
magnetic field; for
Problem 4.38.
Problem 4.37.
4.39.
4.38.
Torque between two magnetic
Force and torque on a triangular current loop.
small current loops are
A
space.
rigid
loop in the form of an equilateral
tri-
is situated in a uniform
magnetic
field
of flux density B. The
steady
field
lines
magnetic
are parallel to the plane
of the loop and are perpendicular to one of its
angle of side length a
sides, as
shown
in Fig. 4.49. If a
steady current
m;
The
= mi
loop has a magnetic
first
and
is
centered
ing locations of
of
its
magnetic
moment
mx, and
the net torque
on the
(c)
on each of the
the net force and (d)
loop.
at the origin
,
(e)
m = 0.1 Am
2
0)
(O) of
=
m2
:
1
directions
(a) P(0, a, 0)
and
m 2 = my, (c) P(0,
P(0, 0, a) and m 2 =
and
(d)
P(0, a, a)
and a
P and
center
its
loop sides, as well as
established in the loop, find (a)
moment
torque on the second loop, for the follow-
the force and (b) the torque
is
free
in
the Cartesian coordinate system. Obtain the
m 2 = m z, (b) P(0, a
0, a) and m 2 =mz,
of intensity I
Two
dipoles.
positioned
and
ni 2
= mz.
Take
10 m.
I
I
I
Magnetostatic Field
Media
in Material
Introduction:
O
ur study of magnetostatics has so far been
restricted to the magnetic field
due to steady
vacuum and other nonmagwe shall introduce
phenomena associated with the mag-
electric currents in a
netic media. In this chapter,
and discuss
netostatic field in the presence of magnetic materials.
Many
of the basic concepts, physical laws,
and mathematical techniques constituting the analysis of materials in the magnetic field are entirely
analogous to the corresponding concepts, laws, and
techniques in electrostatics, which makes our discussions in this chapter much easier. The most
important difference, however, with respect to the
analysis of dielectric materials
linear behavior of the
is
the inherent non-
most important
class
of
magnetic materials, called ferromagnetics. This is a
class of materials with striking magnetic properties
(many orders of magnitude stronger than in other
materials), with iron as a typical example.
We shall start with a qualitative characterization
of microscopic magnetic
phenomena
in
substances
and describe the behavior of different types of
magnetic materials based on the classical atomic
model. By analogy with the polarization vector
in electrostatics, the magnetization vector will
be used to describe the magnetized state of a
material on a macroscopic scale. Magnetization
volume and surface current density vectors will
be defined as macroscopic equivalents to a vast
collection of tiny electric currents that are micro-
scopic sources of the magnetization of a magnetic
These current densities will enable us
due to magnetized
materials using free-space formulas and techniques
from the preceding chapter. We shall derive and discuss Maxwell’s equations and boundary conditions
material.
to evaluate the magnetic field
for magnetostatic systems that include arbitrary
media. The concept of permeability of a material
will
allow for additional macroscopic character-
ization of magnetic materials. Finally, a section
on magnetic
cores
circuits (consisting of
of different
ferromagnetic
shapes with current-carrying
222
Chapter 5
windings)
will
Magnetostatic Field
in
Material
Media
represent a culmination of the theory
netic materials.
be applied to perform the analysis of such
which essentially resembles the dc analysis
will
of the magnetostatic field in the presence of mag-
circuits,
Most of the work of the chapter
of nonlinear electric circuits.
MAGNETIZATION VECTOR
5.1
According to the elementary atomic model of matter, all materials are composed of
atoms, each with a central fixed positively charged nucleus and a number of negatively charged electrons circulating around the nucleus in various orbits. Both these
orbital motions and the inherent spins of the electrons about their own axes can
be represented by small current loops, i.e., magnetic dipoles. These tiny currents
are referred to as Ampere’s currents. The magnetic moment of each elementary
loop is given by Eq. (4.134). In the absence of an external magnetic field, the
equivalent magnetic dipoles have random orientations with respect to one another,
resulting in no net magnetic moment. With an applied field, however, the equivalent
current loops experience torques, which lead to a net alignment of microscopic magnetic dipole moments with the external magnetic field [see Eq. (4.180)] and a net
magnetic moment in the material on a macroscopic scale. The process of inducing
macroscopic magnetic moments by an external magnetic field is called the magnetization of the material. This process is practically instantaneous, and the material
in the new magnetostatic state is said to be magnetized or in the magnetized state.
When magnetized (by an external field), a material is a source of its own magnetic
field, and the total field at an arbitrary point in space (inside or outside the material)
is a sum of the external (primary) field and the field due to the magnetized material (secondary field). In the analysis, we can replace the material by a collection
of microscopic Ampere’s current loops (magnetic dipoles) residing in a vacuum,
as the rest of the material does not produce any field. Then, the secondary field
can, in principle, be determined using the expression for the magnetic field due to a
magnetic dipole in free space, Eq. (4.142), and superposition.
Instead of analyzing every single atom and all microscopic magnetic dipole
moments, however, we rather introduce a macroscopic quantity termed the magnetization vector to describe the magnetized state of a material and the resulting
field. Analogously to the definition of the polarization vector (P) for electrically
1
polarized dielectric materials, Eq. (2.7), the magnetization vector, M, is defined as
the density of the equivalent elementary magnetic moments in a magnetic material
at a
magnetization vector
given point:
(Vm).
/in
,
av
(5.1)
jyj
(unit:
dv
A/m)
We
note that
M dv represents the dipole moment of a magnetic dipole that
alent to an elementary volume dv
Ampere’s currents) within
'
A
it.
The
in the material,
i.e.,
to
unit for the magnetization vector
third source of equivalent microscopic
magnetic dipole moments
is
equiv-
the dipoles (microscopic
all
in
atoms
is
is
A/m.
nuclear spin, but
it
provides a negligible contribution to the macroscopic magnetic properties of materials. Note, however,
that nuclear spin represents the basis of magnetic resonance imaging (MRI), used in medicine, and also
in
many
areas of science and engineering.
Section 5.2
223
Behavior and Classification of Magnetic Materials
In any magnetic material, the magnetization vector at a point
is
a function of
the magnetic flux density vector at that point,
M = M(B),
(5.2)
and this relationship is a characteristic of individual materials. It is entirely analogous to the relationship between the polarization vector and the electric field
intensity vector, Eq. (2.9), in electrostatics.
5.2
BEHAVIOR AND CLASSIFICATION OF MAGNETIC
MATERIALS
A thorough understanding and precise quantitative characterization of microscopic
magnetic phenomena in materials require a full quantum mechanical treatment.
Here, however, we describe qualitatively the behavior of different types of magnetic materials based on the classical atomic model. Generally, materials can be
classified according to their magnetic behavior into diamagnetic, paramagnetic,
ferromagnetic, antiferromagnetic, ferrimagnetic, and superparamagnetic materials.
In diamagnetic materials, the magnetic moments of electrons orbiting about
their nuclei are dominant compared to the magnetic moments attributed to electron
spin. In order to describe diamagnetic behavior, which is present to a greater or
lesser extent with all magnetic materials, we consider first a model of an atom with
a single electron that circulates about the nucleus along an orbit of radius a with a
uniform angular velocity wo = wq z, as shown in Fig. 5.1(a). In the absence of an
external magnetic field, the outward centrifugal force on the electron, given by
Fc f =
m
e vn
o
= m e wla,
(5.3)
a
where
mt
and
vo are the
mass and velocity of the electron
(vo
=
woa), respectively,
balanced by the centripetal (attractive) electric (Coulomb) force,
nucleus and electron. Thus, the balance equation reads
is
Ft = m e Wo«.
Fe
,
between the
(5.4)
The orbiting electron is equivalent to a small current loop (magnetic dipole),
where the current I of the loop is given by Eq. (4.153) with Q = — e and is
Figure 5.1
Atom
with a single electron orbiting about the nucleus,
an applied magnetic
opposite to
it
(c).
field,
whose
in
the absence of an external magnetic
field (a),
direction coincides with the direction of the electron angular velocity vector (b) or
and with
is
m
224
Chapter 5
Magnetostatic
Field in Material
Media
directed oppositely to the direction of electron travel (because the electron charge
is
From Eq.
negative).
(4.134), the
magnetic
moment
of the dipole
—
m = lna 2 {- z) =
()
is
2
In the presence of an external magnetic field, there
(5.5)
is
an additional force on the
electron - the magnetic force, Eq. (4.144),
= Qv
Em
B e xt = —
X
B ex
X
(5-6)
t)
with Bext denoting the magnetic flux density vector of the applied field. For
the situation in Fig. 5.1(b), where the electron angular velocity vector is in the
same
B ex =
new balance equation
direction as the applied field
(inward), so that the
the magnetic force
z,
t
is
Fe + ewaB ex = m t wa,
w — wq + Aw.
i
(5.7)
is compensated
Combining Eqs.
Here, the force unbalance created by the magnetic force
increase
Aw
centripetal
is
(Aw >
of the orbital angular velocity
0).
for by an
and
(5.4)
(5.7) gives
m
e
(w
2
-
= ewB ext
Wg)
Since the perturbation of the electron velocity
strongest applied magnetic
fields,
9
9
w — Wq =
we can
(w
+
is
(5.8)
.
small,
i.e..
Aw
« wq, even for the
write
—
wq)(w
2 wAw.
wq)
(
Substituting this into Eq. (5.8), the increase of the angular velocity
Aw =
new
m
(5.10)
ewa 2
„
_
z
momentum
L=
where
O
(|r|
r is
—-
ea,2
- w.
(5.11)
2
of the electron,
mer
x v
= Xw,
(5.12)
the instantaneous position vector of the electron with respect to the origin
The increase of
the dipole
the electron, in Eq. (5.10),
— L = — —2 m
2m e
moment due
is
Xw.
to the increase of the orbiting velocity of
given by
called the induced magnetic
antiparallel
(i.e.,
in the
(5.13)
e
Am = — — XAw — - ( —— ^
2me
\2 e )
is
is
= a) andX = m e a 2 is the rotational inertia of the electron, Eq. (5.11) becomes
m=
and
moment
‘
=
2
Introducing the angular
to be
e
equivalent magnetic dipole
m
.
^®ext
2
In place of Eq. (5.5), the
comes out
5 9)
moment
(5.14)
XBext-
of the electron.
We
opposite direction) to the applied magnetic
note that
For an electron whose orbital angular velocity vector and the
opposed, as
in Fig. 5.1(c),
the force
F m on
and the force unbalance is compensated
new magnetic dipole moment is smaller
moment
Am
is
again
in
the
—z
the electron
is
in the
Am
B ex
field B ex
field
is
t-
t
are
outward direction,
by a reduced velocity, i.e.. Aw < 0. The
magnitude than m so that the induced
for
in
direction, that
() ,
is,
still
antiparallel to the applied
Section 5.2
Behavior and Classification of Magnetic Materials
Am
and given by the same expression in Eq. (5.14). This same expression for
obtained
also in the case of an arbitrary mutual position of the electron orbit and
is
electrons,
the applied magnetic field, as well as for the model of an atom with
field,
N
travel along orbits that are arbitrarily oriented with respect to the applied
which
magnetic
field,
where
X
should be replaced by the average rotational inertia of
electrons in the atom, X — Xav
With no external magnetic
all
.
random
field applied, the
orientations, so that the net magnetic
in the material are zero.
ferential induced
magnetic dipoles of the atoms have
moment and the magnetization vector
With an applied field, however, each atom acquires a difgiven by Eq. (5.14), with X = Xav and all these moments
moment
,
are antiparallel to the magnetic flux density vector, B, in the material.
the net magnetic
moment
of the atoms opposes the applied
diamagnetic material placed in a magnetic
field.
field will orient itself
As
a result,
Thus, a rod of a
perpendicular to
the applied field, i.e., across the field lines. The term diamagnetism originates from
the Greek word “dia” meaning “across.” If a diamagnetic specimen is brought near
either pole of a strong bar magnet, it will be repelled by the magnet. From Eqs. (5.1)
and (5.14), the magnetization vector is
-~- B,
M = iV Am = — Nwe
4 mg
(5.15)
v
Av
being the concentration (number per unit volume) of atoms in the material.
We
note that this equation represents a special form of Eq.
customary to write
M=
where Xm
is
—
Mo
(5.2). In addition,
B,
it is
(5.16)
a dimensionless proportionality constant given
by
/xoAv e^Xav
(5.17)
4 nil
It is
rials
medium, Xm
=
0.)
We
vacuum, which
is
see that the diamagnetic mate-
are linear magnetic media, because the relationship
M(B)
in
Eq. (5.16)
is
a
linear one. Typical diamagnetic materials are bismuth, silver, lead, copper, gold,
silicon,
germanium, graphite,
sulfur,
hydrogen, helium, sodium chloride, and water.
Substituting typical values for the quantities involved in the expression in Eq. (5.17)
-4
-6
indicates that Xm of diamagnetic materials is of order — 10
(water) to — 10
(bis-
We see that the magnetic susceptibility of diamagnetic materials is negative
and very small. The macroscopic magnetic effect in diamagnetic substances is very
weak and negligible in most practical situations. In general, the diamagnetic effect
is present in all materials when placed in a magnetic field, because it arises from
an interaction of orbiting electrons in atoms with the external field. However, in
other types of magnetic materials, the diamagnetic reaction is completely masked
by other effects.
In paramagnetic materials, the atoms have a small permanent net magnetic
dipole moment associated with them, due almost entirely to spin magnetic dipole
muth).
moments
of electrons. In the absence of an external magnetic
orientation of the atoms produces an average magnetic
finite
volume.
When
an external
susceptibility of
diamagnetic materials
called the magnetic susceptibility of the material. (For a
the only truly nonmagnetic
magnetic
field is applied,
field,
moment
however, there
is
the
random
of a zero in a
a small torque
given by Eq. (4.180) on each atomic moment, which tends to align the moment
in the direction of the applied field. This alignment acts to increase the value of
225
226
Chapter 5
Magnetostatic
Field in Material
Media
the magnetic flux density vector, B, within the material over the external value.
However, the external
field also causes a diamagnetic effect of the orbiting elecmagnetic materials exhibit diamagnetic behavior), which counteracts
the increase in B. The alignment process is also impeded by the forces of random thermal vibrations, and the resulting increase in B is quite small. The overall
macroscopic effect is equivalent to that of a small positive magnetization. The mag-
trons
(all
netization vector,
M,
is
in the
same
direction as the vector B. Eq. (5.16), then, tells
us that the magnetic susceptibility of paramagnetic materials
small.
itself
When a rod of a paramagnetic material
along the
field lines,
Greek means “along”).
is
is
positive
placed in a magnetic
field,
and very
it
orients
and hence the term paramagnetism (the word “para”
paramagnetic substance is brought
be attracted to it. Typical paramagnetic materials are palladium, aluminum, and oxygen, and typical values of Xm
-6
are of orders 10
(oxygen) to 10~ 3 (palladium). Note that air is a paramagnetic
in
Additionally,
near a pole of a strong bar magnet,
medium
as well. Obviously, the
it
if
a
will
paramagnetic effect
is
also very
agnetic materials can be treated as nonmagnetic media (x m
=
weak and param-
0) in
most practical
applications.
The remaining four
classes of magnetic materials (ferromagnetic, antiferromag-
and superparamagnetic) all have strong atomic magnetic dipole
moments caused primarily by uncompensated electron spin moments. The interaction of adjacent atoms leads to an alignment of the atomic moments in either an
aiding (parallel) or opposing (antiparallel) manner.
The macroscopic magnetization of ferromagnetic materials can be many orders
of magnitude larger than that of paramagnetic materials. Interatomic forces in a ferromagnetic sample cause the atomic moments to line up in a parallel fashion over
regions containing large number of atoms (e.g., 10 15 atoms). These regions, called
the magnetic domains (or Weiss' domains), range in size from a few microns to
several centimeters in each dimension. With no external field applied, the domain
moments vary in direction from domain to domain, as indicated in Fig. 5.2 for a polynetic, ferrimagnetic,
Figure 5.2 Magnetic
domains
in a polycrystalline
ferromagnetic specimen
with no external magnetic
field applied.
crystalline ferromagnetic specimen.
Due
to overall vector cancelations, the material
whole has no net magnetization. Upon the application of an external magnetic
field, however, the volumes of the domains that have moments aligned or nearly
aligned with the applied field grow at the expense of their neighbors, and the magnetic flux density vector of the secondary field (that due to the material) increases
greatly over that of the external field alone. For weak applied fields, the movements
of magnetic domain walls are reversible, i.e., the domains go back to their initial
as a
is turned off. Above a certain (not large) field magnitude,
however, this process becomes irreversible. Additionally, the domains start rotating toward the direction of the applied field, so that completely random domain
orientations are no longer attained after the external field has been removed. In
other words, there is a residual or remanent net magnetization of the material that
states after the field
remains after the complete removal of the primary field. This means that the magnetization of the material lags behind the field producing it, and also that the
magnetization state of the material at an instant of time is a function not only of
the magnetic field, i.e., its flux density vector, at that instant, but also of the magnetic
history of the material. This phenomenon is called hysteresis (which is derived from
a Greek word meaning “to lag”). As the applied field becomes even much stronger,
a total alignment of
all
the
domain moments with
point, the ferromagnetic material
is
the applied field occurs.
At
this
said to be saturated. In the state of saturation,
further increase of the external magnetic flux density vector
increase of the magnetization vector.
no longer causes an
Section 5.3
Magnetization Volume and Surface Current Densities
227
Typical ferromagnetic materials are iron, nickel, and cobalt, and their alloys.
The name ferromagnetic comes from the Latin word for iron, “ferrum.” Ferromagnetics are the most important magnetic materials in engineering. They are widely
used in cores of inductors and transformers, and in electric motors, generators, electromagnets, relays, and other devices that use magnetic forces and torques, as well
as in magnetic heads and tracks of computer hard disks and other magnetic storage
(recording) devices. Note that above a certain (very high) temperature, called the
Curie temperature, the thermal vibrations of atoms completely prevent the coupling
(parallel alignment) of atomic magnetic moments, so that ferromagnetic materials
lose all their ferromagnetic characteristics and revert to paramagnetic materials.
The Curie temperature
for iron
is
Illllllllltlll
(a)
770° C.
In antiferromagnetic materials, the forces between adjacent atoms cause the
atomic magnetic moments to line up in opposite directions so that the net magnetic moment of a specimen is zero, even in the presence of an applied magnetic
field.
as well as many oxides, sulfides, and chlorides,
Examples are manganese oxide (Mn02), ferrous
(FeS), and cobalt chloride (C 0 CI2 ). Antiferromagnetism is not of practical
sulfide
this class of materials.
importance.
aligned antiparallel, but the
moment
A
of a specimen.
therefore occurs, although
moments
moments
of adjacent atoms are also
are not equal, so there
is
a net magnetic
large response to an externally applied magnetic field
it is
substantially less than in ferromagnetic substances.
comparison of the atomic magnetic dipole moment
structure for ferromagnetic, antiferromagnetic, and ferrimagnetic materials. The
most important subclass of ferrimagnetics are the ferrites, which have much lower
Fig. 5.3 depicts schematically a
electric conductivity (a)
7
compared with 10 S/m
material
why
when
than the ferromagnetics (for instance, 10
for iron).
Low
many
for high-frequency transformers,
shifters, in spite
-4
to 10
2
S/m,
conductivity limits induced currents in the
alternating (ac) fields are applied (so-called
the ferrites are used in
eddy currents), and
this
applications at high frequencies, such as cores
AM,
short-wave, and
FM
antennas, and phase
of their weaker magnetic effects (compared to ferromagnetics).
The reduced eddy currents lead
to lower Joule’s (ohmic) losses in the material
examples of ferrite substances are iron ferrite (FesCL), nickel ferrite (NiFe 2 C> 4 ), and cobalt ferrite (CoFe 2 C> 4 ). Ferrimagnetism also disappears at
temperatures above a critical value - the Curie temperature.
Finally, superparamagnetic materials are composed of ferromagnetic particles
suspended in a nonmagnetic (dielectric) matrix. Each particle contains many magnetic domains, but the domain walls cannot penetrate the matrix material and there
is no coupling with adjacent particles. Thin dielectric (plastic) tapes with suspended
ferromagnetic particles can store large amounts of information in magnetic form
because the particles are independent from each other and it is possible to change
the state of magnetization along the tape abruptly in very small distances. Such
superparamagnetic tapes are widely used as audio, video, and data recording tapes.
(core). Typical
MAGNETIZATION VOLUME AND SURFACE CURRENT
DENSITIES
In this section,
we
shall obtain the expressions for evaluation of the
macroscopic
volume and surface current densities equivalent to microscopic
Ampere’s currents in a magnetized body - from a given distribution of the
distribution of
|i|i|i|i|i|i|i
(c)
In ferrimagnetic materials, the magnetic
5.3
(b)
Chromium and manganese,
belong to
is
Illllllllltlll
Figure 5.3 Schematic
structures of atomic
magnetic dipole moments
for (a) ferromagnetic, (b)
antiferromagnetic, and (c)
ferrimagnetic materials.
Chapter 5
Magnetostatic Field
in
Material
Media
magnetization vector, M. The vector M,
in turn,
moments
is
obtained by averaging the
Ampere’s current
The equivalent macroscopic current is called the
magnetization current, and the corresponding volume and surface current densities
are denoted by J m and J m s, respectively. The expressions for these current densities
are similar to the expressions for calculating the volume and surface bound (polarmicroscopic magnetic dipoles (magnetic
of microscopic
loops) in the magnetic material.
and p ps ) from the polarization vector (P) in a polarized
be used later for free-space evaluations of the magnetic
flux density vector (B) due to magnetized bodies.
Let us first find the intensity of the total magnetization current Im c enclosed
by an arbitrary imaginary contour C situated (totally or partly) inside a magnetized
magnetic body, as depicted in Fig. 5.4. We note that Im c is actually the current that
passes through any surface S bounded by C. Let the magnetic moments of small
Ampere’s current loops in a vacuum that constitute the magnetization current be
expressed as
ization) charge densities (p p
dielectric body.
They
will
(5.18)
where S m
=
|S m
is
|
the surface area of the loop.
the loops that pass through
S
It is
obvious in
Fig. 5.4 that all
do not pass through S
7 m c Only loops that pierce S
twice, as well as those that
at all, contribute
with zero net current intensity to
only once, that
the loops that encircle C, contribute actually to the total current
is,
intensity flowing through S.
To evaluate /m c
the loops that are strung along
C
in the general case,
(like pearls
the contribution of such loops as either / or
.
on a
—/
string). In
(note that
/,
we
therefore count
doing
that,
we count
generally, differs
from
loop to loop), by inspecting whether the direction of the loop current traversing 5
agreement or disagreement with the reference orientation of S.
Consider an element d / of C and the case when the angle fi between the vector
(or the average of elementary dipole moment vectors m near d/) and vector dl,
which is oriented in accordance to the orientation of C, is less than 90°, as shown in
Fig. 5.5(a). Note that centers of loops that are positioned close to dl and encircle it
are inside an oblique cylinder with bases S m and height
in
is
M
dh
so that the total
number
=
d/cos^5,
(5.19)
of these loops equals the concentration of loops (magnetic
dv = S m d h. The refcontour
C by means of
erence orientation of the surface S is related to that of the
normal
vector
ii on S in
the right-hand rule, so the reference direction of the unit
dipoles), 7V v [see Eq. (2.6)], times the
Figure 5.4 Contour C in a
magnetized magnetic body.
volume of the
cylinder,
C in
Figure 5.5 Element of the contour
and
dl: (a)
0
< p <
90°
and
(b) 90 °
Fig. 5.4, in
229
Magnetization Volume and Surface Current Densities
Section 5.3
two cases with regards
between
to the angle p
M
<p< 180°.
from S upward. We then realize that all the loop currents encircling
dl pass through S in the positive direction (direction in agreement with the orientation of S ) and contribute to the total current with the intensity / (we assume
that all the loops near dl, which is differentially small, have the same moments
and currents). Hence, the corresponding contribution to the magnetization current
intensity through S is given by
Fig. 5.5(a)
is
= Ny Sm dl cos p I
dlm
when p >
In the case
and are taken
all
90°).
=
dl cos(7r
—
P)
=
d/(—
cos/3),
(5.21)
the loop currents encircling dl pierce S in the negative direction
as —I,
dlm
= NvS m
dl
(—cos /l) (-/)
(90°
<p <
which is the same result as in Eq. (5.20).
= y m, where
Eq. (5.1) can be interpreted here as
and hence, for an arbitrary p (0 < p < 180°), we have
M N
d/m
[note that the
(5.20)
90°, Fig. 5.5(b),
dh
and, because
<p <
(0
boundary
= Nw m dl cos p =
case,
p
=
yVv
m
•
dl
=
M
180°),
m
•
is
(5.22)
given by Eq. (5.18),
dl
90° (the contour element dl
(5.23)
is
tangential to the
S m of the loops) and d/m = 0, is also properly included in this formula].
by integrating the result for d/m along the entire contour C, we get
surfaces
Finally,
/mC
= £M-dI.
(5.24)
total
magnetization current
enclosed by a contour
This
is
an integral equation similar
in
form
to
Ampere’s
law, Eq. (4.48).
It tells
us that the circulation (line integral) of the magnetization vector along an arbitrary contour in a magnetostatic system that includes magnetic materials
tion of the current flow
right-hand rule.
is
is
equal
by that contour. The reference direcrelated to the reference orientation of the contour by the
to the total magnetization current enclosed
C
230
Chapter 5
Magnetostatic
Field in Material
Eq. (5.24)
is
Media
true for any contour C. Let us apply
an elementary surface
AS
it
to the contour
C enclosing
inside a magnetic material:
(An)through
§c M
AS
AS
•
dl
(AS ->
AS
(5.25)
0),
where both
sides of the equation are divided by AS. The expression on the left-hand
above equation represents the component of the magnetization volume
current density vector normal to AS,
side of the
.
»
n
(n
—
•
—
(An)through
AS
..
(5.z6)
AS
the unit vector normal to the surface AS), while the expression on the right-
is
hand
is, by definition [Eq.
n curl M. Hence,
side of the equation
M along n, that
is,
n Jm
•
and, since this
is
component of
(4.87)], the
the curl of
•
true for any n
=
n curl M,
(5.27)
•
and thus for
all
components of the vector J m
,
it
implies that
magnetization volume
Jm
--
curl
M = V x M.
(5.28)
current density vector
This
is
material
If
form of the integral relationship in Eq. (5.24). It tells us that
volume current density vector, in A/m 2 at an arbitrary point in a
a differential
the magnetization
is
,
equal to the curl of the magnetization vector at that point.
M = const inside the magnetic material (uniformly magnetized material),
M are zero, and from Eq. (5.28),
all
spatial derivatives of
M — const
no volume magnetization
Jm
— 0.
(5.29)
current in a uniformly
magnetized material
Physically, the currents of adjacent
Ampere’s current loops
that flow in opposite
directions cancel everywhere in the interior of a uniformly magnetized material, and
there
is
no net volume
current. If
magnetization current exists only
M ^ const, however,
if
then volume macroscopic
the magnetization vector varies throughout the
volume of the material (nonuniformly magnetized material)
nonzero, otherwise J m
=
in a
way
that
its
curl
is
0.
On the surface of a magnetized magnetic body, there always exists surface
macroscopic magnetization current (there are parts of small Ampere’s current loops
pressed onto the surface that cannot be compensated by oppositely flowing currents
and the
of neighboring loops). The only exception are parts of the surface where
microscopic magnetic dipoles are normal to the surface, that is, the surfaces S m of
the Ampere’s current loops are laying in the surface of the body. To determine the
magnetization macroscopic surface current density vector, J ms (in A/m), equivalent
to the microscopic currents, we apply Eq. (5.24) to a narrow rectangular elementary
contour, with length A / and height Ah (Ah -> 0), shown in Fig. 5.6. The contour
being differentially small, the magnetic moments m of Ampere’s current loops near
the contour are all parallel to each other. The resultant of all the corresponding
Ampere’s currents, and thus the vector Jm S is perpendicular to the local magnetizalies in the plane
tion vector, M. The contour C in Fig. 5.6 is positioned such that
of the contour and Jms is perpendicular to that plane. There is no magnetization
(M = 0) in free space (a vacuum, air, or any other nonmagnetic medium) surroundAl
in Eq. (5.24) is reduced to
ing the body, so that the circulation of vector
M
,
M
M
M
•
along the lower side of C. By the definition of the surface current density vector
.
Magnetization Volume and Surface Current Densities
Section 5.3
231
Figure 5.6 Elementary contour
used for deriving the boundary
condition for the vector
M on
the surface of a magnetic body.
[see Eq. (3.13)], the total current enclosed
Eq.
(5.24), equals
/ms A/
where a
the angle that
is
M
=
•
Al
=
= MA/sina,
MA/cos/i
(5.30)
M makes with the normal on the surface directed from the
magnetic body outward (a
or, in a
by C, appearing on the left-hand side of
Jms Al, and we have
=
90°
—
Hence,
/$).
Jms
=
M sin a,
(5.31)
Jms
=
Mx
(5.32)
vector form,
n,
magnetization surface current
density vector; n outward
with n standing for the outward normal unit vector on the surface. This
ary condition for the vector
is
the bound-
M on the surface of a magnetic body, connecting the
normal on a magnetic body
surface
magnetization vector in the body near the surface and the magnetization surface
current density vector
M contributes to Jms
on the
surface.
Note
that only the tangential
component
of
.
The expressions in Eqs. (5.28) and (5.32) can be used for determining the disvolume and surface magnetization current densities, J m and J ms of a
magnetized body, assuming that the state of magnetization of the body is described
by a given distribution of the magnetization vector, M, inside the body. We can
tribution of
,
then, considering these macroscopic currents to reside in a
vacuum, calculate the
magnetic flux density vector, B, due to the magnetized body (and any other related
quantity of interest) using the appropriate free-space equations (e.g., various forms
of the Biot-Savart law and Ampere’s law) and solution techniques suitable to
specific geometries and current distributions.
Example
Nonuniformly Magnetized Ferromagnetic Cube
5.1
The magnetization vector
in a
ferromagnetic cube shown in Fig. 5.7
M(x,
where Mq
is
a constant.
y)
=
M ~
a
0
The surrounding medium is
given by
(5.33)
z,
1
air.
is
Find the distribution of magnetization
currents of the cube.
Solution
inside the
From
cube
Eqs. (5.28) and (4.81), the magnetization
is
3M
Jm
3y
Z
A
x
dM z
dx
„
y
M
o
(xx
volume current density vector
-y y).
Figure 5.7 Ferromagnetic
(5.34)
cube with magnetization
Mpc, y); for Example 5.1
232
Chapter 5
Magnetostatic Field
Material
in
Media
The magnetization surface current density vector
Jmsi
=
x x
=
on the front
side
a
sin
=
=
\
(5.35)
a
Uniformly Magnetized Ferromagnetic Disk
thin ferromagnetic disk of radius a
its
M(*, a~) x y
M
[in
uniformly magnetized throughout
bases and
=
(5.32),
respectively, and J ms = 0 on the remaining
= 0 on the back side and left-hand side,
terms of Eq. (5.31),
0 on the top and bottom sides of the cube].
Example 5.2
A
J ms2
by means of Eq.
and right-hand side of the cube,
four sides of the cube
whereas
and
y
a
is,
magnitude
is
and thickness d (d
a)
is
situated in
volume. The magnetization vector
its
M. Determine
(a) the distribution of
is
air.
The
disk
is
normal to disk
magnetization currents of
the disk and (b) the magnetic flux density vector at an arbitrary point along the disk axis
normal to the bases.
Solution
Eq. (5.29)
(a)
tells
us that there
According to Eq. (5.32) and
no magnetization volume current
is
Fig. 5.8, a
inside the disk.
magnetization surface current flows circumfer-
entially along the side disk surface, with density
=
J ms
Mx
fii
= Mi
x
r
=
0,
because
while on the upper and lower disk bases, J ms
and
=
(5.36)
M<j>.
M
is
collinear with both n 2
thin, the circumferential sheet of
magnetization current
113.
Figure 5.8 Magnetization
surface current on a
(b) Since the ferromagnetic disk
uniformly magnetized
ferromagnetic
current intensity
Example
disk; for
its
is
surface can be replaced by an equivalent circular current loop with radius a and
over
Assuming
is
that the loop
is
given by Eq. (4.19), that
in a
= -Ans d —
Md.
(5.37)
vacuum, the magnetic
flux density vector along the z-axis
An
5.2.
is.
IxoMda 2
2(z 2 +«2)
Example 5.3
A
Cylindrical Bar
(5.38)
3/2
Magnet
magnet of radius a and length / is permanently magnetized with a uniform
The magnetization vector, M, is parallel to the bar axis. The medium around
air. Compute the magnetic flux density vector at the center of the magnet.
cylindrical bar
magnetization.
the
magnet
is
the magnet is the same as that
and the vector B both inside and outside the magnet can
be found as that due to a cylindrical current sheet of radius a and length / in a vacuum with the
surface current density given in Eq. (5.36). This sheet, on the other hand, is equivalent to the
solenoid in Fig. 4.10, so that the flux density along the magnet axis is given by Eq. (4.36) with
NI/l substituted by/ms =
[see Eq. (4.30)]. Specifically, B at the center of the magnet equals
Solution
The magnetization surface current density over
over the magnetized disk
in Fig. 5.8,
M
R =
^° l
y/l2
Nonuniformly Magnetized
Fig. 5.9(a)
shows an
tization given
by
—M
(5.39)
+ 4a 2
Infinitely
Long Cylinder
infinitely long cylinder of radius a in air,
M = A/o(l
-
2
r /a
2
)
i,
where
Mq
is
having a nonuniform magne-
a constant. Find (a) the distribution of
233
Magnetization Volume and Surface Current Densities
Section 5.3
magnetization currents of the cylinder and (b) the magnetic flux density vector inside and
outside the cylinder.
Solution
(a)
Applying the formula for the curl
in cylindrical coordinates,
the streamlines of the magnetization
centered at the cylinder
axis, as
volume current
Eq. (4.84),
we
obtain that
inside the cylinder are circles
indicated in Fig. 5.9(b), with the following current density
vector:
L - V x M = - dM
dr
(b)
2M5“
—
-
0
z
r
(5.40)
4>-
There
is
no magnetization surface current over the surface of the
M(a~)
=
0
cylinder, because
in the cylinder close to the surface.
We note that the vector J m in Eq.
(5.40)
is
of exactly the
same form
as the current density
vector J inside the rotating charged cylinder in Fig. 4.20 computed in Eq. (4.69).
The only
Eq. (4.69) vs. 2Mo/a 2 in Eq. (5.40).
Consequently, the magnetic flux density vectors in the two systems are also of the same
difference are the multiplicative constants,
pw
in
form, the only difference being those two multiplicative constants. [The magnetic field
was obtained by visualizing the current
and applying Ampere’s law,
Eq. (4.49).] By substituting, therefore, pw in Eq. (4.71) by 2Mo/a 2 the flux density due
to the magnetized cylinder in Fig. 5.9 turns out to be
due to the rotating charged cylinder
in Fig. 4.20
distribution J as a series of coaxial infinitely long solenoids
,
B=
inside the cylinder (0
<
r
<
p.qMq
and
a)
B=
1
~
[
J
0 outside
i
= mqM
(5.41)
(b)
it.
Figure 5.9 Magnetization
vector (a) and
A ferromagnetic sphere of radius a
M. The sphere
magnetization volume
Uniformly Magnetized Ferromagnetic Sphere
Example 5.5
is
surrounded by
vector at an arbitrary point
is
air.
currents (b) inside a
uniformly magnetized, and the magnetization vector
is
Calculate (a) the magnetization surface current density
on the sphere surface and
(b) the
magnetic flux density vector
at
the sphere center.
Solution
(a)
at the
sphere center and the z-axis par-
in Fig. 5.10,
M = Mz and the magnetization
For a spherical coordinate system with the origin
allel to
the magnetization vector, as
shown
surface current density vector over the sphere surface at a point
9
P
defined by an angle
is
J ms
=
M x n = Msin Z(M, n) = Msin0,
0
<
9
<
n.
(5.42)
been removed, we apply the superposition principle
due to the spherical sheet of current described by the function in
analogous to the computation of the electric field due to a uniformly
(b) After the ferromagnetic material has
to find the
B
field
Eq. (5.42), which
is
2.7. We subdivide the sphere surface into thin rings
ad9, which is depicted in Fig. 5.10. Each such ring can be viewed as an
equivalent circular wire loop with the same current. The current of the ring containing
polarized dielectric sphere in Fig.
of width d/r
the point
=
P equals
d/m
Using Eq.
(4.19), the
magnetic
= /ms d/ = Masin9 d9.
(5.43)
r
flux density vector of this ring at the
sphere center
is
nonuniformly magnetized
infinitely
long ferromagnetic
cylinder; for
Example
5.4.
1
i
234
Chapter 5
Magnetostatic Field
in
Media
Material
Figure 5.10 Ferromagnetic
sphere with a uniform
I
magnetization; for
Example
= asm#
with a r
5.5.
standing for the ring radius. Hence, the resultant flux density comes out
to be
PoM
B
„
2
where the
/*
tt
sin
jf
Problems'.
MATLAB
integral in 0
5.
3
p
#d0 =
1-5.6;
J
is
tt
sin
/
2
3
Q d6
=
- /zqM,
(5.45)
J0=O
evaluated as [see also Eq. (2.32)]
,
/
.
^1
— cos 2 9j sin0 d6 = (—cos 0-1
—
£os^ 0 \
Conceptual Questions (on Companion Website):
Exercises (on
Companion
(5.46)
5.
1-5.3;
Website).
I
!
GENERALIZED AMPERE'S
5.4
We now
i
I
consider a general magnetostatic system where,
lent magnetization currents
surfaces,
LAW
we have conduction
all
addition to equivatheir
currents (free currents) flowing through conductors
(including conducting magnetic materials).
vector, B.
in
(bound currents) inside magnetic bodies and over
these currents act as
if
As sources
they were
in a
of the magnetic flux density
vacuum, and Ampere’s law,
I
Eq. (4.48), becomes
Ampere's law
for
with conductors
B
a system
•
dl
=
po ( Ic
+ /me).
(5.47)
and
magnetic materials
with Ic and
/m c
standing for the total conduction current and the total magnetiza-
by an arbitrary contour C. Dividing this equation
by po, moving Im c to the left-hand side of the equation, then substituting it by the
circulation of the magnetization vector, M, from Eq. (5.24), and finally joining the
tion current, respectively, enclosed
two
integrals along
C into a single
integral,
we
•
get the equivalent integral equation:
dl
=
Ic
,
(5.48)
I
which
generalized Ampere's law
is
conveniently written as
l
H
dl
= lc
.
(5.49)
I
Section 5.4
This equation
is
referred to as the generalized Ampere’s law.
Generalized Ampere's Law
235
The new quantity on
the left-hand side of the equation,
H=
B
M
(5.50)
Ho
is
(unit:
called the magnetic field intensity vector
to the electric flux density vector,
The generalized Ampere’s law
and
is
easier to use than the
D,
magnetic
and
field intensity
vector
A/m)
measured in A/m. It is analogous
which is defined by Eq. (2.41).
is
in electrostatics,
valid for magnetostatic fields in arbitrary media,
is
form
in Eq. (5.47)
because
it
has only free (true)
is
that
currents on the right-hand side of the integral equation.
The most general representation of conduction current
volume current density, J, which yields
X
*3 II
JC
by means of the
f J dS,
(5.51)
•
Js
generalized Ampere's law
in
terms of the volume current
where S is a surface of arbitrary shape bounded by the contour C (orientations of
C and S are in accordance to the right-hand rule). Since this integral relation holds
for any choice of C, applying Stokes’ theorem, Eq. (4.89), results in the following
density
differential relation:
V
x
H = J,
(5.52)
generalized differential
Ampere's law
namely, the differential form of the generalized Ampere’s law. Eqs. (5.51) and (5.52)
represent, respectively, the integral and differential Maxwell’s second equation for
the magnetostatic field in an arbitrary
Toroidal Coil with a Ferromagnetic Core
Example 5.6
A uniform
/.
If
field intensity
Solution
N turns of wire
and dense winding with
netic core of length
magnetic
Due
medium.
there
is
l
(4.66)].
placed over a thin toroidal ferromag-
a steady current of intensity I through the winding, find the
vector in the core.
to symmetry, magnetic field lines in the core are circular as in the air-filled
toroid in Fig. 4.18. Moreover, the field in the core
Eq.
is
ferromagnetic
Applying generalized Ampere’s
along the toroid
axis, as
shown
H=
—
is
uniform, because the toroid
law, Eq. (5.49), to a circular
in Fig. 5.11, gives
HI
(thin toroid with
=
contour
is
thin [see
C of length
NI, from which,
an arbitrary core).
(5.53)
Figure 5.11 Evaluation of
Note that
this result
holds true for a core
(including nonlinear and
made from an
arbitrary magnetic material
inhomogeneous media).
the magnetic
vector
a
Prove that the
Closed Surface
flux of the
in a
magnetic
Uniformly Magnetized Material
field intensity
vector through a closed surface situated
inside a uniformly magnetized ferromagnetic material equals zero.
Solution
According to our experience so
far with vector calculus
and
different forms
of Maxwell’s equations, the flux of a vector through a closed surface in integral notation
corresponds, in differential notation, to the divergence of that same vector (with similar cor-
respondence between the circulation along a contour and the curl). Let us therefore consider
the divergence of the magnetic field intensity vector, H, in the material. Since the material is
uniformly magnetized, the divergence of the magnetization vector, M, at an arbitrary point
with
ferromagnetic core; for
Example
Example 5.7
field intensity
in a toroidal coil
5.6.
236
Chapter 5
Magnetostatic
Field in Material
in the material
zero
is
on the other hand,
field
Media
(M =
const).
The divergence of
the magnetic flux density vector, B,
always zero [Eq. (4.103)]. Consequently, the definition of the magnetic
intensity vector, Eq. (5.50), gives
is
V-H =
— V-B — V-M =
(5.54)
0,
Mo
i.e.,
the divergence of
H
is
also zero in the material, which in integral notation [see
Eq. (1.173)] reads
HdS = 0,
j)
for any closed surface
S
(5.55)
situated inside the material.
Problems 5.7-5.11; Conceptual Questions (on Companion Website):
:
MATLAB
Exercises (on
Companion
5. 4-5.7;
Website).
PERMEABILITY OF MAGNETIC MATERIALS
5.5
We now introduce the concept of permeability for macroscopic characterization of
magnetic materials, which is analogous to the permittivity concept in electrostatics.
Substituting Eq. (5.2) into Eq. (5.50), we obtain that the magnetic field intensity
vector (H) at a point in any magnetic material is a function of the magnetic flux
density vector (B) at that point, or vice versa,
constitutive equation of
B = B(H).
an
(5.56)
arbitrary (nonlinear)
This
magnetic material
the general constitutive equation for characterization of magnetic materials,
is
For linear magnetic materials, the magnetizaand hence also to H [see Eq. (5.50)],
parallel to Eq. (2.46) in electrostatics.
tion vector,
which
is
M,
is
linearly proportional to B,
customarily written as
M = X mH.
Here, Xm
is
(5.57)
the magnetic susceptibility of the material, that
is,
the
same dimension-
less quantity defined in Eq. (5.16). Note, however, that Eq. (5.16) holds only for
diamagnetic and paramagnetic materials, where, as
B ^ mqH,
while Eq. (5.57) holds for
B = m0
where p T
—
+ Xm
1
is
(
all
we shall
see later
linear magnetic media.
in this section,
Thus we have
H + M) = mo( + Xm)H = MoMrH,
1
a dimensionless proportionality constant called the relative
permeability of the material, which
e r in electrostatics. In practice,
pT
is
entirely analogous to the relative permittivity
(or
Xm)
for a given material can be
experimentally. In analogy to Eq. (2.50) in electrostatics,
we
determined
also introduce the
permeability (or absolute permeability) of the material.
permeability (unit:
M =
H/m)
(5.59)
/ZrAO),
where the value of the permeability of a vacuum, po,
is
given in Eq. (4.3), with
which,
(5.60)
constitutive equation of a
linear
magnetic material
The
unit for
media,
constitutive equation for free
space
pr =
p
1
is
henry per meter (H/m). For free space and other nonmagnetic
and
B = Mo H.
(5.61)
Section 5.5
For diamagnetic materials (see Section
than the permeability of a vacuum
5.2),
(e.g., /z r
=
the value of
pr is
Permeability of Magnetic Materials
237
slightly smaller
0.999833 for bismuth, a substance
which shows diamagnetism more strongly than most materials), whereas p, of paramagnetic materials is slightly greater than p-o (e.g., p, r = 1.0008 for palladium, one
of the strongest paramagnetic materials). Because p, differs only insignificantly from
li o, it is very common to assume p, = p-o for diamagnetic and paramagnetic materials, as well as for antiferromagnetic substances, in most practical applications. Thus,
these three classes of materials are commonly said to be nonmagnetic.
For ferromagnetic, ferrimagnetic, and superparamagnetic materials, on the
other hand, p is much larger than /xq. These materials, especially ferromagnetic
ones, often exhibit permanent magnetization, highly nonlinear behavior, and hys,
teresis effects, as
the function
we shall discuss in more
B(H
)
in
Eq. (5.56)
is
detail in a later section. In ferromagnetics,
in general nonlinear
and has multiple branches.
The magnetization properties of the material depend on the applied magnetic field
intensity, H and also on the history of magnetization of the material, i.e., on its pre,
vious states. In other words, the value of
not unique, but
is
a function of
p,
for a ferromagnetic material generally
is
H and the previous history of the material. Typically,
maximum
value of p r is around 250 for cobalt, 600 for nickel, and 5000 for iron
0.4% impurity), whereas it is as high as about 200,000 for purified iron (0.04%
impurity) and 1,000,000 for supermalloy (79.5% Ni, 15% Fe, 5% Mo, 0.5% Mn). In
many applications involving ferromagnetics, we assume
the
(with
(5.62)
perfect magnetic conductor
(PMC)
and such media are customarily referred to as perfect magnetic conductors (PMC).
As we shall see in the next section, the magnetic flux density vector, B, in a nonmagnetic medium near the surface of a ferromagnetic (or PMC) body is always
normal to the surface, the same as for the electric field intensity vector, E, near
the surface of a perfect electric conductor (PEC), with a -> oo. Shown in Table 5.1
are values of the relative permeability of an illustrative set of selected materials,
Table 5.1.
Relative permeability of selected materials
Material
Mr
Material
Mr
Bismuth
0.999833
Titanium
1.00018
Gold
Mercury
0.99996
Platinum
1.0003
0.999968
Palladium
1.0008
Silver
0.9999736
Manganese
1.001
Lead
0.9999831
Cast iron
150
Copper
0.9999906
Cobalt
250
Water
0.9999912
Nickel
600
Paraffin
0.99999942
Nickel-zinc ferrite (Ni-Zn-Fe 203 )
Wood
0.9999995
Manganese-zinc
Vacuum
1
Steel
Air
1.00000037
Iron (0.4% impurity)
Beryllium
1.0000007
Silicon iron
Oxygen
Magnesium
1.000002
Permalloy (78.5% Ni, 21.5% Fe)
1.000012
Aluminum
1.00002
Mu-metal (75% Ni, 14% Fe, 5% Cu,
Iron (purified - 0.04% impurity)
Tungsten
1.00008
Supermalloy (79.5% Ni,
ferrite
650
(Mn-Zn-Fe 203 )
1200
2000
(4%
5000
7000
Si)
7 x 10
4% Mo, 2% Cr)
15% Fe, 5% Mo, 0.5% Mn)
10
4
5
5
2 x 10
10 6
238
Chapter 5
Magnetostatic
Field in Material
Media
of considerable theoretical and/or practical interest - in terms of their magnetic
properties.
A
magnetic material
homogeneous when
said to be
is
do not change from point to point
material is inhomogeneous [e.g., p
=
its
magnetic properties
region being considered. Otherwise, the
in the
p(x,y, z) in the region]. Finally, some magNamely, an applied magnetic field
netic materials, such as ferrites, are anisotropic.
(vector B) in one direction can produce magnetization
becomes
direction in a material. Accordingly, Eq. (5.60)
~
BX~
=
By
- permeability tensor of
an anisotropic magnetic
|/i|
LB Z ]
material
where
[yu] is
~
ftxx Mjry l^xz
Mz* Mzy Mzz
the permeability tensor.
_
vector
M)
in
another
a matrix equation.
Hx ~
Hy
Pyz
l^yx fLyy
_
(i.e.,
(5.63)
-
LtfzJ
analogous to the permittivity tensor,
It is
[e],
defined by Eq. (2.52). However, Eq. (5.50) remains valid (it represents the definition of the
field), although B, H, and
are no longer parallel at a point. The
M
H
anisotropy of ferrites
is
used
in a
number
of microwave devices, including
some
types of gyrators, directional couplers, and isolators.
and homogeneous magnetic material, we can
Ampere’s law and bring 1 /p (because it
constant) outside the integral sign in the integral form of the law, Eq. (5.49), or
Note
substitute
is
that in a linear, isotropic,
H
by B/p
in the generalized
outside the operator (curl) sign in
<j)
respectively.
B
•
dl
=
differential form, Eq. (5.52), yielding
its
VxB = pi,
and
pic
(5.64)
We realize that these equations are identical to the corresponding free-
space laws, Eqs. (4.48) and (4.83), except for po being substituted by p. In the same
way, replacing pg with p in Eq. (4.7) gives the version of the Biot-Savart law for a
volume conduction current
in a
homogeneous magnetic medium
4it
x
f (J dv)
p
of permeability p:
R
(5.65)
&
l
H = B Ip. Additionally, Eq. (4.132) and the concept of magnetic permeability
imply that the Laplacian of the magnetic vector potential of a volume current in a
with
homogeneous magnetic medium
satisfies the
V2A =
Finally,
following differential equation:
-//J.
(5.66)
dv
f J
(5.67)
Eq. (4.108) becomes
A= p
inLir
From Eq. (5.50), we find that the magnetization vector in a linear and isotropic
magnetic material can be expressed in terms of the magnetic field intensity vector as
M= — -H= (— If
the material
vector, J m at
,
=
J
H = (/* - 1)H.
(5.68)
r
/
homogeneous, then the magnetization volume current density
a point in the material can be obtained directly from the conduction
is
also
volume current density
Jm
1
V Mo
Mo
vector, J, at that point:
VxM = Vx l(p
T
— 1 )H =
1
(p T
—
1 )
V
X
H=
(p r
—
1
)J.
(5.69)
We
see that there cannot be magnetization volume current (J m
neous linear magnetic medium with no free current (J = 0).
Problems'. 5.12-5.15; Conceptual Questions (on
5.6
From
239
Maxwell's Equations and Boundary Conditions for the Magnetostatic Field
Section 5.6
=
0) in a
Companion Website):
homoge-
5.8.
MAXWELL'S EQUATIONS AND BOUNDARY
CONDITIONS FOR THE MAGNETOSTATIC FIELD
the Biot-Savart law for the magnetic flux density
dB
of a single conduction
current element J dv in free space and the superposition principle,
we proved
in
Section 4.8 the law of conservation of magnetic flux, Eq. (4.99), for the magnetostatic field in free space. We can repeat here that same proof for a magnetization
current element, J m dv (or J ms dS), and get the same result, which means that the
net magnetic flux (the flux of the vector B) due to any distribution of conduction and
magnetization currents (which, equivalently, reside in a vacuum) through a closed
we conclude
surface always equals zero. In other words,
that the law of conserva-
tion of magnetic flux (Maxwell’s fourth equation) holds for structures that include
arbitrary magnetic materials.
We now
write
down
the
of Maxwell’s equations governing the mag-
full set
netostatic field in an arbitrary
medium, together with the associated
constitutive
equation:
fc H-dl = /s J-dS
Maxwell's second equation,
static field
$s B dS =
0
B = B(H)
[B
•
The corresponding
(5.70)
= fiH]
Maxwell's fourth equation
constitutive equation for
B
differential Maxwell’s equations are:
V
x
H=J
V B=
and
•
(5.71)
0.
In addition to integral and differential Maxwell’s equations for any field, bound-
Maxwell's differential
equations, magnetostatic
field
ary conditions always represent the third form of field equations, and they are
derived from the respective integral equations (differential equations apply only
Let us derive here the boundary conditions for the magnetostatic field
on the boundary surface between two arbitrary media. Let J s be the density vector
of a surface conduction (free) current that may exist on the boundary. We apply
the generalized Ampere’s law in integral form, Eq. (5.49), to a narrow rectangular
elementary contour C positioned such that J s is normal to the plane of the contour,
as depicted in Fig. 5.12. Having in mind Eqs. (2.79) and (5.30), we obtain
at a point).
<j>
H
•
dl
= Hn A/ -
H
A = Ic = J
2t
l
s
Al,
(5.72)
which yields
H\t
~
H2 = Js-
(5.73)
t
In vector form [also see Eq. (2.84)],
n x Hi — n x
H2 = J
s
,
(5.74)
where n is the normal unit vector on the surface, directed from region 2 to region
If no surface conduction current exists on the boundary, Eq. (5.73) becomes
H\t
— H2
t
(J s
=0
),
1.
(5.75)
boundary condition
for
directed from region
2
region
1
H
to
t
;
n
240
Chapter 5
Magnetostatic Field
in
Material
Media
Figure 5.12 Deriving the
boundary condition for
tangential components of
vector H on the boundary
surface between two arbitrary
magnetic media.
that
is,
the tangential
component of
H
is
continuous across the boundary free of
conduction current.
Noting, on the other hand, that the law of conservation of magnetic
flux,
Eq. (4.99), and the continuity equation for steady currents, Eq. (3.40), have exactly
the same form, we conclude that the corresponding boundary conditions must
have exactly the same form as well. Therefore, from the boundary condition for
normal components of the vector J in the steady current field, Eq. (3.55), we
directly write the boundary condition for normal components of the vector B in
the magnetostatic
boundary condition
for
field:
n B]
Bn
•
It tells
us that
At an
Bn
is
interface
—
n
•
B2 =
holds that
or
B\ n
= B2 n
(5.76)
-
always continuous across a boundary.
between two
linear magnetic
with no surface conduction current (J s
field lines
0
is
=
0),
media of permeabilities
/xj
and
\x 2
the law of refraction of the magnetic
entirely analogous to the corresponding laws in electrostatics,
field, Eq. (3.56). With aq and 012 denoting the angles
and region 2 make with the normal to the interface, as
Eq. (2.87), and steady current
that field lines in region
shown
in Fig. 5.13,
1
we have
tanoq
/j.\
tan
[i 2
law of refraction of magnetic
(5.77)
field lines
012
field lines are bent farther away from the
normal in the medium with the higher permeability. Bending of field lines is basically due to unequal magnetization surface currents on the two sides of the interface.
This relationship indicates that magnetic
Figure 5.13 Refraction of
magnetic field lines
magnetic-magnetic
at a
interface.
Section 5.7
Image Theory
for the
Magnetic
241
Field
Note that if region 2 is filled with a ferromagnetic material and region 1 with a
nonmagnetic material, then q. 2
pi and p-i/p.2 ~ 0, and Eq. (5.77) results in
»
oq^O
(|n 2
»/ri)
(5.78)
any a 2 (except for a 2 = 90°, i.e., for field lines in region 2 parallel to the intermeans that magnetic field lines in a nonmagnetic medium near the interface with a ferromagnetic medium (or PMC) are always normal to the interface.
Finally, using Eq. (5.32) and adding up the magnetization surface current density vectors that accumulate on the two sides of a magnetic-magnetic interface,
analogously to deriving Eq. (2.89) in electrostatics, we arrive to the following
boundary condition for the tangential components of the magnetization vector:
for
face). This
n x
where n
total
is
directed from
Mi - n
medium 2
to
M
x
2
medium
= J ms
(5.79)
,
1 (Figs. 5.12
and
5.13)
and J ms
is
the
magnetization surface current density vector at the interface.
Problems'. 5.16; Conceptual Questions (on
Companion Website):
5.9-5.12;
MATLAB Exercises (on Companion Website).
5.7
IMAGE THEORY FOR THE MAGNETIC FIELD
Magnetostatic systems often include current conductors in the presence of large
flat
ferromagnetic bodies.
described in Section 1.21,
By utilizing image theory, similarly to the procedure
we can remove the ferromagnetic body from the system
and replace it by an image of the original current distribution. The equivalent problem is then much simpler to solve because it consists of a known current distribution
image) in free space.
Consider a straight current conductor in a nonmagnetic half-space in the vicinity of an infinite planar interface with a ferromagnetic (or PMC) half-space. Let
the conductor be parallel to the interface. Eq. (5.78) tells us that the ferromagnetic
material in the lower half-space, i.e., the induced magnetization current in the material, influences the resultant magnetic flux density vector, B, such that it has no
tangential component at the upper side of the interface, as shown in Fig. 5.14(a).
This condition remains unaltered, however, if we replace the ferromagnetic block
by another conductor parallel to the interface that is positioned symmetrically with
respect to the original conductor and carries a current of the same intensity and
the same direction, Fig. 5.14(b). We conclude thus that, as far as the magnetic field
in the upper half-space is concerned, systems in Fig. 5.14(a) and Fig. 5.14(b) are
equivalent. This is an example of the image theory (theorem) for the magnetic field.
The theory is not restricted to conductors parallel to the material interface only.
It states that an arbitrary current configuration above an infinite ferromagnetic (or
PMC) plane can be replaced by a new current configuration in free space consisting
from the original current configuration itself and its positive image in the ferromagnetic plane. The equivalence is with respect to the magnetic field above the
ferromagnetic plane, the component of that field due to the induced magnetization current in the ferromagnetic material being equal to the field of the image.
As another example, Fig. 5.15 shows the image of a current conductor consisting of three segments with different orientations with respect to the ferromagnetic
interface in Fig. 5.15(a), where it is a simple matter to conclude that the vector B in
the plane of symmetry in Fig. 5.15(b) has no component tangential to the plane.
(original plus
boundary condition
for
M
t
242
Chapter 5
Magnetostatic Field
in
Media
Material
®
©
/
/
i
MO
Mo
'
;
(a)
Figure 5.14
X
I
®
i
i
i
i
(b)
(a) Straight
PMC)
current conductor parallel to the interface of a
image theory, the influence
on the magnetic field in the upper
half-space can be represented by a positive image of the original
ferromagnetic (or
half-space, (b) By
of the ferromagnetic material
M» Mo
current.
Example 5.8
A
Force on a Conductor above a Ferromagnetic Plane
very long and thin wire
(a)
parallel to
surface,
its
the wire per unit of
its
By image
Solution
wire system
in air, as
of
is
length
its
and
is
situated in air at a height h
carries a current of intensity
Here,
B
theory, the system wire-ferromagnetic
shown
in Fig. 5.16.
The
force
due
i.e.,
(b)
Figure 5.15 Image theory
magnetic field due to
a current conductor with
arbitrarily oriented segments
above a ferromagnetic (or
for the
PMC)
plane: (a) original
system with the material
interface
and
(b) equivalent
free-space system.
An
infinitely
is
I\
=
/2
=
/
(original)
and d
conductor per unit
2 h, that
is,
(5.80)
gj.
along the upper conductor due to the lower
PMC)
We
see
block always attracts a current conductor running
Strip
Conductor between Two
PMC
Planes
long thin strip conductor of width a carries a steady current of intensity
PMC planes, as shown in Fig. 5.17(a).
the space between the PMC planes, assuming that
placed between two parallel
flux density vector in
Solution
—
it.
Example 5.9
strip
equivalent to a symmetric two-
to the magnetization current in the ferromagnetic material.
that a large ferromagnetic (or
parallel to
is
on the upper
then obtained using Eq. (4.166) with
the magnetic flux density
is
above a ferromagnetic half-space,
Determine the magnetic force on
length.
n, = IB =
conductor,
/.
By
/.
The
Find the magnetic
it is
multiple applications of the image theory for the magnetic
air-filled.
field,
we
obtain
the equivalent system in Fig. 5.17(b), which represents an infinite planar current sheet with
surface current density 7S
=
I /a in free
thus uniform, with flux density
B=
space.
ijlqJs/2
The magnetic
= ^1 /(2a)
field
between the PMC planes
and Eq. (4.47)].
[see Fig. 4.21
is
B
B
/
243
Magnetization Curves and Hysteresis
Section 5.8
i
<•>-
h
-F'
A m
Mo
M0
M»M0
M0
Figure 5.16 Evaluation of the
h
force
on
a horizontal straight
wire with a current above a
ferromagnetic plane, using
0
/
Uniformly Magnetized Hemisphere on a
Example 5.10
A
PMC
uniformly magnetized ferromagnetic hemisphere of radius a in
The magnetization vector
(/r r
oo) plane.
shown
in Fig. 5.18. Calculate the
is
M, and
We
first
is
for
Example
5.8.
Plane
air
is
lying
on a
PMC
perpendicular to the plane, as
magnetic flux density and
center of the bottom surface of the hemisphere (point
Solution
it
image theory;
field intensity vectors at the
O in the figure).
evaluate the distribution of equivalent magnetization currents of the
hemisphere. The magnetization volume current density vector inside the material and the
magnetization surface current density vector on the bottom surface of the hemisphere are
both zero. There
is,
however, a hemispherical sheet of magnetization current over the upper
surface of the hemisphere given by Eq. (5.42),
ory for the magnetic
field,
where now 0 < 8 <
we then supplement
this sheet,
8
Using the image the-
Till.
considered to be in a vacuum,
with another hemispherical sheet below the plane of symmetry, which
is also described by
symmetric with respect
which corresponds to a positive image of the current, exactly as required by the
the function in Eq. (5.42), but with
to 6
= jr/2,
n /2 <
0
<n
(note that sin#
is
image theory). We thus obtain the full spherical sheet of magnetization current in Fig. 5.10,
and conclude that the magnetized hemisphere on the PMC plane is equivalent to the magnetized sphere of Fig. 5.10, in free space. Hence, the magnetic flux density vector at the sphere
center, as well as that at the point O in Fig. 5.18, is given by Eq. (5.45). Finally, by means of
Eq. (5.50), the magnetic field intensity vector at the point O is
MO
8/
a
8
PMC
8
H= — -M= ^M-M = —
3
8
PMC
(5.81)
3
(b)
(a)
Problems 5.17-5.19; Conceptual Questions (on Companion Website):
:
5.13;
MATLAB Exercises (on Companion Website).
Figure 5.17
(a) Strip
conductor between two
parallel
PMC
planes and (b)
equivalent infinite planar
5.8
MAGNETIZATION CURVES AND HYSTERESIS
current sheet
for
In this section, we consider in more detail the B-H relationship, Eq. (5.56), for ferromagnetic materials. This relationship, being nonlinear in general, is usually given
as a graph showing B (ordinate) as a function of
(abscissa).
curve representing
H
A
Figure 5.18 Ferromagnetic
hemisphere with
a
uniform
magnetization lying on a
plane; for
Example 5.10.
PMC
Example
in
5.9.
free space;
Chapter 5
Magnetostatic Field
in
Material
Media
Figure 5.19 Simple apparatus
for
measurement
of
magnetization curves.
the function 77(77) on such a diagram
is called a magnetization curve and is obtained
by measurement on a given material specimen.
Fig. 5.19 shows the simplest apparatus for measurement of magnetization
curves, where a uniform winding is placed over a thin toroidal core (ring) cut from a
ferromagnetic sample we want to measure. If a current 7 is established in the toroid
(primary coil), the magnetic field intensity 77 in the core is given by Eq. (5.53), where
N is the number of wire turns of the coil and I is the mean length of the toroid. By
varying 7, therefore, we directly vary 77 ( 77 is proportional to 7). The magnetic flux
density B in the core is then measured, as a response to 77, by a ballistic galvanometer (BG) connected to another winding (secondary coil) that is placed over the core.
Namely, the galvanometer measures the charge that passes through the secondary
circuit, as a consequence of the change of magnetic flux through the secondary coil.
This charge, as
we
shall see in the next chapter,
is
proportional to the flux change,
so that the galvanometer actually serves as a fluxmeter. Thus, by changing, step by
primary coil and field 77 in the core, and measuring the corresponding values of the flux density B, we obtain, point by point, the magnetization
curve of the material.
step, the current 7 in the
Shown
in Fig. 5.20 is a typical initial
sample, where the material
Figure 5.20 Typical
initial
magnetization curve for a
ferromagnetic material.
0
is
magnetization curve for a ferromagnetic
completely demagnetized and both
B and
77 are
zero
d
Magnetization Curves and Hysteresis
Section 5.8
is applied. As we begin to apply a current in the primary circuit in
magnetic flux density also rises, but not linearly. Moreover, the value of
rapidly at first and then more slowly. The first part of the initial magnetization
before a field
Fig. 5.19, the
B rises
curve (roughly up to the point
P
in Fig. 5.20) represents the region of easy (steep)
magnetization. In the upper section of the curve, the increase in magnetization due
domains in the material not already parallel to H
and the curve tends to become flat. This is the region of hard (flat)
magnetization. Very strong magnetic fields are usually required to reach the state of
saturation, where all the moments of magnetic domains in the material are parallel
to H and the magnetization curve flattens off completely.
The permeability at any point on the magnetization curve is given by
to gradual rotations of magnetic
is
more
difficult,
B
=
H
where
B
is
(5.82)
jj,
the ordinate of the point (in T) and 77
relative permeability
is
on the curve with the
then
/z r
=
is
the abscissa (in A/m).
n/no- The maximum permeability
largest ratio of
B
to 77,
i.e.,
at the point of
The
at the point
is
tangency with
the straight line of steepest slope that passes through the origin and intersects the
magnetization curve
(Fig. 5.20).
Note
that the
maximum
the steepest slope of the magnetization curve, because
slope of the curve
B/ d 77), but equals
(
the ratio
n
\i
is
does not correspond to
not proportional to the
B/H.
Having reached saturation, let us now turn to Fig. 5.21, where we continue our
experiment - by reducing 7 and 77. As we do so, the effects of hysteresis begin to
show, and we do not retrace the initial-magnetization curve. Hysteresis means that
B lags behind 77, so that the magnetization curves for increasing and decreasing the
applied field are not the same. Even after 77 becomes zero, B does not go to zero,
but to a value B = B r termed the remanent (residual) magnetic flux density. Note
that the existence of a remanent flux density in a ferromagnetic material makes
permanent magnets possible. We then reverse 77 (by reversing the polarity of the
,
battery in Fig. 5.19), and increase
zero at 77
= — 77c
,
where 77c
is
it
the negative direction further, the material
negative polarity.
It
B comes
in the negative direction, so that
the so-called coercive force.
As
77
to
increased in
is
becomes magnetized even more, with
passes through the stages of easy and hard magnetization, the
magnetization curve flattens off, and negative saturation is reached. The end of the
curve on the left-hand side of the diagram is for the field 77 = — 77m after which
,
Figure 5.21 Typical hysteresis
loop for a ferromagnetic
material.
245
246
Chapter 5
Magnetostatic
Media
Field in Material
Figure 5.22 Demagnetization
sample by
H;
of the normal
of a ferromagnetic
reversals of the applied field
definition
magnetization curve.
we
start
gives
B
reducing H. The next intercept of the curve with the
B = -B
x
(negative remanent flux density).
At
this point,
axis (for
we
H — 0)
reverse the bat-
H in the positive direction. This
makes the flux density zero at a positive field H = H (coercive force). With a further increase in H, the material reaches positive saturation (at H = Hm and a full
and continue - by increasing
tery polarity again
c
),
cycle in the
curve during
BH
diagram
is
completed. The loop traced out by the magnetization
this cycle is referred to as the hysteresis loop.
Having carried our ferromagnetic specimen to saturation
magnetization curve in Fig. 5.21, we now move on to Fig. 5.22,
the applied field
H
,
but over successively smaller ranges
(
±H
at
both ends of the
to continue to cycle
is
brought to smaller
and smaller amplitudes on each reversal). We obtain thus a series of hysteresis loops
that decrease in size, and the residual B for H = 0 eventually becomes zero, that is,
the material is left in a demagnetized state. Such a process is used for demagnetization of objects that have a residual magnetization (i.e., remanent flux density)
under conditions of zero applied magnetic field. In practice, the process can be
carried out by inserting the object to be demagnetized inside a coil with a lowfrequency ac current, and then gradually reducing the current amplitude in the coil
or slowly removing the object from the coil with the current amplitude constant.
The curve connecting
the tips of the hysteresis loops in Fig. 5.22
known
is
another charac-
normal magnetization curve. For a
particular ferromagnetic material, the normal and initial magnetization curves are
teristic of
ferromagnetic materials
as the
very similar.
H
row
Ferromagnetic materials having small coercive forces
c and therefore nar(thin) hysteresis loops (with small loop areas), as illustrated in Fig. 5.23, are
referred to as soft ferromagnetics.
As we
enclosed by the hysteresis loop
proportional to energy loss per unit volume
is
shall see in a later chapter, the area
of the material in one cycle of field variation. This
is
so-called hysteresis loss,
which corresponds to the energy lost in the form of heat in overcoming the friction encountered during the movements of magnetic domain walls and rotations
of the domains in a ferromagnetic material. Soft ferromagnetic materials have a
large magnetization for a very small applied field and exhibit low hysteresis losses.
Additionally, they have very large values of initial permeability [/u. in Eq. (5.82)
for a very small H], For example, supermalloy, a typical soft ferromagnetic, has
an initial relative permeability on the order of 10 5 B r = 0.6 T, and c = 0-4 A/m.
Soft ferromagnetic materials are used for building transformers and ac machines
,
H
Section 5.9
Magnetic
Circuits
- Basic Assumptions
Figure 5.23 Hysteresis loops
for soft
and hard ferromagnetic
materials.
(motors and generators), where the material
is
permanently exposed to alternating
magnetization.
Hard ferromagnetic materials, on the other hand, have large coercive forces
and
hence broad (fat) hysteresis loops (Fig. 5.23). They have small initial perc
meabilities, and are used for building permanent magnets and dc machines. In
these applications, fields do not change frequently, and hysteresis losses, therefore,
H
are not significant, in spite of large hysteresis loop areas.
of permanent-magnet materials
teristic
are permanently magnetized even with
is
a high
remanent
no magnetic
The
field applied),
important that their coercive force be large, so that the material
ily
essential charac-
B
flux density
but
may
it
(they
r
is
also
not be eas-
demagnetized. Alnico, an aluminum-nickel-cobalt alloy with a small amount of
is a typical hard ferromagnetic, having an initial [x x around 4, B r around 1 T,
copper,
and
H
c
on the order of 50,000 A/m.
Conceptual Questions (on Companion Website): 5.14-5.16.
MAGNETIC CIRCUITS
THE ANALYSIS
5.9
- BASIC
ASSUMPTIONS FOR
Magnetic circuit in general is a collection of bodies and media that form a way
along which the magnetic field lines close upon themselves, i.e., it is a circuit
of the magnetic flux flow. The name arises from the similarity to electric circuits.
In practical applications, including transformers, generators, motors, relays,
magnetic recording devices,
etc.,
cores of various shapes, that
wound about
windings
magnetic
may
or
may
circuits are
not have
formed from ferromagnetic
air gaps,
parts of the cores. Fig. 5.24
with current-carrying
shows a
typical
magnetic
circuit.
The
analysis of magnetic circuits with steady currents in the windings (dc
netic circuits)
in arbitrary
is
media, Eqs.
(5.70). Given, however, great complexity of the rigorous
magnetic circuits, we introduce here a set of approximations
conjunction with Maxwell’s equations, make the analysis much simpler
analysis of practical
that will, in
mag-
based, of course, on Maxwell’s equations for the magnetostatic field
and yet accurate enough for engineering applications.
for the Analysis
247
1
248
Chapter 5
Magnetostatic
Field in Material
Media
Figure 5.24 Typical magnetic
circuit.
Firstly, we assume that the magnetic flux is concentrated exclusively inside the
magnetic circuit, i.e., in the branches of the ferromagnetic core and air gaps. This
is never exactly true. For a toroidal ferromagnetic core with a uniform winding
shown in Fig. 5.25(a), however, the flux is restricted to the interior of the toroid
as a consequence of the geometrical symmetry of the structure, provided that the
wire turns are very tightly
wound over the core.
age flux between the wire turns. In addition,
In reality, there
in circuits
is
always some leak-
containing coils placed only
over parts of the core, some flux lines bridge the space between the core sections
through
field
Boundary conditions
as indicated in Fig. 5.25(b).
air,
for the magnetostatic
applied to the interface between the ferromagnetic and air in Fig. 5.25(b)
tell
us that magnitudes of the magnetic flux density vector at points a and b in the figure
are related as
so
J3 a
we conclude
%
/z r £b-
As
/z r is
very large for ferromagnetic materials,
that practically the entire magnetic flux
is
Ba
B b,
restricted to the ferro-
magnetic core, and the larger the q, r the more accurate this assumption. Note that
in electric circuits this is always true for the current flow because the conductivity
of air is zero, whereas the permeability of air is not.
Secondly, we assume that air gaps in the magnetic circuit are narrow enough so
that the fringing flux near the gap edges can be neglected. For more precise analysis,
formulas for an effective length and cross-sectional area of the gap may be used to
incorporate the fringing effects into the basic equations for the
6
6
(b)
Figure 5.25 Toroidal
ferromagnetic core with
a
(a)
uniform winding and (b)
concentrated winding.
we assume
circuit.
uniform throughout the volume of
each branch of the circuit. Then, in applying Maxwell’s equations, every cross section of a branch is assumed to have the same area, and the path of every flux line
along the branch is assumed to be of the same length, equal to the mean length
of that part of the circuit. This is true only for thin magnetic circuits. In many
applications involving thick cores, however, the error due to thin magnetic circuit
approximations is acceptable, which is illustrated in the following examples.
Finally,
Example
Assume
5.1
that the magnetic field
is
Thick Toroid with a Linear Ferromagnetic Core
that the toroidal coil in Fig. 4.18
is
wound about
a core
made
of a linear ferromag-
Determine the magnetic flux through the core, (b) Find
the error made in the flux computation if the magnetic field in the core is assumed to be
uniform, specifically for b — a = 0.1a (thin toroid) and b - a = a (thick toroid).
netic material of permeability
/u.
(a)
Solution
(a)
The magnetic
From Eq.
// inside a thick toroidal coii
field intensity vector,
H,
is
=
-
the
same
as in the air-filled toroid in Fig. 4.18.
(4.65),
H(r)
—
2nr
,
a
<
r
<
b.
(5.83)
.
Magnetic
Section 5.9
Circuits
- Basic Assumptions
for the Analysis
Figure 5.26 Evaluation of the
magnetic
flux
through
a cross
section of a thick toroidal coil
with a linear ferromagnetic
core; for
The magnetic flux through the core
is
—
where dS
(b)
Under
is
f
B(r)
5.1
1
then obtained by integrating the flux density B(r)
/xH(r) over a cross section of the toroid, as
<t>
Example
=
dS
shown
/xNIh
2n
r
=
in Fig. 5.26,
b
b
—— In-,
dr
fjiNIh
2n
r
Ja
(5.84)
a
the surface area of a thin strip of length h and width dr.
the assumption of a uniform field distribution in the core, the path of every flux
assumed to be of the same length, equal to the mean length of the toroid, / =
n(a + b ), and the magnetic field intensity is given by Eq. (5.53). The approximate flux is
then obtained by multiplying the constant flux density B = ixH by the surface area S of
line
is
the core cross section,
= BS =
$ approx
FF
The associated
(iH(b -a)h
tc
relative error in the flux
l^approx
~
computation
2 (b/a
Q|
<5<j>
(1
For b
b
—
a
—a=
=
a
case), the
(
0.1 a, that
b/a
= 2),
is,
5<t>
=
is
a
computed
-
(5.85)
+b
as
1)
(5.86)
+ b/a) In (b/a)
b/a = 1.1, the error turns out to be 5<j> = 0.075%, while for
3.8%. We see that even for a quite thick toroid (the latter
=
approximate expression
in
Eq. (5.85)
is
reasonably accurate.
Thick Toroid with a Nonlinear Ferromagnetic Core
Example 5.12
be wound around a core made from a nonlinear ferroshows the cross section of such a structure, with a = 2 cm,
6 = 4 cm, 6 = 1 cm, N = 200, and / = 1 A. Assume that the initial magnetization curve
of the material can be approximated by the piece-wise linear curve shown in Fig. 5.27(b).
Calculate the magnetic flux through the core (a) rigorously and (b) assuming that the field in
Let the toroidal
coil of Fig. 4.18
magnetic material.
the core
is
Fig. 5.27(a)
uniform.
B
B
/
0
Figure 5.27 Analysis of a thick toroidal
for
initial
Example
coil
1000
with a nonlinear ferromagnetic core:
magnetization curve of the material, and
5.1 2.
-)
Hk =
A/m
(b)
(a)
idealized
H
Ilk
—
(c) distribution of
a
1
c
i-
b
(c)
(a) cross section of
the structure, (b)
the magnetic flux density along the radial
axis;
n
:
250
Chapter 5
Magnetostatic
Field in Material
Media
Solution
(a)
From Eq.
(5.83), the
tfmin
minimum and maximum magnetic
=
=
H(b)
respectively. Since the field value
tization curve in Fig. 5.27(b)
which
is
A/m
and
Hk =
1000
796
H max = H(a) =
A/m,
satisfies the
where Ua
is
the
initial
Ma
=
^
no
=
permeability of the material.
= Hk
H(c )
we
in the linear
is
regime and above
(5.88)
Hk
is
in saturation
[
B(r )
= Sm
],
in the linear regime. In this latter case,
is
= Ma H(r),
B(r)
(5.87)
condition
which H(r ) >
that the part of the core for
whereas the remaining part of the core
A/m,
Hmax*
^min < Hk <
we conclude
1592
core are
representing the limit on the magne-
up to which the material
(“knee” value),
in saturation
field intensities in the
0.001
From
H/m,
(5.89)
the condition
(5.90)
,
obtain the radial distance c that represents the boundary between the two parts of
the core:
C=
[see Fig. 5.27(a)].
Sketched
2^ =3 2Cm
(5 9,)
'
in Fig. 5.27(c)
is
'
the distribution of the magnetic flux density,
B(r), along the radial axis.
The magnetic
through the core
flux
is
(see Fig. 5.26)
b
f
<&=
B(r) d S
= B m h(c - a) +
,
Ja
b
/
—dr
Jc
r
fi*NIh
—
f
2n
.
*
saturation
linear regime
= Bm h
(b) If
we assume
where
1061
NI
a)
(c
is
5.10
r mea
Hk
5.20;
=
-
191.4
c
n Wb.
,
=
is
(a
+ b)/2 =
(5.92)
3 cm. That
implies that the entire core
= B m (b — a)h —
relative to the result in Eq. (5.92)
given that the core
Exercises (on
ln
is,
H = H(rmean —
)
is in
saturation.
The
hence
^approx
Problems
-
which, being greater than
magnetic flux through the core
The error
b
,
a uniform field distribution in the core, then the field intensity every-
in the core equals that for r
A/m,
+ 2nH
tt
k
is
<5<p
200
=
gWb.
(5.93)
4.5%, and
this is
very reasonable
both thick and nonlinear.
Conceptual Questions (on Companion Website):
5.17;
MATLAB
Companion Website).
KIRCHHOFF
S
LAWS FOR MAGNETIC
CIRCUITS
With the assumptions made (in the previous section) that the field is restricted to
the branches of the magnetic circuit (flux leakage and fringing are negligible) and
is uniform in every branch, we now specialize general Maxwell’s equations for the
magnetostatic field to obtain the laws analogous to Kirchhoffs laws in the electric
circuit theory. Thus, applying the law of conservation of magnetic flux, Eq. (4.99), to
,
l
,
Section 5.10
Figure 5.28
A
Kirchhoff's
Laws for Magnetic
251
Circuits
closed surface S
about a node and
a closed
path
C along
the axes of branches of
a magnetic circuit - for the
formulation of Kirchhoff's laws
for
a closed surface
as
shown
magnetic
S placed about a node (junction of branches)
circuits.
in a
magnetic
circuit,
in Fig. 5.28, yields
M
B dS =
•
£
where Si
and Bi (i
(i
=
m
1,2, ...
M
,
)
J2BiSi
= 0,
(5.94)
Kirchhoff's "current"
magnetic
M)
1,2,
0
law
for
law
for
circuits
are cross-sectional areas of the branches in the junction
are the magnetic flux densities in these branches.
Applying, on the other hand, the generalized Ampere’s law, Eq. (5.49), to a
contour C placed along a closed path of flux lines in the circuit (Fig. 5.28) gives
P
£
=
H
•
dl
= Ic
Q
E^lE^
j
=l
(5.95)
P) standing for the lengths of the branches along the path
P) for the magnetic field intensities in the branches, whereas
2,
Q) are the numbers of wire turns and current intensities,
respectively, in the coils that exist along the path. The product Nkh, expressed in
ampere-turns, is termed a magnetomotive force (mmf), in analogy to an electromo-
with
lj
(j
1, 2,
.
and Hj (j = 1, 2,
Nk and Ik (k = 1,
.
tive force
.
.
.
,
.
,
.
.
.
(emf) in electric
circuits.
Eqs. (5.94) and (5.95) are referred to as Kirchhoff’s laws for magnetic
cuits.
In addition to these circuital laws,
we need
cir-
the “element laws” describing
individual parts of the circuit, analogous to current-voltage characteristics for
elements
(e.g., resistors)
tionships
B=
in the analysis of electric circuits. These laws are the relaB(H), i.e., the magnetization curves, for the branches of the circuit,
including air gaps (where B = hqH). Because of the nonlinear nature of the ferromagnetic portions of the magnetic circuit, and nonlinearity of magnetization curves,
the analysis of magnetic circuits often resembles the analysis of nonlinear electric
circuits which contain diodes and other elements with nonlinear current-voltage
characteristics. This is, at the same time, the most important difference between the
analysis of magnetic circuits and the electric circuit theory, which primarily deals
with linear electric circuits.
where the ferromagnetic materials in the circuit can be considered
however, an equivalent electric circuit with linear resistors and timeinvariant voltage generators can be introduced. By solving the equivalent circuit
using some of the standard circuit-theory techniques, we obtain thus all required
In cases
as linear,
Kirchhoff's "voltage"
magnetic
k=
circuits
252
Chapter 5
Magnetostatic Field
in
Material
Media
quantities for the original magnetic circuit.
As an
note that the mag-
illustration,
netic flux through a simple magnetic circuit in Fig. 5.29(a) can be expressed, having
in
mind Eq.
(5.53), as
O = BS = ix HS =
NI
/jlNIS
(5.96)
H'
/
where
reluctance (unit:
H
1
(5.97)
J
is
the so-called reluctance of the core.
It is
defined generally as the ratio of the mag-
netomotive force (ampere-turns) to the flux. The SI unit for reluctance is ampere
_1
per weber ( A/Wb) or inverse henry (H ). We note that the final expression for 0 in
Eq. (5.96) is analogous to the expression for the current in a simple electric circuit in
which an ideal voltage source of emf NI is connected with a resistor of resistance IZ,
as indicated in Fig. 5.29(b). Eq. (5.97),
moreover, has the same form as the expres-
sion for the resistance of a resistor with uniform cross section and conductivity
a
in
Eq. (3.85). The concept of reluctance can be used for the analysis of arbitrary linear
magnetic circuits, where the equivalent electric circuit is obtained by replacing the
individual parts of the core and the air gaps by resistors with resistances calculated
from Eq.
(5.97)
and representing the coils by voltage generators with electromotive
magnetomotive forces NI.
forces equal to the
O
Simple Nonlinear Magnetic Circuit with an Air Gap
Example 5.13
Consider a magnetic
an
air
R=
gap shown
50
£2.
emf
is Iq
coil
has
—
4
mm,
N=
ferromagnetic core with a
is
is
£
=
coil
and
1000 turns of wire with the total resistance
circuit
is
the cross-sectional area of the toroid
of the generator in the coil circuit
curve of the material
and
The
The length of the ferromagnetic portion of the
(width) of the gap
the
circuit consisting of a thin toroidal
in Fig. 5.30(a).
200 V. The idealized
given in Fig. 5.30(b). Find the magnetic
=
/
is
S
1
=
m, the thickness
2
So = 5 cm and
initial
,
magnetization
field intensities in the
core
in the air gap.
Figure 5.29 Reluctance
concept: a simple linear
magnetic
circuit (a)
and the
Solution
We
adopt standard approximations for the analysis of magnetic
circuits
and
neglect the field nonuniformity in the ferromagnetic core, as well as the flux leakage from
H) and Bq Hq)
equivalent electric
the core and fringing around the air gap edges (Section 5.9). Let (B.
circuit (b).
ignate the magnetic flux density and field intensity in the core and the gap, respectively, as
Figure 5.30 Analysis of a
simple nonlinear magnetic
circuit
with an
(a) circuit
air
gap:
geometry, with flux
density and field intensity
vectors
and
in
the circuit,
(b) idealized
initial
magnetization curve of the
material, with the load line
and operating point for the
circuit; for Example 5.1 3.
(
,
des-
Section 5.10
indicated in Fig. 5.30(a).
From
Kirchhoff's
Laws
for
Magnetic
Circuits
253
Kirchhoff’s laws for magnetic circuits, Eqs. (5.94) and (5.95),
we have
—
BS = B 0 S0
B = £0
»
(5.98)
and
HI
where
I
= £/R =
4
A
is
+ HqIq =
AT,
the current intensity of the
(5.99)
coil.
The above equations, combined with
the constitutive equation for the air gap,
Bo
=
(5.100)
between the
result in the following relationship
+ B— =
HI
and
flux density
field intensity in the core:
NI.
(5.101)
Mo
With the numerical data
substituted,
H + 3183B = 4000
and
(H
in
A/m;
BinT),
equation of the load line for the magnetic circuit (the locus of
this represents the
possible combinations of values
B
and
all
H for this given circuit configuration and excitation,
not yet taking into account the characteristics of the core material).
in the
(5.102)
BH chart [Fig. 5.30(b)], we conclude that
its
Upon
plotting this line
intersection with the magnetization curve
of the core material belongs to the second part of the curve (hard magnetization section
-
also see Fig. 5.20). This intersection (point P) represents the operating point for the circuit,
i.e., it
determines the actual position of the ferromagnetic core on the magnetization curve
and excitation (NI). To
for the given circuit configuration
ordinate of the point
P,
we
and
find numerically the abscissa
composed of the load
solve the system of equations
line equation,
Eq. (5.102), and the equation of the line describing the hard magnetization segment of the
magnetization curve, read from Fig. 5.30(b),
(5.103)
+ 2.5 x 10" 5 (//- 1000),
and what we get is B = 0.44 T and H = 2600 A/m. By means of Eq. (5.98), Bo = 0.44 T as
well, and Eq. (5.100) then gives Hq = 350 kA/m (field intensity in the air gap). We see that
B=
Hq
0.4
This can also be concluded without actually solving the circuit, by introducing the
-4
P, m = B/tt = 1.7 x 10
H/m, into Eq. (5.98) and
tt.
permeability of the material at the point
expressing the magnetic field intensity in the gap as
tt0
=
—
tt
=
(5.104)
135.3tt.
M0
(Note that
/r
here
is
a function of 77, not a constant.) It
air gaps, in general, that
in the
is
typical for
the magnetic field intensities in the gaps are
magnetic
much
circuits
with
larger than those
ferromagnetic material.
Example 5.14
Simple Linear Magnetic Circuit with an Air Gap
n
o
Assume
that the core of the circuit of Fig. 5.30(a)
material of relative permeability
Solution
Combining Eq.
4>
where the reluctances [Eq.
72
=
—— =
q, r
=
1000, and
m {jrs + 7kY
(5.97)] of the core
B=
flux
=HAAAAArn
through the core.
iiH we can write now
,
NI
=
7Z
(5.105)
+ 72o
and the gap are
Figure 5.31 Equivalent
l
1.6
x 106
HT
1
and
72 0
MrMOO
and the equivalent electric
-4
the core amounts to 4> = 5 x 10
Wb.
respectively,
a linear ferromagnetic
compute the magnetic
(5.101) with the relationship
= BS =
made from
is
=
=
6.37 x 10
6
H -1
(5.1
,
06)
MoSo
circuit
is
shown
in Fig. 5.31.
electric circuit for the
magnetic
Hence, the
flux
through
circuit of
Fig. 5.30(a),
assuming that
the ferromagnetic material
is
linear; for
Example 5.14.
Chapter 5
Magnetostatic
Field in Material
Media
Figure 5.32 Analysis of a
nonlinear magnetic circuit with
three branches: (a) circuit
geometry, with the adopted
independent node and closed
paths in the circuit, and (b)
idealized
initial
magnetization
curve of the material, with
computed operating
points for
the branches; for Example 5.15.
Nonlinear Magnetic Circuit with Three Branches
Example 5.15
Dimensions of the magnetic
5i
= £2 =
^3
=
1
circuit
cm 2 The mmf
.
shown
in the first
in Fig. 5.32(a) are
branch of the
=
l\
circuit is
l3
=
N1 =
2lj
=
20
cm and
400 ampere-turns.
The
initial magnetization curve of the core can be linearized as in Fig. 5.32(b). Calculate the
magnetic flux densities and field intensities in the three branches of the circuit.
Solution Let us orient the branches of the circuit as in Fig. 5.32(a). There is a total of two
nodes and three closed paths in the circuit. Thus, in complete analogy with the analysis of
electric circuits, Kirchhoff’s “current” and “voltage” laws for magnetic circuits, Eqs. (5.94)
and (5.95), are to be applied to one node (independent node) and two closed paths (independent closed paths), respectively. We choose the node ]\f\ and closed paths C\ and C2 in
Fig. 5.32(a).
The corresponding equations
-B
1
S1
are:
+ B 2 S2 + B 3 S3 =
—»
0
H + H2 = NI
-H + H =
l l1
Depending on whether
3 l3
(5.107)
,
,
(5.108)
0.
(5.109)
l2
2 l2
Bi=B 2 + B 3
the operating points for the individual branches of the circuit belong
to the linear part or to the saturation part of the magnetization curve in Fig. 5.32(b),
we may
have
B,=
li a
Hi
initial
them
is
A/m
for Hi
> 900 A/m
is /r a
=
0.9/900
H/m =
0.001
H/m.
So, there
combinations for the magnetization stages of the branches and only one
true.
Suppose,
in all
T
permeability of the material
exist a total of eight
of
Hi < 900
(5.110)
0.9
where the
for
first,
that
none of the magnetizations
becomes
in the
branches
is
in saturation
(
B=
[x a
H
three branches). Eq. (5.107) then
Hi
=H
2
+ H3
,
(5.111)
H
and the solution of the system with Eqs. (5.111), (5.108), and (5.109) is H\ = 1500 A/m, 2 =
1000 A/m, and 3 = 500 A/m. This is contradictory to the assumption of linearity in all three
branches, since both H\ and
2 appear to be larger than 900 A/m. We conclude that our
initial guess is not correct, and the circuit has therefore to be solved again, with another
assumption, i.e., that some of the branches are in saturation.
To reduce the number of additional trials (out of remaining seven combinations) to a
minimum, we note first that the adopted reference directions of the magnetic flux density
vectors in Fig. 5.32(a) likely reflect the actual flux flow in the branches, meaning that B 1,
B 2 and B 3 are all positive. Eq. (5.107) then tells us that B\ must be larger than both B 2
and 63 individually, so it is logical to expect that the branch with the coil would first reach
saturation ( B\ — 0.9 T). The other two branches, however, must remain in the linear regime
(B 2 = a H2 and B 3 — /x a H3 ), because B 2 = 0.9 T would imply B 3 = 0 and vice versa, which
H
H
,
)JL
is
impossible.
With
this,
Kirchhoff's
=
H2 — 600 A/m,
Eq. (5.107) becomes
H2 + H3 = 900 A/m,
and the solution of the
Laws
Section 5.1 0
new system
of three equations
(5.112)
is
H\
1700 A/m,
H
and 3 = 300 A/m. Obviously, these values are consistent with the new assumption made
about the magnetization stages in the branches, and this is the true solution for the circuit.
The actual positions of the operating points for the branches on the magnetization curve are
indicated in Fig. 5.32(b). The remaining two flux densities in the circuit are B 2 = 0.6 T and
B3 =
0.3 T.
Reverse Problem
Example 5.16
For the magnetic
circuit
shown
the Magnetic Circuit Analysis
in
in Fig. 5.33,
=
l\
l3
=
20 cm,
l2
=
l'
2
+ 1% =
10 cm,
Iq
= 1 mm,
S3 = 1 cm S2 = 2 cm and N2 I2 = 1000 A turns. The core is made from the material
whose initial magnetization curve can be approximated analytically by the function
5i
=
2
2
,
,
1
B=
If
500
C f_T
+ H’
H ~°
the magnetic flux in the central branch of the circuit
downward reference
Solution
This
is
is
$2
=
(5.113)
)-
200 n Wb with respect to the
direction, find N\I\.
a reverse
problem
in the
magnetic
(magnetic flux in one or more branches of the
or
5inT
(HinA/m;
more magnetomotive
forces) that produces
circuit analysis: for a given
circuit), find the
it.
unknown
response
excitation (one
In the analysis of magnetic circuits with
nonlinear magnetization curves (given either graphically or analytically), reverse problems
are generally
much
simpler to solve than direct problems, because in the latter case
to simultaneously solve the full set of equations written in
we have
accordance to Kirchhoff’s circuital
laws together with nonlinear material characteristics for individual parts of the circuit. In
reverse problems,
on the other hand, we
with the given
start
field quantities for
branches and then applying the circuital and material equations one at a time
one or more
we
solve for
other field quantities, one by one, and for the required magnetomotive forces.
With reference to the notation
in Fig. 5.33, the flux densities in the
of the second (central) branch of the circuit
B2
From
the magnetization curve,
i.e.,
( B 2 ) and in
=B 0 =
ferromagnetic section
the air gap (Bo) are
^ = 1T.
(5.114)
by solving the magnetization equation, Eq. (5.113), for
the corresponding field intensity in the material,
h2 =
500B 2
1.5
- B2
= 1 kA/m,
(5.115)
Figure 5.33 Magnetic circuit
with a nonlinear magnetization
curve given analytically; for
Example
5.1 6.
for
Magnetic
Circuits
256
Chapter 5
Magnetostatic
Field in Material
whereas
Media
in the gap.
//o
~=
=
795.8
kA/m.
(5.116)
MO
Kirchhoff’s “voltage” law for the closed path
N
H„ =
2 I2
~
H2
l2
C2
now
in Fig. 5.33
H0 =
521
~
lo
3
yields
A
A/m,
,
(5.117)
h
with the corresponding flux density in the third branch given by
B3 =
I.5//3
500
0.765 T.
+ //3
(5.118)
The flux density in the first branch is next obtained using Kirchhoffs “current” law
node M\ in Fig. 5.33, and from it the associated H value,
B
=
1
—— =
B
2 S2 — A 3 S 3
-----
1.235
T
Hi
=
we apply Kirchhoffs “voltage” law to the
N\I\
=
H\l\
l
in a
±_
=
2.33
kA/m.
Magnetic
(5.119)
-fli
closed path C\ and obtain the
+ H2 2 + H0 lo =
Demagnetization
Example 5.17
500 Ai
1.5
5i
Finally,
—
for the
1362
A
turns.
mmf we seek:
(5.1
20)
Circuit
The ferromagnetic core of the circuit shown in Fig. 5.34(a) has the cross-sectional area
5 = 1 cm 2 mean length / = 20 cm, and air-gap thickness /o = 1 mm. The number of wire
turns of the coil is N = 1000, the resistance of the winding is R = 20 £2, and the emf of the
generator is £ = 20 V. The core does not have residual magnetization and the switch K is in
,
the off position (open).
By
turning the switch on, the
mmf
is
applied to the circuit and the
magnetic flux through it rises following the initial magnetization curve of the material. This
curve can be considered as linear, as shown in Fig. 5.34(b), where the initial permeability is
= B m in the stationary
fj a = 0.001 H/m. The magnetic flux density in the core becomes B
.
state.
Figure 5.34 Magnetization and
demagnetization in a magnetic
circuit with an air gap: (a)
circuit
initial
geometry, (b) idealized
magnetization and
demagnetization curves of the
material, (c)
magnetic
flux
density and field intensity
vectors
in
the
circuit,
and
operating points for the
two stationary
Example 5.1 7.
in
(d)
circuit
states; for
The switch
is
then turned off (opened) and a
new
stationary state established in the
Section 5.10
circuit.
Kirchhoff's
Laws for Magnetic
257
Circuits
The demagnetization curve for the material can be approximated by two straight-line
[Fig. 5.34(b)], where the coercive force is given by H c = Hm What is the magnetic
segments
.
gap
field intensity in the
in the
new
state?
Solution After the switch K is turned on (closed), the current in the winding is [Fig. 5.34(c)]
= £/R = 1 A, and the relationship between the flux density B and field intensity in the
core is the same as in Eq. (5.101). Moreover, as
H
I
B = p a H,
we
obtain [see also Eq. (5.105)]
B=
Nl(— + — )
\Ma
and
1
(5.121)
maximum
this is the
=
1 T,
(5.122)
MO/
B = Bm —
value for the flux density,
T. Hence,
1
Hm = 5 m //r =
a
kA/m.
After the switch
K is opened, Eq.
(5.101)
HI
becomes
+ B— =
(5.123)
0,
Mo
and this represents the equation of the load line for the magnetic circuit in this state. We
suppose first that the intersection of this line with the demagnetization curve in Fig. 5.34(b)
belongs to the horizontal segment of the curve, i.e., that B = B m = 1 T. From Eq. (5.123),
= — 4 kA/m, which is impossible because < —Hc ( c = m = 1 kA/m). We conclude
thus that the operating point for the circuit belongs to the vertical segment of the demagnetization curve, as indicated in Fig. 5.34(d). The magnetic field intensity in the core is therefore
H
H
H = —H = —1 kA/m.
c
B=
0.25 T.
For the magnetic
=
Eq. (5.123) then gives the corresponding flux density in the core,
field intensity in the gap is Hq = 5/mo = 200 kA/m.
The magnetic
Example 5.18
N\
Complex Linear Magnetic
shown
circuit
made can be considered
Solution
5.35(a),
in Fig.
h
—h=
as linear, with relative permeability
parts of the core
all
Circuit
10 cm,
=5 cm,
I3
Iq
=
1
mm,
N2 = 500, and 7i = I2 = 1 A. The ferromagnetic material from which the core
1000,
area of
H
H
Fig. 5.35(b)
S
is
=
1
cm 2
.
Find the magnetic
shows the equivalent
=
ju. r
1000.
The
is
cross-sectional
field intensity in the air gap.
electric circuit for the
problem, with reluctances,
Eq. (5.97),
(a)
72i
=
72 2
=
272. 3
=
—
=
8 X 10
5
H -1
,
Ko =
figure also
8 x 10
6
H _1
.
24)
(5.1
M o5
MrMCP
The
=
!R2
22.1
shows the adopted loops for the loop analysis of the
(electric) circuit.
The
corresponding loop equations are
+ 72-3 + 72o)4>i + 1Z\ 2 = N\I\,
72i 4*1 + (72i + 722) ^2 — 7Vi/i + A^2?2,
(72i
and
their solution for the flux
722
4>i
72q72i
The magnetic
through the central branch of the
M
(5.125)
<t>
— IZ 1 N2 H
I]
+ 72o72-2 + 72i72.2 + 72-i723 + 722723
field intensity in the
=
(5.126)
circuit
2.84 x 1(T
5
Wb.
(5.127)
Figure 5.35 Analysis of a
gap comes out to be
linear
magnetic
circuit
with
three branches: (a) circuit
H0 =
= 226 kA/m.
M0 5
(5.128)
geometry and
analysis; for
Problems 5.21-5.27; Conceptual Questions (on Companion Website); 5.18-5.21;
:
MATLAB
Exercises (on
Companion Website).
(b) equivalent
electric circuit for
loop
Example
5.1 8.
258
Chapter 5
Magnetostatic Field
in
Material
Media
MAXWELL'S EQUATIONS FOR THE TIME-INVARIANT
ELECTROMAGNETIC FIELD
5.1 1
Maxwell’s equations for the magnetostatic field, Eqs. (5.70), represent a general
mathematical model for determining the magnetic field (H) from a distribution of
steady electric currents
(J),
which
considered to be known. The distribution of
is
currents, however, for the geometry, material properties,
and excitation of a given
structure can be obtained from Eqs. (3.59), which also yield the solution for the
electric field (E) in the system. In addition, Eqs. (3.60) provide a
means
for evalu-
ating the associated distribution of charges (p) in the system from the electric field
distribution. All these equations together represent a full set of Maxwell’s equations
that govern both the electric and the magnetic field due to steady electric currents.
These two fields, moreover, can be considered as components of a more complex
field - the electromagnetic field - produced by steady currents. The combined field
is
called the time-invariant electromagnetic field,
Maxwell’s equations
Maxwell's
first
form for
in differential
equation,
and we summarize here the four
this field:
V x E=0
V xH=J
V D=p
static field
Maxwell's second equation,
static field
(5.129)
•
Maxwell's third equation
V-B =
0
Maxwell's fourth equation
What
very
is
important,
the
and magnetic
electric
constituting
fields
the
and independent
from each other, and can be analyzed separately, as we have done in Chapters 3
(electric part) and 5 (magnetic part). This is not the case, however, with the
electromagnetic
time-invariant
field
are
we
unrelated
entirely
Under
same form
whereas the first two equations have additional terms on the
equations, which are responsible for the coupling between the
time-varying electromagnetic
field, as
shall see in the following chapters.
nonstatic conditions, the third and fourth Maxwell’s equations retain the
as in Eqs. (5.129),
right-hand side of
and magnetic fields that change in time. In chapters that follow, the electric
and magnetic fields will always be treated together, as related integral parts of the
electric
time-varying electromagnetic
field.
Continuity Equation from Ampere's
Law
Starting from Maxwell’s second differential equation for the time-invariant electromagnetic
field
derive the corresponding form of the continuity equation.
Solution
Let us take the divergence of both sides of the differential form of Maxwell’s
second equation (generalized Ampere’s law) for
V
The second-order vector
any vector
2
field,
2
The divergence of
is
if
A
=V
•
(5.130)
J.
derivative on the left-hand side of this equation
the curl of an arbitrary vector function (A)
V
obvious
[Eq. (4.103)].
(V x H)
is
equal to zero for
so the right-hand side of the equation must be zero as well,
operator.
This
•
static fields, in Eqs. (5.129):
is
•
(V x A)
=
it
for
A
representing any vector
a
(b
x
c)
=
(a
always zero,
or, in
•
J
=
0,
and
terms of the del
0.
the magnetic vector potential, because then
To prove
is
V
field,
x b)
B=V
x
A
[Eq. (4.1 16)] and
V B=
•
0
however, we apply formally the identity
c
L
259
Problems
the differential form of the continuity equation under the static assumption,
this is exactly
We
Eq. (3.41).
see
now
that Eq. (3.41)
is
actually included implicitly in the set of Maxwell’s
equations in Eqs. (5.129).
Problems
:
5.28;
MATLAB Exercises (on Companion Website).
Problems
5.1.
Nonuniformly magnetized parallelepiped.
rectangular ferromagnetic parallelepiped
uated in
the
air in
A
is sit-
octant of the Cartesian
first
coordinate system (x,y,z >0), with one vertex at the coordinate origin,
lengths a, b, and
y,
and
c,
z, respectively.
+
Figure 5.36
Uniformly
The magnetization vector
magnetized square
given by M(x, y, z) =
sin(7rz/c)y + sin(^x:/fl) z],
in the parallelepiped
Mo[sin(7ry/h) x
and the edges, of
parallel to coordinate axes x,
ferromagnetic
is
plate; for
Problem
where Mo is a constant. Compute (a) the
volume magnetization current density vector
in the parallelepiped and (b) the surface magnetization current density vector over
5.2.
Hollow
drical
cylindrical bar
magnet.
its
sides.
<
b ), and
permanently magnetized with a uniform magnetization and situated in air. The
magnetization vector, of magnitude M, is parallel to the bar axis. Find (a) the distribution
of magnetization currents of the magnet and
(b) the magnetic flux density vector along the
length
/,
is
axis.
5.3.
M=
assume that the magnetization vector is
Mo y (Mq — const). Neglecting end effects,
center (point O).
5.5.
Nonuniformly magnetized ferromagnetic disk.
A thin ferromagnetic disk of radius a and thickness d (d <$C a) in air has a nonuniform mag= Mo(r/a) 2 z (Fig. 5.37),
netization, given by
where Mo is a constant. Calculate (a) the dis-
M
tribution of magnetization currents of the disk
and
(b) the
magnetic flux density vector along
the z-axis.
Uniformly magnetized square ferromagnetic
plate. A uniformly magnetized square ferromagnetic plate of side length a and thickness d (d <^a) is situated in air. With reference to the coordinate system in Fig. 5.36, the
magnetization vector in the plate is given by
= Mo z, where Mo is a constant. Determine
the magnetic flux density vector at an arbitrary
Figure 5.37
Nonuniformly
magnetized thin
ferromagnetic
M
point along the z-axis.
5.4.
(in
disk; for
Problem
5.6. Infinite cylinder
Magnetization parallel to plate faces. Consider
the square ferromagnetic plate in Fig. 5.36, and
An
infinitely
product
V (V x
V (V
•
•
(cross product of a vector with itself
the expression for the curl (of
be zero.
is
with circular magnetization.
long ferromagnetic cylinder of
change the
A), and obtain
x A)
=
(V x V)
•
A=0
always zero). Alternatively, the divergence, using Eq. (1.167), of
any vector)
in the
5.5.
radius a in air has a nonuniform magnetization.
a scalar triple product, the cyclic permutation of the order of the three vectors does not
result) to the scalar triple
find
the magnetic flux density vector at the plate
A hollow cylin-
bar magnet of radii a and b (a
5.3.
Cartesian coordinate system in Eq. (4.81) turns out to
260
Chapter 5
Magnetostatic
Field in Material
In a cylindrical coordinate system
r < a), where Mo is a constant. Find (a)
volume magnetization current density vec-
5.13.
Thin toroidal
coil
A
with
with a linear ferromagnetic
N
zation current density vector on the cylinder
uniformly and densely over a thin toroidal
surface, (c) the magnetic flux density vector in
ferromagnetic core of permeability
the cylinder, and (d) the magnetic flux density
steady current of intensity /
vector outside the cylinder.
the
Find the magnetic field intensity vector, H, (a) along the
axis of the magnetized disk from Example 5.2,
netic field intensity vector, (b) the magnetic
Magnetic
field intensity vector.
magnet
coil
turns of wire
is
fx.
If
a
established in
the circulation of (a) the mag-
coil, find
and
flux density vector,
(c) the
vector through the core, along
5.14. Solenoidal
wound
is
coil
magnetization
its
mean
length.
with a linear ferromagnetic
from Example 5.3, and (c) inside and outside
the nonuniformly magnetized infinitely long
cylinder from Example 5.4.
core.
Total (conduction plus magnetization) current
ber of wire turns per unit length of the core
in a
tion
The magnetic
is
flux density vector, B,
ferromagnetic material
of spatial
There is a uniform and dense solenoidal
winding wound over a very long cylindrical ferromagnetic core of permeability [x. The num-
a
is
coordinates,
known
func-
Prove that
(a)
ume
J
current density in the material, J to t
+ Jm
,
can be obtained as J to
(b) Specifically,
compute J to
t
t
=V
for
B
=
B(x, y, z) = {[(2x + z)/y
+ 2y) z] T (x, y, z in m).
nates:
(x
]
x
+
(2 /y) y
+
face magnetization current density vector over
the surface of the core.
5.15.
volume current density
vector, J m
=
in air.
dl
=
situated
vector, (b) the magnetic flux density vector,
and
magnetization vector inside and outside
the cylinder, as well as (d) the distribution of
magnetization currents of the cylinder.
5.16.
boundary
Magnetic-magnetic
Assume that the plane z =
(z > 0) and medium 2
1
=
tive
0.
permeabilities
ix T \
=
conditions.
0 separates
(z
<
0),
600 and
medium
with rela/x T 2
= 250,
The magnetic field intensity vector
+
in medium 1 near the boundary (for z = 0 ) is
Hi = (5 x — 3 y + 2 z) A/m. Calculate the magrespectively.
magnetic
The magnetization
vector,
current density vector,
J,
flux density vector.
M, and conduction
are known at every
netic field intensity vector in
point of a magnetic body. Find the expression
for the circulation of the
magnetic
vector along a closed path
C
boundary
flux density
magnetization and conduction current.
Consider an imaginary closed path C inside
a homogeneous magnetic material of relative
(for z
=
and
Js
5.17.
(b) there
= 3y A/m
is
H2
medium
2 near the
no conducon the boundary (J s = 0)
0~),
tion current exists
situated entirely
inside the body.
5.12. Total
/x r is
a uniform conduction current of
distribution of (a) the magnetic field intensity
(c) the
const), in a current-free (J to t
5.11. Circulation of the
is
density J flowing along the cylinder. Find the
0) region, prove the following vector identity:
§c
There
.
Closed path in a uniform field. Considering an
arbitrary contour in a magnetic field with no
spatial variation of the magnetic flux density
vector (B
Ferromagnetic cylinder with a conduction current. A very long ferromagnetic cylinder of
radius o and relative permeability
Constant flux density vector in a magnetic
region. In a certain magnetic region, the
magnetic flux density vector does not vary with
spatial coordinates. The conduction volume
current density vector is J. Find the magnetization
steady current of intensity / flows
density vector in the core, as well as (e) the sur-
x B/ixq.
given as
the following function of Cartesian coordi2
A
N'.
through the winding. Determine (a) the magnetic field intensity vector, (b) the magnetic
flux density vector, (c) the magnetization vector, and (d) the volume magnetization current
the total (conduction plus magnetization) vol-
5.10.
conduction current
total
core.
density.
5.9.
The
Ic What is the total magnetization current /m c enclosed by C?
is
tor in the cylinder, (b) the surface magneti-
(b) at the center of the cylindrical bar
5.8.
/x r .
C
enclosed by
<
the
5.7.
permeability
2 -axis
M = Mo(r/a)<\>
coincides with the cylinder axis,
(0
whose
Media
,
if
(a)
a surface current of density
on the boundary.
Force on a conductor above an infinite PMC
corner. Two PMC (/x T -> 00 ) half-planes are
connected together
at
an angle of 90° with
.
261
Problems
respect to each other, as
A
shown
in Fig. 5.38.
/J-T—^OO
very long and thin wire in air runs parallel
from each
to the half-planes, at a distance h
of them.
If
a steady current of intensity I
M
is
established in the wire, find the per-unit-length
magnetic force on
it.
h
r
’
Figure 5.40
Uniformly
a
Figure 5.38 Cross
magnetized cylinder
between two
0
section of a system
consisting of a
parallel
Mr-— oo
and an
Problem
5.18.
corner; for
5.21.
5.1 7.
Uniformly magnetized hollow disk on a PMC
plane. A uniformly magnetized hollow ferromagnetic disk surrounded by air is lying on a
PMC plane. The magnetization vector is M, and
it is normal to the plane, as shown in Fig. 5.39.
The disk radii are a and b, and thickness is
d (d
a, b). Obtain the magnetic flux density
«
vector along the z-axis for z
>
planes;
infinitely large
PMC
90°
PMC
Problem 5.19.
for
current-carrying wire
0.
Simple nonlinear magnetic circuit. Fig. 5.41
shows a magnetic circuit consisting of a
thin magnetic core of length / = 40 cm and
2
cross-sectional area S = 2.25 cm and an air
gap of thickness Iq = 0.25 mm. The winding has
TV = 800 turns of wire with a steady current
of intensity I = 1 A. The core is made from
a nonlinear ferromagnetic material whose initial magnetization curve can be linearized in
parts as in Fig. 5.27(b). Find the magnetic field
intensities in the core
and
in the air gap.
S
Figure 5.39 Hollow ferromagnetic disk with a uniform
magnetization lying on a
PMC
plane; for Problem 5.18.
Figure 5.41 Simple nonlinear magnetic circuit
with an
5.19,
Magnetized cylinder between two
A ferromagnetic cylinder of radius a and height
h
placed between two parallel PMC planes,
5.22.
is
as portrayed in Fig. 5.40.
The
The medium around
is uniformly
magnetized throughout its volume, with the
= Mq z, where
magnetization vector given as
Mo is a constant. Find the magnetic flux density
the cylinder
is
air.
cylinder
5.20.
gap; for Problem 5.21
PMC
Complex
nonlinear
magnetic
circuit.
Dimensions of the magnetic circuit shown in
Fig. 5.42(a) are l\ = / 3 = 2/2 = 20 cm and S\ =
S2 = S3 = 2 cm 2 The magnetomotive forces in
the circuit are N\Ii = 100 ampere-turns and
.
M
vector in the space between the
air
PMC planes.
TV2/2
=
300 ampere-turns. The idealized
magnetization curve of the core
planes,
Fig. 5.42(b).
Compute
both inside and outside the cylinder.
sities
Magnetic
flux
branches of the
Example
5.12 but for the piece-wise linear
through a thick toroid. Repeat
approximation of the initial magnetization
curve of the core material given in Fig. 5.30(b).
5.23.
and
Magnetic
is
field intensities in
each of the three
circuit.
circuit
5.42(a),
in
the magnetic flux den-
with
a
zero
flux
branch. Referring to the magnetic
Fig.
initial
shown
let
l\
=
I3
= 30
cm,
I2
in
one
circuit in
- 10 cm,
H
262
Chapter 5
Magnetostatic Field
l\,S\
in
/
3,
Material
Media
53
Figure 5.43 Nonlinear magnetic circuit with
(a)
three branches,
two
gaps, and
air
two mmf's;
for
Problem 5.24.
B[ T]
k
5.25.
1
1
1
1
—
//[A/m]
the material whose initial magnetization curve
can be approximated analytically by the func> 0.
tion B = arctan(///250) T (
in A/m),
1
1000
0
Reverse problem with a nonlinear magnetic circuit. Refer to the magnetic circuit shown in
Fig. 5.43 and described in the previous problem, and assume that the core is made out from
H
(b)
Under
Figure 5.42 Nonlinear magnetic
three branches and
geometry and
two mmf's:
(b) idealized
circuit
these circumstances, find the
in the third
with
through the
(a) circuit
branch such that the magnetic flux
first branch (branch with the given
=
mmf 7Vi/i) is
initial
magnetization curve of the material; for
mmf N2 I2
upward reference
i
125 ^uWb with respect to the
direction.
Problem 5.22.
I
5.26.
Si=S3 = 10 cm 2 S2 = 20 cm 2 Ni = 1500,
and N2 = 1000. Let the idealized initial magne,
,
tization curve of the core be that in Fig. 5.27(b).
If
the current of the winding in the second
branch of the
current
I\
circuit
is
I2
= 0.5
of the winding in the
A,
find the
first
branch
such that the magnetic flux in that branch
5.24.
Remanent flux in a circuit with zero mmf.
Repeat Example 5.17 but for / = 50 cm and
is
/o
5.27.
=
mm.
0.5
Linear magnetic circuit with three branches.
Assuming
that the ferromagnetic material out
of which the core of the magnetic circuit from
Problem 5.24
linear,
is
made can be considered
with relative permeability
/x r
=
as
1000,
N2 I = 500 A turns, (a) find the reluc-
zero.
and that
Nonlinear magnetic circuit with two air gaps.
For the magnetic circuit shown in Fig. 5.43, l[ =
l'[ = /
= /3 = 12 cm, l2 = 5 cm, Iq = 0.6 mm,
3
and N\I\ = 1100 A turns. The cross-sectional
area of each of the branches is S — 2.5 cm 2 The
initial magnetization curve of the core can be
approximately represented as in Fig. 5.27(b).
Determine the mmf N2 I2 of the second coil
such that the magnetic flux through that coil is
tances of the individual parts of the core and
2
the air gaps and generate an equivalent electric circuit for
the problem, (b)
electric circuit in (a), find the
By
solving the
magnetic
field
intensities in each of the air gaps.
.
zero.
5.28. Continuity
equation from Ampere’s law in
from Maxwell’s second
form for the time-invariant
integral form. Starting
equation
in integral
electromagnetic
integral
field,
derive the time-invariant
form of the continuity equation.
1
Slowly Time-Varying
Electromagnetic Field
Introduction:
W
e
now
introduce time variation of electric
and magnetic
fields into
our electromagnetic
model. The new field is the time-varying electromagnetic field, which is caused by time-varying
charges and currents. In the new model, all the
quantities, in general, change in both space and
time. As opposed to static fields, the electric and
magnetic fields constituting the time-varying electromagnetic field are coupled to each other and
cannot be analyzed separately. Namely, a magnetic field that changes in time produces (induces)
an electric field and thus an electric current, and
this
phenomenon
is
known
as
electromagnetic
induction. Additionally, a changing electric field
induces a magnetic
field.
As we
shall see in a
mutual induction of time-varying
and magnetic fields is the basis of time
of electromagnetic waves and of electromagnetic
radiation.
The time-retardation concept is one of the most
important phenomena in electromagnetics.
cally tells us that there
is
a time lag
It
basi-
between a
change of the field sources, i.e., of time-varying
charges and currents, and the associated change of
the
fields,
so that the values of field intensities at
a distance
from the sources depend on the values
of charge and current densities at an earlier time.
In other words,
it
takes
some time
for the effect
of a change of charges and currents to be “felt”
The time lag equals the time
needed for electromagnetic disturbances to propaat distant field points.
gate over the corresponding distance.
We
shall see
later chapter, this
in a later chapter that the velocity of propagation
electric
of electromagnetic disturbances in a
=3
vacuum or
air
x 10 8 m/s (speed of light
retardation in electromagnetic systems (lagging in
(free space) equals c$
time of time-varying electric and magnetic fields
behind their sources), and as such of propagation
and other electromagnetic waves in free space).
Hence, the time lag in free space is r = R/cq, where
263
264
R
Chapter 6
Slowly Time-Varying Electromagnetic Field
the distance between the source and field
is
(observation) points.
If
the time r for
that
all
combinations of source and
domain of
field points in a
interest
is
much
than the time of change of the sources
shorter
the
[e.g.,
period of change of time-harmonic (steady-state
sinusoidal) charges
/
and currents, T =\/f, where
the frequency of the sources], the retardation
is
new
effect in the
system can be neglected. With
ignating the
maximum dimension
interest (containing all sources
of the
and
D
des-
domain of
all field
points
of interest for the analysis), which most often
is
the
under consideration, so that R < D,
and r the corresponding (maximum) time lag in the
domain, we thus have r = D/co <£ T = 1// as the
condition under which the retardation is insignificant. This means that the system size and the rate of
change of charges and currents are such that electromagnetic disturbances propagate over the entire
system (or the useful part of the system) before
the sources have changed significantly. We refer to
such charges and currents as slowly time-varying
sources and the corresponding electromagnetic
slowly time-varying
indicating that
fields,
is
not present under the static assumption
electromagnetic induction, and the fundamen-
is
governing law of electromagnetics describing
new phenomenon is Faraday’s law of electromagnetic induction. However, this law is probatal
this
most important basic experimental fact of
It is the underpinning of all of
the dynamic-field practical applications, from electric motors and generators, through propagation
of electromagnetic waves, to antennas and wireless
communication.
bly the
electromagnetics.
We
entire system
fields as
feature of the quasistatic electromagnetic field
shall start the study of the slowly time-
varying electromagnetic
induced electric
by introducing the
field
field intensity
vector due to a single
point charge in accelerated motion as an experi-
mental postulate, and then generalize this concept
to the evaluation of the induced electric field intensity vector due to any spatial distribution of slowly
time-varying currents. The
component due
the system, with the
will
Coulomb
electric field
to time-varying excess charge in
same form
as in electrostatics,
be discussed and added to the
field equations.
The concept of the induced electromotive
force
slow when compared to the velocity of travel of electromagnetic
disturbances (waves). In time-harmonic electro-
will
magnetics, a slow time variation corresponds to the
tions for the slowly time-varying electromagnetic
the rate of their change in time
low frequency, and the
retardation
is
field
negligible
is
is
model
which the time
in
On
induced electric field due to motion of conductors
in static magnetic fields will be introduced as an
time-varying
time-harmonic)
field
field.
(e.g.,
the other
high-frequency
cannot be analyzed without
taking into account the travel time of electromag-
from one point
in the
system to
another. In this chapter (and the following one),
we
our attention to slow time variations and
low frequencies of sources and fields.
restrict
The slowly time-varying (low-frequency)
has
time-invariant
the
This
is
why
tromagnetic
ties
field
being
it
is
field.
many formal
(static)
elec-
similarities with
electromagnetic
field.
also called the quasistatic elecIn
be completed, along with the associated
version of the continuity equation. In parallel, the
side, the rapidly
tromagnetic
field will
called accordingly the
low-frequency electromagnetic
netic disturbances
be introduced and Faraday’s law of electromagnetic induction derived using the magnetic
vector potential. The full set of Maxwell’s equa-
addition to
now time-dependent,
6.1
all
the quanti-
the only essentially
field
and discussed
in the
con-
All
new concepts and equations
will
be applied
whole variety of quasistatic
electromagnetic systems. Examples will include systems based on transformer induction (stationary
conductors in changing magnetic fields) and those
involving motional induction (moving conductors
in static magnetic fields), as well as structures in
which both types of electromagnetic induction are
present at the same time (moving conductors in
changing magnetic fields).
to the analysis of a
INDUCED ELECTRIC FIELD INTENSITY VECTOR
We know from
field,
impressed electric
text of Faraday’s law of electromagnetic induction.
Chapter
1
that a point charge
Q in free space is a source of an electric
predicted by Coulomb’s law and described by Eq. (1.24).
the Biot-Savart law (Chapter 4)
tells
On
the other side,
us that there will also be a magnetic
field,
given
Section 6.1
by Eq.
(4.4), if this
charge moves with some velocity v in space.
Induced
Electric Field Intensity
Vector
265
We now introduce
whenever the velocity
deceleration) a = dv/d t of the
a third field, which will exist in the space around the charge
v changes in time,
charge
is
i.e.,
whenever the acceleration
not zero. This
new
an
field is
(or
electric field in its nature. It
is
called the
2
induced electric
field
and
its
intensity vector
Eind(0
is
dv
G Jt_
—
4tt
which
is
an experimental result as
a time-varying field, that
(4.107),
we conclude
is,
Ei nd
well.
is
Of
(6.1)
point charge in an
’
accelerated motion
is V/m. This is
Comparing Eqs. (6.1) and
course, the unit for Ei ncj
a function of time.
1
any instant
actually equal to the negative of the time rate of
is
change of the magnetic vector potential, A,
Eind(0
this
R
that the induced electric field intensity vector at
of time and any point of space
Combining then
given by
—
at that point,
3A
(6.2)
IT'
induced
electric field intensity
vector (unit:
temporal differential relationship between
A
V/m)
and Ej nc with
J, and A in
i
the spatial integral relationship between the current density vector,
Eq. (4.108), we obtain the following integral (more precisely, integro-differential)
expression for the induced electric field intensity vector due to an arbitrary volume
current distribution:
ind
_
“
_mo
4 7t
_9_
f
Jdv
dtjv R
Here, the time derivative operator can be
directly to the current density vector
_ _mo f
~ 4 tv
Jv
(3J/3Q dv
(6.3)
R
Ei n d due to volume current
moved inside the integral sign and applied
because the differentiation with respect to time
spatial integration over the volume v are entirely independent operations and
can be performed in an arbitrary order. Similarly, Eqs. (4.109) and (4.110) yield the
corresponding expressions for a surface current of density J s and line current of
and
intensity v?
f (3J s /3r) dS
4 tt Js
R
Mo
_ _M0
Ejnd(0
-
Eind(0
Mo [
f ( di/dr)
4tc Ji
(6.5)
Ejnd due to line current
dl
R
'While using the notation Ej n d(0 for the induced electric field vector to emphasize its time dependence,
and similarly designating other time-varying field quantities (field intensity vectors, flux density vectors,
potentials, charge and current densities, etc.) that will be introduced in this chapter, we always keep
mind
that
all
these quantities, in general, are functions of spatial coordinates as well,
e.g.,
Ei„d
=
Eind (x,y,z,t).
2
When
dealing with time-varying spatially distributed quantities,
respect to time
(e.g.,
we
use the partial derivative with
3 A/3 1) rather than the ordinary (total) derivative (dA/dr) to
emphasize their
multivariable character.
3
As mentioned in Chapter 3, for circuit-theory quantities (e.g.,
we use lowercase notation (e.g., and v) to distinguish
in time,
i
constant (dc) regime, which are capitalized (/ and V).
to surface current
Ei n d
We conclude from Eqs. (6.3)-(6.5) that an induced electric field will exist in a system
whenever time-varying electric currents (representing accelerated or decelerated
motion of electric charges) exist in the conductors. Such currents are said to induce
in
due
(6.4)
current intensity and voltage) that vary
it
from the same quantities
in the time-
x
.
266
Chapter 6
Slowly Time-Varying Electromagnetic Field
the
field.
On the other side,
it
will
be zero
time
at all instants of
constants with respect to time (time-invariant currents) at
means
Eqs. (6.3)-(6.5) represent a general
when
J s and / are
J,
,
points of the system.
all
for evaluating (analytically or
numerically) the electric field due to any spatial distribution of slowly time-varying
we
currents in free space. In a later chapter,
in these
shall introduce certain corrections
expressions to extend their validity to rapidly time-varying current dis-
wave propagation
tributions. Essentially, these corrections introduce
computation of
Example
due to rapidly time-varying
fields
Induced
6.1
effects in the
currents.
Electric Field of a Straight
Wire Conductor
1
5
Find the expression for the electric
by a slowly time-varying current
wire structure in
We
Solution
field intensity
vector at an arbitrary point in space induced
segment of length
in a straight
i(t)
/
representing a part of a
air.
use the integral expression in Eq. (6.5) and refer to Fig. 6.1. Let P' and P
field point positioned arbitrarily
denote, respectively, a source point along the segment and a
The coordinate defining the position of the point
in space.
and *2
(*2
—x\ =
/)
P'
is
x, x\
< x < * 2 where
,
x\
are the coordinates of the starting and ending point of the segment,
Note that both x\ and X 2 can be either positive or negative, as well as zero,
depending on the actual position of the point P with respect to the segment. Finally, let d
respectively.
/
Figure 6.1 Evaluation of the
induced
electric field
due
mark
induced electric
(Fig. 6.1), the
to a finite straight wire
conductor with
d/
/no
a slowly
dt
4tt
time-varying current; for
Example
finite straight
f dl
J,
=
is
/x o
di
the segment.
„
C
*2
An d t * JX[
y/x 2
+ d2
and
given by
dx
(6.6)
'
2
jx + d2
where di/dt can be brought outside the integral sign because the current intensity i(t) does
not change along the wire (given that the current is slowly time-varying). 4 The solution of
6.1
this integral
Ejnd -
field intensity
R ~
As R
vector at the point P
P from
the perpendicular distance of the point
= dxx
dl
I
is
(6.7)
wire
conductor
We
note that the expression
electric field
due
in
Eq. (6.7) can be combined for computing the induced
any number of straight wire segments with a slowly
to structures containing
time-varying current.
Example 6.2
A
Induced
Electric Field of
source of electromagnetic interference (EMI) can be approximated by a square current
contour of side length a
=
current whose intensity,
i(t), is
cm
5
in free space, as
a pulse function
electric field intensity vector at the point
Solution
the
is
4
an EMI Source (Square Contour)
We note
domain of
M.
for electromagnetic disturbances to propagate over
interest (from source points at the
on the order of
As we
needed
that the time
shown in Fig. 6.2(a). The contour carries a
shown in Fig. 6.2(b). Compute the induced
r
=
q/cq
=
0.167 ns (cq
=
contour to the
3 x 10
8
field
point
m/s). Since this time
M)
is
in Fig. 6.2(a)
much
shorter
shall see in a later chapter, the intensity of a rapidly time-varying current in a wire conductor, in
general, changes along the conductor.
5
Note
that
2
f Ax/Jx +
easily verified
2
cl
=ln(x+
by differentiation.
\l
1
+ d2 j + C
(C being
the integration constant), which can be
1
I
Induced
Section 6.1
Electric Field Intensity
267
Vector
all
M
r[ns]
all
L
(a)
X
Eind [V/m]
Figure 6.2 Evaluation of the
induced electric field near
square current contour:
333-
Ejndi
Ejnd
(a)
1
o
E
Ejnd2
M
©
,111
Ejnd4
y
0
5
10
f[ns]
and
-333
(d)
By means
(rise
or
fall)
of the current intensity
i(t)
At
in Fig. 6.2(b),
=
5 ns, the
of the superposition principle, the total induced electric field intensity vector
[Fig. 6.2(c)]
Ejnd
=
+ Ej n(j2 + Ej n d3 + Ej n d4>
Ej nlji
(6.8)
where the field intensity vectors due to individual sides of the contour, Ej n di,
Ej n d 4 are
obtained from Eq. (6.7). Due to mutual antisymmetry of the second and fourth current segments with respect to the point M, E; nd 2 = — Ej n d 4 The distances d in the expressions for
computing E; n di and E; n d 3 equal all and 3a/2, respectively, whereas xi = —a /I and *2 = a /I
are the same in both expressions. Hence, the resultant Ej nd at the point
comes out to be
.
.
.
,
,
.
M
Eind
—
=
Ej n di
i
T Ej n d3 —
7
—1.11 x 10
In
,
4 7Z dt
— xV/m
+ Vi
i + 72
+ VTo
— i + vTo
l
In
(di/df in A/s),
(6.9)
dt
Ejnd
Fig.
6.2(b),
dr/d/
is
nonzero only during the
rise
and
fall
intervals
when di/dt = Ai/At = ±15 A/(5 ns) = ±3 x 10 9 A/s, which
= ±333 V/m. The function Ej n d(0 is plotted in Fig. 6.2(d). We see that very
current
pulse,
of
yields
strong
pulses of the induced electric field in the form of “spikes” are generated in the vicinity of
the contour. This field thus
may cause
a very strong undesirable interference
(EMI)
into the
operation of neighboring circuits in the system.
Example 6.3
Induced
Electric Field of a Circular
Wire Segment (Arc)
Consider a wire conductor in the form of an arc representing a part of a wire contour with
a slowly time-varying current of intensity
a and angle a, as
shown
in Fig. 6.3(a).
intensity vector at the arc center (point
Solution
the same,
(d) resultant field intensity
Example
current can be considered as slowly time-varying and the system as quasistatic.
From
to
as a function of time; for
(c)
the
due
individual sides of the contour,
®
given by
of the problem,
contour current intensity
intensity vectors
Ejnd3
is
geometry
(b)
as a function of time, (c) field
15
1
than the time of change
i(t)
in free space.
The
arc
is
defined by
a [see Fig. 6.3(a)],
O) due
we can
its
radius
Find the expression for the induced electric
bring
field
to this current conductor.
Since the distance of the field point from the source point in this case
R=
a
it
is
always
outside the integral sign in Eq. (6.5), which
6.2.
-
268
Chapter 6
Slowly Time-Varying Electromagnetic Field
Figure 6.3
(a) Electric field
induced by
a slowly time-
varying current along a circular
wire segment and (b) application of the head-to-tail rule
for vector addition to solve the
integral of dl along the
segment; for Example 6.3.
leads to
MO d /
—
Ejnd
dl,
(
A rca dr
6 10 )
.
M and N are the starting and ending points of the conductor, respectively. Using the
where
we observe from
head-to-tail rule for vector addition,
dl
Fig. 6.3(b) that
=MN
(
6 11
.
)
(note that this result holds true not only for a path in the form of an arc but for an arbitrarily
M and N). Finally, as the distance between
shaped path, planar or nonplanar, between points
= 2asin(a/2),
points
and N equals
M
MN
r
Emd
Ejnd - circular wire segment
Example 6.4
=
—Ana
—
Mo
A
dt
7
-
—
2n
«
- Mo sin-
MN =
dr
di
.
x.
(
2 dr
6 12 )
.
Current Contour of Complex Shape
Consider the contour consisting of two semicircular and two linear parts in Fig. 1.51, and
assume that it carries a time-harmonic current of intensity i(t) = cos(9 x 10 6 r) A (r in s) with
respect to the clockwise reference direction, as well as that a = 10 cm and b = 20 cm. Under
these circumstances, determine the induced electric field intensity vector at the contour
center (point O).
Solution Since the period of change of the current intensity i(t), T = 2n/u) = 0.7 gs
(where a> = 9 x 106 rad/s is the angular frequency), is much longer than the time
z = b/co = 0.667 ns needed for electromagnetic disturbances to propagate from source
points at the larger semicircle to the field point O in Fig. 1.51, the time-harmonic current
in the
contour can be considered as a low-frequency (slowly time-varying) current.
Referring to Fig. 6.4 and employing the superposition principle, the resultant induced
electric field intensity vector
c
Ejndl
that
is,
—
Ejndi
Mo d / -i
QR
An a d/
,
is
Mo
2n
—
,
+ Ej n d3 =
0.
Ej n d2
given by Eq.
-
Eind 3
—
M0 dt"i
An dt
~z
(6.7),
=
(6.12),
-J^
Anb
SP
— (-x),
=
2n
(6.13)
at
on the other hand,
+ y/b 2 + 0
a + ya 2 + 0
b
In
From Eq.
dt
df
From Eq.
=
(6.8).
„
X
—
—
An
= - M0
b
dz
dt
In
a
„
x.
(6.14)
Section 6.2
269
Slowly Time-Varying Electric and Magnetic Fields
Figure 6.4 Evaluation of the
induced
electric field
a current
due to
contour with two
and two linear
Example 6.4.
semicircular
parts; for
Due
to symmetry, Ej n d 4
Ej nc
i
=
Ej ncj 2 so that the total Ej nc equals
= 2Ej n d2 = — rr^-r
27r at
Problems
:
6.
1-6.9;
i
,
In
- x=
1.25sin(9 x 10
6
f)
x
V/m
(6.15)
(fins).
a
Conceptual Questions (on Companion Website):
6. 1-6.8;
MATLAB Exercises (on Companion Website).
6.2
SLOWLY TIME-VARYING ELECTRIC AND MAGNETIC
FIELDS
The slowly time-varying
electric field, in general,
is
composed of two components:
the induced electric field, given by Eqs. (6.3)-(6.5), and the
field
- the
field
field intensity
due
to excess charge,
vector
is
which we denote here as
Coulomb
E q The
.
electric
total electric
thus
E(0
= Ein d (f) + E*(0.
(6.16)
total electric field
plus
The Coulomb
field
component has
the
same form
due to excess volume charge
1
E q (t)
47T£o
R2
Jv
induced
by
now time-dependent. For
in free space
f PC) dv
=
fields
as in electrostatics, given
Eqs. (1.37)-(1.39), however, the charge densities are
instance, the field
Coulomb
^
is
obtained as
(6.17)
Equivalently,
E qit) = -W(t),
(6.18)
electric field
due
to excess
charge
where the
electric scalar potential
V(t)
is
expressed as
f Pit)
4 Tteo Jv
[see Eqs. (1.101)
and
(1.82)].
R
dv
(6.19)
slowly time-varying electric
potential
270
Chapter 6
Slowly Time-Varying Electromagnetic Field
As
E q (t)
the spatial distribution of the field
electrostatic field,
we can
has
the properties of the
all
write
E At)
[Eq. (1.75)] or
V
E
x
=
dl
•
(t)
=
(l
0
(
0
6 20 )
.
(6.21)
[Eq. (4.92)], which are the mathematical expressions of the conservative character
of the field
E q (t).
and
In accordance to Eqs. (1.88)
(1.90), the time-varying potential
M and N in space
difference (voltage) between points
given by
is
/•N
vmn(0
The
tial
due
field
= ^m(0 -
=
E^fr)
/
•
(6.22)
dl.
JM
to excess time-varying charge also obeys Gauss’ law. In differen-
notation [Eq. (1.166)],
V E q (t) =
However,
•
Ej nc
i
(6.23)
'
so
and (4.119) leads
a combination of Eqs. (6.2)
v
Pit)
—
3A\
= -V
to
3
=
A)
(V
=
(6.24)
0,
3t
31
where the time derivative operator can be brought outside the divergence operator (which implies spatial differentiation) because these two operations are entirely
independent from one another and can be performed in an arbitrary order. Hence,
the divergence of the vector E(r) in Eq. (6.16) equals
V-E(f)
=V
•
+V
Ei nd (0
•
E,(f)
=
P{t)
(6.25)
So
which means that Gauss’ law holds for the
total electric field as well.
In the case of dielectric materials in the time-varying electric
ization vector [Eq. (2.7)]
and bound charge densities [Eqs.
the material are time-dependent.
Eq. (2.44),
is
field,
the polar-
and
(2.23)] in
(2.19)
The generalized Gauss’ law
in integral
form,
written as
<j)
D(l)
dS
•
=
J
pit) dv,
(6.26)
where D(r) is the total electric flux density vector in the material, given by
Eqs. (2.41) and (2.47) with E representing the total electric field intensity vector.
On the other side, the slowly time-varying magnetic field has the same form as
the magnetostatic field. The magnetic flux density vector in free space can thus be
obtained using the time-varying version of Eqs. (4.7)-(4.9). For instance, the
due to a volume current distribution in free space is given by
[J(r)
P-0
Bit)
field B(/)
R
‘
4tt
The
dv] x
field
l
(6.27)
can also be obtained indirectly, via the magnetic vector potential, as
[Eq. (4.116)]
slowly time-varying magnetic
B(f)
=V
x
A
it).
(6.28)
field
where
slowly time-varying magnetic
potential
A(l)
_
.lit)
/fo
f
4?r
Jv
dv
R
with analogous expressions for surface and line currents.
(6.29)
Section 6.3
Faraday's
Law
of Electromagnetic Induction
In magnetic materials, the magnetization vector [Eq. (5.1)] and magnetization
current densities [Eqs. (5.28) and (5.32)] are now time-dependent. The integral form
of the generalized Ampere’s law, Eq. (5.51), can be rewritten as
<j>
H(f)
•
dl
= J J(r)
•
dS,
(6.30)
that of the law of conservation of magnetic flux, Eq. (4.99), as
and
<j>
B(f)
•
dS
=
(6.31)
0.
Problems'. 6.10.
6.3
FARADAY'S
We now
LAW
OF ELECTROMAGNETIC INDUCTION
introduce Faraday’s law of electromagnetic induction, as the most impor-
and the explicit
change in time. Following
Oersted’s discovery in 1820 that electric currents produced magnetic fields, Michael
Faraday was convinced that the reverse was also possible - that a magnetic field
could produce an electric current. In 1831, Faraday set up an apparatus consisting of an iron toroidal core (ring), like the one in Fig. 5.19, with two coils wound
on it. The primary coil was connected through a switch to a battery (voltaic cell)
and the secondary coil was short-circuited by a wire running above a compass, as
sketched in Fig. 6.5. Thus, any electric current in the secondary coil would, by means
of its magnetic field, deflect the compass needle. Upon closing the switch, Faraday
observed a momentary deflection of the needle, indicating a brief surge of current
induced in the secondary coil. The same happened when the switch was opened,
terminating the current in the primary coil, but the needle deflection was opposite
in polarity with respect to the previous one. In steady states, however, i.e., once
the current in the primary coil reached its final value (equal to the battery voltage
divided by the resistance of the primary circuit or to zero), there was no current in
the secondary coil and the compass needle was at its zero position. Faraday realized that a current was produced (induced) in the secondary coil by a changing
magnetic field in the iron core (the field was changed from zero to a final steady
value, corresponding to a steady current intensity established in the primary circuit,
and then back to zero when the current was terminated). This extraordinary discovery led to the formulation of the law of electromagnetic induction, named after
tant governing law of the slowly time-varying electromagnetic field
relation
between the
electric
and magnetic
fields that
Faraday.
The mathematical statement of Faraday’s law of electromagnetic induction
describes the time variation of the magnetic flux through an arbitrary contour as
Figure 6.5 Sketch of the
apparatus used
1
in
Faraday's
831 experiment that led
to the discovery of
electromagnetic induction.
272
Chapter 6
Slowly Time-Varying Electromagnetic Field
HISTORICAL ASIDE
when
member
permanent magnet was moved in and
coil, a current was induced in the
coil. He then demonstrated that a continuous current could be generated by rotating a copper disk
between the poles of a large permanent magnet
and taking leads off the rim and the center of
of the Royal Society at
the disk. This invention, referred to as Faraday’s
Michael Faraday (17911867), an English physicist and chemist, selfeducated from books he
was binding to earn a living,
became
a
He was
age 34.
unfamiliar
with
that
a
out of a wire
mostly
wheel (see
mathe-
tric
Fig. 6.35),
was the
first
dynamo
(elec-
generator). Faraday used his concept of lines
matics, but at the
same
was an enor-
of force to explain the principle of electromag-
time
netic induction observed in his experiments.
experimen-
talist
explained that an electric current was induced in
a conductor only when magnetic lines of force cut
of the greatest scientists ever.
across
he
mously
gifted
and imaginative thinker, and certainly one
Upon a recommendation by Sir Humphry Davy (1778-1829),
the discoverer of six chemical elements, Faraday
was appointed Chemical Assistant at the Royal
Institution on March 1, 1813. After repeating Oersted’s (1777-1851) 1820 experiment, he
demonstrated in September of 1821 that the magnetic field around a straight wire with an electric current was circular, and, in the same set
of experiments, went a large step further - by
making a current-carrying wire suspended above
a permanent magnet circle around the magnet,
he invented the first electric motor. Generalizing
from the patterns formed by iron filings around
magnets, he introduced the concept of electric
and magnetic field lines (he called them lines of
force) as a new approach to studying electricity
and magnetism. Faraday was elected to the Royal
Society in 1824 and was made Director of the
Laboratory at the Royal Institution in 1825. In
1826, he started Christmas Lectures for children
at the Institution, which not only continue, but
now
and
it,
this
He
could be either because the lines
of a changing magnetic field expanded and col-
lapsed in space cutting thus across the conductor
(transformer induction) or because the conductor
moved
induction).
across the static field lines (motional
He
that the
realized
magnitude of
the induced current was dependent on the
num-
ber of lines of force cut by the conductor in
unit time, which is a true equivalent to a more
mathematical formulation of what is now known
as Faraday’s law of electromagnetic induction. It
was the discovery of electromagnetic induction
in 1831, more than any other, that allowed electricity to be turned, during the remainder of the
19th century, from a scientific curiosity into a
powerful technology. In the 1830s, Faraday also
studied the relationships between the
amount of
material deposited on electrodes of an electrolytic
cell,
the
amount
of electricity passed through the
and chemical properties of different elements,
and formulated fundamental principles of electrocell,
chemistry (Faraday’s laws of electrolysis).
He
also
over the
discovered that light could be affected by a mag-
world. In his famous August 29, 1831 experiment,
netic force - he demonstrated in 1845 that a strong
he wound two
magnetic
are
televised to giant audiences
coils of
all
wire on the same iron ring
and discovered that a current in
one coil if changed in time induced a current in
the other coil. He concluded that an electric current could be produced by a time-varying magnetic
field and thus discovered electromagnetic induction. Faraday’s induction ring was the world’s first
electric transformer. In subsequent experiments
in the fall of 1831, he attempted to create a current using a permanent magnet. He discovered
(see Fig. 6.5)
field
could rotate the plane of polar-
- which later became
Faraday magneto-optical effect. In
his famous lecture “Thoughts on Ray-vibrations”
at the Royal Institution in April of 1846, he suggested that the propagation of light through space
ization of polarized light
known
as the
consisted of vibrations of lines of force, which,
intuitively,
was not
far
from Maxwell’s (1831—
1879) explanation, given
much
later
(in
1865)
and based on rigorous mathematical derivations,
Section 6.3
light was an oscillatory electromagnetic
disturbance - electromagnetic wave. In a series of
that
studies
from 1846
Faraday evolved his
model based
to 1850,
global theoretical electromagnetic
on the “force
- the
field”
in tension filling the space
ies,
field of lines of force
around charged bod-
current-carrying conductors, and permanent
magnets, and thus established the
field
theory of
electromagnetism. These concepts of electric and
magnetic force
fields
were put into a mathematical
Faraday's
Law
of Electromagnetic Induction
form a generation
made
it
later
273
by Maxwell, who himself
very clear in his texts that the basic ideas
for his classical electromagnetic field equations
came
from Faraday. Despite his epochal
achievements and far-reaching contributions to humanity, Faraday remained a modest
and humble person throughout his life. We honor
Faraday also by using farad (F) as the unit for
directly
scientific
capacitance.
Edgar Fahs Smith
(Portrait:
Collection,
University of Pennsylvania Libraries)
the cause of the induced electromotive force along the contour. In developing the
general electromagnetic model,
it is
usually taken as an experimentally based postu-
However, because we started our study of electromagnetic induction by taking
the mathematical expression for the induced electric field intensity vector due to a
point charge in accelerated motion as an experimental postulate, we are now able to
actually derive Faraday’s law from the facts that we already know about the induced
late.
electric field.
In a region with free charge carriers
electric
intensity vector,
field
[Eq. (4.143)],
where
line integral of Ej nd
a conducting wire), the induced
(e.g., in
E, nci, acts on the carriers by the force <2Ej nd
Q is the charge of a carrier (e.g., a free electron). Therefore, the
M and N in space represents
along a line joining any two points
the electromotive force (emf) induced in the line:
/N
e ind
The induced emf
is
measured
=
in volts
M
Eind
'
(6.32)
dl.
J/
induced electromotive force
(emf), in volts
and defined
in the
same way
as the
emf
of a
voltage generator in Eq. (3.112). In fact, the line can be replaced by an equivalent
voltage generator
whose emf is
Cj n d,
as
shown
in Fig. 6.6.
For a closed line (contour) that does not change or move
use Eq. (6.2) and write
e ind
We now recall
=
j>
E ind
•
dl
=£
^
•
dl
=
£
in time, Fig. 6.7,
A
we
(6.33)
dl.
that the circulation of the magnetic vector potential along a contour
equals the magnetic flux through the contour, Eq. (4.121), so that
(6.34)
Faraday's law of
electromagnetic induction
This equation
is
known
as Faraday’s
law of electromagnetic induction.
the most important experimental pillars of electromagnetics.
shows,
It is
first
one of
of
all,
one another under nonstatic
conditions. More specifically, it states that a magnetic field that changes with time
produces (induces) an electric field and an electromotive force, as well as an electric current in conducting media [by virtue of Ohm’s law in local form, Eq. (3.18)].
that the electric
and magnetic
It
fields are related to
274
Chapter 6
Slowly Time-Varying Electromagnetic Field
it quantifies the induced emf in an arbitrary contour as being equal to the
negative of the time rate of change of the magnetic flux through the contour, i.e.,
Finally,
through a surface of arbitrary shape spanned over the contour and oriented in accordance to the right-hand rule with respect to the orientation of the contour. This rule
tells
us that the flux
is
in the direction
defined by the
thumb of the
right
hand when
the other fingers point in the direction of the emf, as indicated in Fig. 6.7.
Expressing the magnetic flux using the flux density vector [Eq. (4.95)] leads to
M
£E
oN
ind
-dl
—^jfB-dS,
(6.35)
where the reference directions of dl and dS are interconnected by the right-hand
rule: fingers - dl, thumb - dS (see Fig. 6.7). Since the circulation of the field intensity
vector due to excess charge is zero [Eq. (6.20)], we have that
^ind
lE
d,=
£ Ejnd
‘
which means that Faraday’s law can be expressed
am
(6.36)
dl,
in
terms of the total electric
intensity vector as well. In addition, the time derivative
Figure 6.6 Induced emf
in
two points
in
a line joining
=
space.
on the right-hand
field
side of
Eq. (6.35) can be moved inside the surface integral, provided that the surface S
does not change or move in time. Combining these two conclusions, we obtain the
following version of Faraday’s law of electromagnetic induction in integral form:
Faraday's law in integral form
(6.37)
We note
the formal similarity between this equation and the generalized Ampere’s
law in integral form, Eq. (6.30), where — 3B/3 1 in Eq. (6.37) stands for J in Eq. (6.30)
in the flux integrals, while the electric and magnetic field intensity vectors, E and H,
appear in the corresponding line integrals on the left-hand side of equations.
The contour C in Fig. 6.7 can be an imaginary (nonmaterial) contour, i.e., it
does not need to be a conducting wire loop for Faraday’s law of electromagnetic
induction to be true. The electric field and emf are induced by a magnetic field
that changes with time regardless of whether or not conducting wires are present.
However, in the case when C does represent a conducting wire contour, there is a
current of intensity
induced current
find
in the wire, as
shown
in Fig. 6.8,
and of the equivalent closed
emf
ejnd
[this,
basically,
—
^ind
(6.38)
~R~
where R
is
the total resistance of the contour
circuit including the ideal voltage
generator with
comes from the version of Kirchhoff’s voltage law
in
Eq. (3.118)]. This current is called the induced current. We can say thus that, in
general, time-varying magnetic fields, B (t), induce electric currents in conducting
media, which also change with time.
On the other hand, if the conducting wire loop is not closed (e.g., there is a small
gap in the loop), there is no current flowing through it, 6 and the loop behaves
like an open-circuited generator with emf n d, as in Fig. 6.6. The voltage across the
gap equals the induced emf, i.e., vnm( 0 = ^ind(0 [ see Eq. (3.115)].
air
Figure 6.7 Arbitrary
contour
in
magnetic
a time-varying
field
<?j
- for the
formulation of Faraday's law
of electromagnetic
6
induction.
current can exist even
As opposed
to dc
and slowly time-varying cases, as we shall see in a later chapter, a rapidly time-varying
in open-ended wire conductors that do not form closed current circuits.
Section 6.3
Faraday's
Law
of Electromagnetic Induction
275
HISTORICAL ASIDE
As
Heinrich Friedrich Emil
Tartu), Estonia, then a part of Russian Empire.
Lenz
geophysical scientist, he traveled around the world
Russian
a
a
(1804-1865),
was
physicist,
of
professor
physics
at the University of St.
Petersburg.
of the
entists,
He was one
three
great sci-
together
with
Faraday (1791-1867) and
Henry (1797-1878), who
independently from each
other investigated electromagnetic induction at
about the same time at three remote places on the
Lenz was born and educated in Dorpat (now
and made extremely accurate measurements of the salinity, temperature, and specific
in the 1820s
gravity of sea waters.
He
also studied electricity
and magnetism, and discovered,
in 1833, that the
conductors increases with a
rise in temperature [see Eq. (3.22)]. In 1834, he
discovered that an induced current always proresistivity of metallic
duces effects that oppose
cause. This
its
became
to
be known as Lenz’s law. From 1840 to 1863, Lenz
was the Dean of Mathematics and Physics at the
University of
St.
Petersburg.
globe.
emf in the contour is
magnetic flux through the contour
that caused the emf in the first place. This fact, contained in Faraday’s law, is an
experimental result referred to as Lenz’s law. To illustrate it, assume that at an
instant t the flux in Fig. 6.8 increases in time, i.e., d<h/ dt > 0. From Eq. (6.34), the
emf £j n d(0 at that instant is negative, and so is the induced current intensity ij n d(0
in Eq. 6.38. The induced current produces a secondary magnetic field, whose reference direction is determined by another application of the right-hand rule: fingers current, thumb - field [for example, see Fig. 4.31(b)]. Hence, given the reference
direction of /; nd in Fig. 6.8, the reference direction of the secondary magnetic field,
and its flux, will be the same as that of the primary (original) magnetic field B(t) and
flux 0(0- So, with respect to that reference direction, the secondary flux is negative,
because q n d(0 is negative at the time instant considered. We conclude thus that the
magnetic field due to the induced current opposes the change (increase in this case)
in the primary magnetic field, which caused the induced emf and current in the first
place, and this is the statement of Lenz’s law. Generally, Lenz’s law represents a
rule for (quickly) determining the actual direction of an induced current in a loop
(circuit), without fully applying Faraday’s law. This direction is always such that the
magnetic field due to the induced current opposes (tends to cancel) the change in
the magnetic flux that induces the current. 7
By virtue of Stokes’ theorem, Eq. (4.89), or simply by analogy with the differential form of the generalized Ampere’s law, Eq. (5.52), we obtain the differential
equivalent of Eq. (6.37), namely, Faraday’s law of electromagnetic induction in
The minus
sign in Eq. (6.34) indicates that the induced
in a direction that
opposes the change
in the
Figure 6.8 Induced current
in a
conducting wire loop
situated in a time-varying
magnetic
field.
differential form:
V
x
9B
E=-
(6.39)
Hi'
7
The magnetic
field
much weaker than
due
to the
induced slowly time-varying current
form
in typical thin-wire circuits
the primary magnetic field inducing the current. Therefore,
ble with respect to the primary field
electromagnetic induction.
and
falls far
Faraday's law
it
is
is
usually negligi-
short of fully canceling the flux change causing the
in differential
276
Chapter 6
Slowly Time-Varying Electromagnetic Field
In words, the curl of the time-varying electric field intensity vector existing at any
point of space and any instant of time equals the negative of the time rate of change
of the magnetic flux density vector at that point.
Note
that the differential
form of Faraday’s law can also be derived by taking
the curl of both sides of Eq. (6.2) and using Eq. (6.28). This results in
/3A\
^
=
V x Ei„
d = -V x
(— )
which
is
(6.40)
the version of the law with the induced electric field intensity vector. Since
the curl of the field intensity vector due to excess charge
conclude that
V
x
E= V
x Ej nd
zero [Eq. (6.21)],
is
|VxE? = Vx E
which gives the version of the law with the
Eq. (6.39).
6.4
3B
--9 (V x A) =
ind
we
(6.41)
,
total electric field intensity vector,
MAXWELL S EQUATIONS FOR THE SLOWLY
TIME VARYING ELECTROMAGNETIC FIELD
Faraday’s law of electromagnetic induction, given by Eq. (6.37) or Eq. (6.39), represents Maxwell’s first equation for the time-varying electromagnetic field. It is
essentially different
magnetic
from Maxwell’s
first
equation for the time-invariant electro-
field in Eqs. (5.129). It tells us that a
magnetic
field
changing with time
The remaining three Maxwell’s equations are given by
Eqs. (6.30), (6.26), and (6.31), and they retain the same form as in the time-invariant
case. We now summarize the full set of Maxwell’s equations in differential form
gives rise to an electric field.
along with the constitutive equations for the slowly time-varying electromagnetic
field in a linear isotropic
Maxwell's
first
equation
Maxwell's second equation,
medium:
V
x E(f)
V
x
=
H(0 =
J(f)
quasistatic field
Maxwell's third equation
v
Maxwell's fourth equation
V B (t)
•
D(r)
•
= pit)
- 0
constitutive equation for
D
D(t)
= eE(t)
constitutive equation for
B
B(0
=
J (0
=
constitutive equation for J
crE(t)
Because of the new term on the right-hand side of the
first
equation, the time-
varying electric and magnetic fields are coupled together and cannot be analyzed
opposed to time-invariant electric and magnetic fields, which are
independent from each other. On the other hand, as we shall see in a later
separately, as
entirely
chapter. Maxwell's equations for the rapidly time-varying electromagnetic field dif-
from the corresponding equations for the slowly time-varying field only in the
generalized Ampere’s law (second equation). Namely, an additional term exists on
fer
the right-hand side of this equation
electric field rapidly
second equation
in
in
the general case, expressing the fact that an
changing with time gives rise to a magnetic field. Therefore, the
the system of Eqs. (6.42) represents the quasistatic version of
1
Section 6.5
Maxwell’s second equation and
time-harmonic) fields.
is
true for slowly time-varying
Computation
(e.g.,
of Transformer Induction
277
low-frequency
Eqs. (6.19) and (6.29) give the expressions for slowly-time varying electromagand (6.18) in Eq. (6.16) leads
netic potentials in free space. Substituting Eqs. (6.2)
to the following expression for the electric field intensity vector, E, in terms of the
potentials:
dA
vm
(6.43)
electric field via potentials
We see that both potentials are needed for E, whereas A alone suffices for the magnetic flux density vector, B, in Eq. (6.28). Eqs. (6.43)
and
(6.28) represent a
for evaluating (by differentiation) the electromagnetic field (E, B)
means
from potentials
(V, A), the evaluation of potentials being, in general, considerably simpler than the
direct evaluation of field vectors.
In addition to Maxwell’s equations for a given class of electromagnetic
we always have
in
mind the associated version of
fields,
the continuity equation, which
represents one of the fundamental principles of electromagnetics - the principle of
conservation of charge, but can also be derived from Maxwell’s equations. Thus,
by taking the divergence of both sides of the quasistatic version of the generalized
Ampere’s law in Eqs. (6.42), as in Eq. (5.130) for the static case, we obtain that
V
which
is
•
J(f)
=
(6.44)
0,
the differential form of the continuity equation for slowly time-varying cur-
We note that it has the same form as the continuity equation for time-invariant
(steady) currents, Eq. (3.41). We also note that it can be regarded as the special
rents.
case of the general continuity equation for time-varying currents, Eq. (3.39), with
dp/dt ^ 0. Namely, in the slowly time-varying field, the rate of the time-variation in
excess charge
slow enough to be neglected while evaluating the current continuity
is
balance. In integral notation,
J(0
l
dS
=
(6.45)
0,
time-varying currents
which is the same as Eq. (3.40) for steady currents.
Eq. (6.45) implies that the slowly time-varying current intensity
conductor, just as the steady current intensity I along a wire,
cross section of the conductor
chapter.
is
It
also
means
- the
fact that
we have
is
i(t)
the
along a wire
same
in every
already used throughout this
that Kirchhoff’s circuital law for slowly time-varying currents
given by
X>(0 = 0,
(6.46)
k=
where
N
is
the
number
of conductors (branches in a circuit) meeting at a node.
see that Kirchhoff’s current law for low-frequency ac circuits has the
We
same form
as
that for dc circuits [Eq. (3.42)].
6.5
COMPUTATION OF TRANSFORMER INDUCTION
devoted to the application of Faraday’s law of electromagnetic induce\ n q, and electric field intensity vector, Ej n d,
in stationary contours due to given slowly time-varying current distributions and
their magnetic fields. This kind of electromagnetic induction is called transformer
This section
continuity equation for slowly
is
tion in evaluating the induced emf,
278
Chapter 6
Slowly Time-Varying Electromagnetic Field
the basis of current and voltage transformation by magnetic
induction, because
it is
coupling between
circuits.
one
Namely,
enables time-varying currents and voltages
it
(primary circuit) to be transformed, by induction, to time-varying
currents and voltages in another circuit (secondary circuit), where the transfer of
in
circuit
energy between the circuits is actually performed by the magnetic field due to the
currents in the primary circuit causing the induced electric field in the secondary circuit. The electromagnetic induction due to motion of conductors in magnetic fields
will be introduced and studied in the next section.
In
some
contour
(e.g.,
applications,
we
flux
emf induced in a
we apply the version of Faraday’s law
Eq. (6.34), where we only need to evaluate the
are interested only in the total
a wire loop). In such cases,
of electromagnetic induction in
magnetic
,
through the contour and compute
a simple task to do.
However,
its
time derivative, which
to find the actual distribution of the
is
usually
emf along
a
contour or to find the distribution of the induced electric field intensity vector in
space, which is necessary in many applications, we have to employ the version of
the law in Eq. (6.35). This equation, although always true, enables us to analytically
solve for the field Ej nc only due to highly symmetrical primary current distributions.
These are the cases where the vector Ej nc is tangential to some (or all) sections
of the contour and has a constant magnitude along such sections, while being
perpendicular to the remaining sections of the contour (if such sections exist). In
other words, these are the cases in which we are able to bring the induced electric
field intensity, find* outside the integral sign on the left-hand side of Eq. (6.35), and
solve for it. Note that Eqs. (6.3)-(6.5), on the other hand, provide general solution
procedures for computing Ej n d.
Because Faraday’s law and Ampere’s law have the same mathematical form,
there is a complete formal analogy in their application. In solving for Ej n d due to B
j
j
in highly
symmetrical situations, therefore,
application of Ampere’s law in solving for
whenever
infinitely
unit of
its
Electric Field of
the solenoid
The solenoid
is air.
is
The magnetic
Find the induced electric
Solution
H) due
to J, discussed in Section 4.5,
an
Infinite
Solenoid
long solenoid with a circular cross section of radius a has N' turns of wire per
length.
wound about
slowly time-varying current of intensity
in the
shall exploit the parallelism with the
(or
possible.
Induced
An
we
B
field
/(f)
a ferromagnetic core of permeability
flows through the winding.
Because of symmetry, the
dE- nd and
P
due to currents induced in the core can be neglected.
vector inside and outside the solenoid.
lines of the
induced electric
inside the solenoid, at a distance r
dE" d be
the field intensity vectors at this
elements denoted as
/
dl'
A
field intensity
solenoid winding are circles centered at the solenoid
arbitrary point
/x.
The medium outside
and
/
dl"
and shown
axis.
field
due
To show
to the current
this,
consider an
from the solenoid axis (Fig. 6.9). Let
point due to two symmetrical current
in Fig. 6.9. In
accordance to Eq.
(6.5), these
dE"nd is tangential to the circular contour C
of radius r centered at the solenoid axis. The same is true for any other pair of symmetrical
current elements, which can also be in a plane that does not contain the point P. As all current
two vectors are such
that their
elements constituting the current
we conclude
sum dEj nd
/(f)
in the
-F
winding can be grouped
that the resultant vector Ej ncj at the point
P
is
in
such symmetrical
same conclusion holds
for a point outside the solenoid. In addition, £j n d
the magnitude of Ei n<j
depends only on the radial coordinate
the solenoid axis. Hence,
system whose z-axis
is
Ejnd
where
<j>
is
—
£j ntj(r
the circular unit vector in the system.
>
0
pairs,
tangential to the contour C. This
=
const along C,
r of the cylindrical
i.e.,
coordinate
(6.47)
j
r
Section 6.5
Computation
of Transformer Induction
279
Figure 6.9 Evaluation of the
induced
electric field
due to
a slowly time-varying current
in
an
infinite
solenoid
(cross-sectional view); for
Example
6.5.
The magnetic field due to any spatial distribution of slowly time-varying currents has the
same form as that due to the same spatial distribution of steady currents. Therefore, we can
use the result of the analysis of a solenoid with a steady current of intensity I in the winding,
performed
Example
in
4.13. This
(field lines are parallel to the
means
that the magnetic field inside the solenoid
H = H{t) i = N'i(t)
while
H=0
magnetic
is
axial
solenoid axis) and uniform, given by
(6.48)
z,
outside the solenoid. Note that this result does not take into account the
field
due
to currents induced in the ferromagnetic core,
which can be neglected.
We now use Faraday’s law of electromagnetic induction, Eq. (6.35), and apply it to the
contour C in Fig. 6.9 in a manner completely analogous to the application of Ampere’s law
in Eqs. (4.54)-(4.56). The circulation of Ej nd along C equals
Find
2nr
d<f>
=
(6.49)
~df~'
The magnetic
flux
through the contour
<J>
(there
is
no
is
=
flux outside the solenoid),
Bnr2
for r
Bn a 2
for r
where
—
/j-N'r
d/
(r
2
<
a)
and
a
(6.50)
a
H
B = iH
solutions for the induced electric field intensity inside
Find
<
>
£ind =
df
and
is given in Eq. (6.48). The
and outside the solenoid come out to be
i
liN'a
2
di
(
2r
>
a).
(6.51)
df
respectively.
We note that the magnetic field inside the solenoid varies synchronously with the current
winding [H(t) a /(f)], whereas the induced electric field varies synchronously with the
in the
time derivative of the current [E; n d(0
d/(f)/df].
Inhomogeneous Conducting Loop around
Example 6.6
a Solenoid
Consider the solenoid from the previous example, and assume that a wire loop of radius b
(b
> a) is placed coaxially around it, as shown in Fig. 6.10(a). The two halves of the loop
made from different materials, with conductivities o\ and 02 The cross-sectional area of
are
both wire parts
.
is
S.
The magnetic
field
due to currents induced
in the
core and in the wire
Find of
an
infinite
solenoid
+
a
280
Chapter 6
a
n
Slowly Time-Varying Electromagnetic Field
loop can be neglected. Calculate the voltage between the junctions of two wire parts (points
M and N).
Solution
The system
C
contour
Having
shown
in Fig. 6.10(b),
R2
R\ and
in Fig. 6.10(a), while
the contour.
can be analyzed from the circuit-theory point of view,
in Fig. 6.10(a)
using the equivalent circuit diagram
mind the second expression
in
,
a
emf in the
emf amounts
to
t di
2
(6.52)
due
field
to the induced currents are
of Eq. (3.85), the resistances are
=
Ri
Employing Eq.
the induced
dr
Electromotive forces generated by the magnetic
By means
is
in Eqs. (6.51), this
d/
neglected.
ex d
—dOT- = -nixN —
=
e ind
where
are resistances of the two wire parts constituting
R2 =
and
(6.38), the current in the circuit,
i.e.,
(6.53)
the induced current in the loop,
is
given by
^ind
'
ind ~~
(6.54)
R + R2
'
{
between points
Finally, the voltage
V MN
Note
=
•
2
obtained from
is
— R2
R\
e ind
~
D
R\l
ind
that in the case of a
M and N
—
nix(a2
+ R2
2{R\
Fig. 6.10(b) as
o\)N'
2
dr
(6.55)
e nd
‘
2 (cti
)
homogeneous wire loop
(p\
=
a2 ),
dr’
cr2 )
there
is
no voltage between
different points along the loop.
(b)
Figure 6.10
composed
(a)
of
parts with
different conductivities
placed around an
infinite
solenoid carrying a slowly
time-varying current and (b)
equivalent circuit diagram;
for
Example
Example 6.7
Wire loop
two
Magnetic
Refer to the system
N =
1
and
1000 turns/m,
ix
=
hq
a
in
—
due to Induced Current
Field
and assume that
2
20 cm, S = 1
o\
6.10(a),
Fig.
10 cm,
=
b
mm
Under
(air-filled solenoid).
Loop
in a
—
i(t)
,
=
57
A
(r
MS/m, a2 =
15
2 cos lOOOr
in
s),
MS/m,
these circumstances, find the magnetic flux density
vector at the loop center due to the induced current in the loop.
Solution
Let us
first
check whether the current
in the solenoid
6.6.
time-varying. Since the solenoid
infinitely long, this
is
problem
one, and the check of whether a low-frequency analysis
is
can be considered as slowly
is
actually a two-dimensional
sufficiently accurate here
should
be performed in the cross section of the structure containing the wire loop. The maximum
dimension of the structure relevant to the field computation in this cross section is 2b, and the
corresponding time of propagation of electromagnetic disturbances is r = 2b /cq = 1.33 ns.
This time is much shorter than the period of change of i(t), which equals T = 2-n/u> — 6.28 ms
(w = 1000 rad/s). We conclude that this structure can indeed be analyzed as a low-frequency
(quasistatic) problem. This
performed
in the
From Eq.
means
that
(6.54), the
induced current intensity
6nd(l)
IXqo\<j2 N'
—
(cri
The magnetic
flux density vector
Eq. (4.19) for the
field
Let us finally compare the
primary
/(/) in
due
=
solenoid
B(/)
S
d/
the results of the analysis of the structure
in the
=
loop
is
1.5 sin 1000/
A.
(6.56)
d/
to this current at the center of the loop
—
,
0 and the contour radius
2
=
2b
By means
is
obtained using
b:
(6.57)
4.7 sin 1000/ z fxT.
field Bj n d(/) to the
the solenoid winding.
field inside the
2
+ o2 )b
point defined by z
Bind (')
current
we can use
previous example.
primary magnetic
field
-
that
due to the
of Eq. (6.48), the flux density vector of the
is
=
nuN'i(t) z
=
2.5 cos 1000/ z
mT.
(6.58)
Computation
Section 6.5
We see that
|Bj n d(?)|/|B(?)|
=
3
x 10
1.9
281
of Transformer Induction
1
,
i.e.,
i
I
Bjnd (0
« |B(0|.
1
(6.59)
due to the induced current in the loop is absolutely neglimagnetic field of the solenoid, which caused the induced current in
also note that the waveforms of these two fields (sin 1000? and cos 1000?)
In other words, the magnetic field
gible with respect to the
the
first
place.
We
are such that B; n d(?) tends to
that actually created
compensate for the change
Of course,
it.
in B(?),
opposes the action
i.e., it
accordance with Lenz’s law.
this is in
I
Open-Circuited Coil around a Solenoid
Example 6.8
i
= 2 m and circular cross section of radius a = 10 cm has
air-filled solenoid of length
= 1750 turns of wire. There is a low-frequency time-harmonic current of intensity
6
rad/s. An open?(?) = Io sin a>f flowing through the winding, where 1$ = 10 A and co = 10
—
10 turns of wire is placed around the solenoid, as shown in
circuited short coil with Nj
An
(a)
/
N\
Compute
Fig. 6.11(a).
the voltage between the terminals of the
coil.
»
Solution The solenoid is very long (/
a), so that the end effects can be neglected while
computing the magnetic field about its center. This means that the solenoid can be considered
as infinitely long while computing the magnetic flux through the short coil in Fig. 6.11(a). As
the coil consists of 2 wire turns, this flux is given by
N
= ^2 ^single turn.
t
<
>
(6.60)
Figure 6.1
where Single
the flux through a surface spanned over any of the turns. In other words,
turn is
the electromotive forces induced in individual turns
in the coil
e ind
is
N2 times that in Eq. (6.52) with n =
= -^2
n HQN\N2a
fl’l’single turn
d?
There
no current
is
2
ncL>iXQN\N2a lQ
cos
dt
because
it is
cot.
(6.61)
l
open-circuited, so that the voltage across the
coil terminals is [Fig. 6.11(b)]
v(?)
= -e ind (?) =
ncotioN\N2a
2
Io
cos
cot
=
3.45 cos 10
?
kV
(?
An
coil
placed around a very long
solenoid carrying a
low-frequency current:
2
d?
I
in the coil,
/jlq
add in series, and hence the total emf
and N' = N\/l, which results in
all
1
open-circuited short
in
(6.62)
s).
three-dimensional view
showing the windings and
(b) cross-sectional view
showing reference
directions for the emf and
voltage; for Example 6.8.
(a)
/
Example 6.9
An
infinitely
Rectangular Contour near an Infinite Line Current
long straight wire carries a slowly time-varying current of intensity
?'(?).
A
dx
same plane with the wire, with two
sides parallel to it, as shown in Fig. 6.12. The distance between the wire and the closer
parallel side of the contour is c. Determine the emf induced in the contour.
rectangular contour of side lengths a and b
lies in
C
i
the
By
=
a
X
;
Mo i(t)
2tc x
integrating this density across the flat surface
(6.63)
spanned over the contour, we obtain the
magnetic flux through the contour:
Figure 6.12 Evaluation of
the emf
in a
contour
in
infinitely
rc+a
3 >(?)
=
B(x,
/
Jx=c
t)
b dx
=
2n
dS
c+a
W(t)b
l
dx
X
—
Mo i(t)b
2n
,
c
+a
In
is
similar to that in Fig. 5.26).
rectangular
the vicinity of an
long wire with a
slowly time-varying current;
,
(6.64)
c
where dS is the surface area of a thin strip of length b and width dx in Fig. 6.12, and the flux
is determined with respect to the reference direction into the plane of drawing (note that this
integration
dS
S
Solution The magnetic flux density vector produced by the current in the wire at any point
plane of the contour is perpendicular to the plane and at a distance x from the wire
(Fig. 6.12) and an instant ?, its magnitude is [see Eq. (4.22)]
?)
S
8
in the
B(x,
B
(h
c
for
Example
6.9.
Chapter 6
Slowly Time-Varying Electromagnetic Field
For the adopted direction of the
direction for the induced
emf
—
^ind(f)
Induced Emf
Example 6.10
A
coil
with
N = 400
core of length
I
=
40
dO
Hob
dr
17
is
function sketched in Fig. 6.13(b)
made
is
c
is
di
(6.65)
di
wound uniformly and
5=1 cm
cross-sectional area
slowly time-varying current whose intensity,
The core
+a
"’—
with a Nonlinear Core
in a Coil
turns of wire
cm and
right-hand rule gives the clockwise reference
flux, the
contour, with respect to which,
in the
i(t), is
densely about a thin toroidal
2
,
as
shown
in Fig. 6.13(a).
A
a periodic alternating triangular-pulse
established in the coil, where Iq
—
0.1
A
and
T=
1
ms.
of a nonlinear ferromagnetic material that exhibits hysteresis effects. In
steady state, the operating point
(
B H)
,
can approximately be represented as
periodically circumscribes a hysteresis loop that
in Fig. 6.13(c),
where B m /Hm
=
hu
=
0.001
H/m. The
resistance of the coil can be neglected. Find the voltage across the coil terminals in the time
interval 0
<
Solution
<
t
T.
The magnetic
the winding [H{t)
cx /(f)]
core varies synchronously with the current in
field intensity in the
and
is
given by
H(t)
=
Ni(t)
(6.66)
I
[see Eq. (5.53)].
From
the core, B(t ),
(for
H = //m
afterwards
The function H(t)
is
plotted in Fig. 6.13(d), where
the hysteresis loop in Fig. 6.13(c),
first
),
it is
varies as a linear function of time
then becomes time-invariant while
Figure 6.13 Analysis of electromagnetic induction
=
v\ 2 ); for
from
H
is
Hm = NIq/1 —
100
A/m.
that the magnetic flux density in
B = -B m
H = 0) to B = B m
H = Hm to H = 0,
(for
reduced from
when H(t) is reversed and
and so on. The amplitude of
again a linear (now decaying) function of time
increased in the negative direction from
material (note that v
we conclude
Example 6.10.
H=0
in a coil
to
H = -H m
with a core
,
made from
a nonlinear ferromagnetic
Section 6.6
the periodic trapezoidal-pulse
waveform of
B
B m = PhHm =
is
0.1 T,
283
Electromagnetic Induction due to Motion
and B(t)
is
plotted in
Fig. 6.13(e).
The magnetic
^ind(O)
is
through the
flux
coil is
= NB(t)S
<t>(f)
and the induced emf
given by Faraday’s law of electromagnetic induction, Eq. (6.34).
of the coil
is
negligible, the voltage across the coil terminals
v{t)
This function
is
=
V 12 (f)
=
-Cind (0
=
d4>(0
_ NS
dB(t)
dt
amplitude
is
We
(6.67)
dt
proportional to the slope of the function B(t) and
proportional to the slope of B(t) in the
NS(2B m )/(T/4) = 8 NSB m /T =
see that the induced
the time derivative of
its
in the coil,
the resistance
is
a periodic alternating rectangular-pulse function with the
It is
As
first
is
plotted in Fig. 6.13(f).
same period ( T ). The pulse
quarter of the period, that
Vq
is,
—
32 V.
emf and voltage of
current,
which
the coil do not vary synchronously with
a consequence of the nonlinearity and hysteresis
is
behavior of the core material.
Problems'.
6. 1
Companion
1—6.17; Conceptual Questions (on
Website): 6.9-6.16;
MATLAB Exercises (on Companion Website).
ELECTROMAGNETIC INDUCTION DUE TO MOTION
6.6
Consider a conductor moving with a velocity v in a
netic field of the flux density B.
The
field exerts the
mag-
static (time-invariant)
magnetic Lorentz force,
Fm
,
given by Eq. (4.144), on each of the charge carriers in the conductor. This force
“pushes” the carriers to move, and separates positive and negative excess charges
We
can formally divide Fm by the charge of a carrier and obtain
quantity, although expressed in V/m, is not a true electric
field intensity vector, because it is not produced by an excess charge [Eq. (6.18)] or
by a time-varying current [Eq. (6.2)]. By definition given by Eq. (3.107), it represents
an impressed electric field intensity vector, which we term here the induced electric
field intensity vector due to motion, and write
in the conductor.
Fm IQ = vxB.
This
new
Eind
=V
X B.
(
6 68 )
.
induced
electric field intensity
vector due to motion (unit:
This field generates an induced electromotive force, as given by Eq. (6.32). Hence,
the
emf along
motion
a line through a conductor
in a time-invariant
magnetic
between points
=
/
r
dl
Ei n d
7m
This
emf is referred to
as the
emf due
For a moving contour (closed
=
N
(v
/
7m
X B)
dl.
(6.69)
to motional induction or simply motional emf.
line).
e ind
—
£
(v
x B)
•
(6.70)
dl.
Note that the velocity of different parts of the contour need not be the same, including cases
when some
V/m)
field is
/N
eind
M and N (Fig. 6.6) due to
parts are stationary while other
In other words, the motion of the contour
may
move
in arbitrary directions.
include translation, rotation, and
deformation (changing shape and size) of the contour in an arbitrary manner.
When a contour moves and/or changes in a static magnetic field, the magnetic flux through the contour changes with time. It is possible to relate the total
motional emf
(unit:
V)
)
Chapter 6
Slowly Time-Varying Electromagnetic Field
Figure 6.14
A contour moving
a time-invariant
in
magnetic
field.
emf induced in the contour to the rate of change of the flux, i.e., to express the
motional emf in terms of Faraday’s concept of changing flux through the contour,
as in Eq. (6.34). To see this, consider the moving contour C in Fig. 6.14. Let v be the
velocity of an element dl of the contour. In a time interval dr, this element moves a
distance dp = vdr (see Fig. 6.14). Eq. (6.70) thus becomes
Cjnd
=
<j)
x B)
(v
=
dl
•
(dp x B)
<j)
•
dl.
(6.71
Applying the identity (a x b) c = (c x a) b (scalar triple product is unaffected by
permutation of the order of vectors) to the scalar triple product (dp x B) dl
and noting that dl x dp equals the vector surface element dS shown in Fig. 6.14,
we have
•
•
cyclic
•
=
—®
last integral
x dp)
=
—
<c
dr
Jc
B
(6.72)
represents the magnetic flux through the strip
AS swept out by the
dr
This
B
dS.
ej n d
contour
C
contour
in Fig. 6.14).
•
(dl
Jc
during the interval dr (tinted strip between the positions
Let us mark
this flux
dO
t
h r0 ugh
1
and 2 of the
a s, so that
d^through A S
—
^ind
by
(6.73)
dr
The
strip
AS
represents the difference in area between the surface S
the contour in position
position 2 (at instant
r
1 (at
+
instant
r)
and the surface
=
S'
dO =
is
bounded by
the contour in
U AS.
Designating the magnetic flux through S and
increment in flux from r to r + dr equals
this
bounded by
dr), as
S
and
S'
O'
-
(6.74)
5'
by
O
and
O', respectively, the
O,
(6.75)
exactly the negative of the flux through the strip AS,
— dO hrough as- Finally, substituting this equation in Eq.
— dO/ dr, which is the same as in Eq. (6.34). We conclude
t
i.e.,
(6.73) leads to
that the
dO =
ej nci
=
same form of
Faraday’s law of electromagnetic induction holds for both transformer and motional
emf
in a
contour.
Moving
A
metallic bar of length a
allel
metallic
rails,
as
Metallic Bar in a Static Magnetic Field
—
shown
2
in
m
slides without friction at a constant velocity
Fig. 6.15.
The bar
is
perpendicular to the
over par-
rails
and the
.
Section 6.6
mechanical force acting on the bar
Electromagnetic Induction due to Motion
285
Fme ch = 4 N. The whole system is situated in a uniform
B = IT. The field lines are perpendicular to the
is connected
of the page. A load of resistance R = 5
is
time-invariant magnetic field of flux density
plane of the
rails
between the
rails.
and directed out
The
and in the rails, as well as the magnetic field due
be neglected, (a) Find the velocity of the bar. (b)
the load and discuss the energy balance in this system.
losses in the bar
to induced currents in the system, can
Evaluate the power of Joule’s losses in
Solution
(a)
The bar moves
field and the emf is induced in it due to motional
and the field is uniform (i.e, the same everywhere),
the motional emf in Eq. (6.69) becomes
in a static
induction. Since the bar
the expression for
magnetic
is
straight
=
eind
where the emf
v
is
directed from point
the velocity of the bar (which
is
and
1
are
all
(v
x B)
•
(6.76)
1,
M to point N along the bar (Fig. 6.15),
1
= MN, and
emf in a
straight conductor
moving in a uniform
magnetic field
to be determined). Moreover, since the vectors v, B,
is
orthogonal with respect to each other and
ejnd
=
|1|
=
a,
we can
write
vBa.
(6.77)
M
The bar and the rails constitute a conducting loop, that is, an electric circuit of the
form shown in Fig. 6.8. Hence, there will be a time -invariant induced current in the loop
of intensity
v Ba
find
_
“ R _
~ R
t
ind
[see Eq. (6.38)], given with respect to the
(6.78)
same reference
direction as the emf,
where we
The
on the other hand, produces a secondary magnetic field, which is neglected
in evaluating the emf in Eq. (6.76). Note that this field opposes the increasing flux of the
primary field B through the loop, the area of which expands as the bar moves to the right,
as yet another example of Lenz’s law.
N
neglect the resistance of the bar and the rails and the corresponding Joule’s losses.
current
7j nci,
We know
presence of a magnetic
that, in the
From Eq.
experiences a magnetic force.
in Fig. 6.15
comes out
(4.163), the
field,
a current-carrying conductor
magnetic force on the metallic bar
to be
Fm —
Jjndl
x
B
— Im &aB.
*
This force opposes the motion of the bar and the generation of the
again
is
in
accordance with Lenz’s law. In other words,
Fm
is
(6.
emf
in
it,
9)
which
directed oppositely to the
Fmec h on the bar. Furthermore, as the velocity of the bar is constant,
these two forces must be exactly equal in magnitude (Newton’s second law), that is,
mechanical force
Fmech
Hence, the velocity we seek
emd
(b)
By
=
10
now
=
F
™2g2
(v
= const).
(6.80)
=
m/s
5
(6.81)
-
give the values for the induced
emf and current
in the loop:
V and /ind = 2 A.
Joule’s law, Eq. (3.77), the
power of Joule’s
P} =
the
—
is
v
Eqs. (6.77) and (6.78)
y g2^2
= Em =
Rlfnd
losses in the load resistor
=
20
W.
is
(6.82)
Note that, on the other side, the power generated by the induced emf in the bar,
power of the equivalent ideal voltage generator of emf Cj n d in Fig. 6.8, equals
Find
—
^ind^ind
—
20
W
i.e.,
(6.83)
Figure 6.15
A
moving
uniform
in a
metallic bar
time-invariant magnetic
field
(elementary electric
generator); for
Example
6.1
1
286
Chapter 6
Slowly Time-Varying Electromagnetic Field
power used
[see Eq. (3.121)]. Finally, the mechanical
to
move
the bar at the velocity v
is
obtained as
^Anech
We
=
^mectA
—
—
= Ph
20
W.
(6.84)
see that, as expected,
^rnech
which
is
in
Fig. 6.15 is a
^ind
(6.85)
compliance with the principle of conservation of energy. The system
in
simple example of an electric generator based on electromagnetic induction,
where the applied mechanical energy is converted into the electric energy and delivered
to the load. The agent by which the energy transfer is carried out is the induced emf in
the moving bar, and the energy of this emf is ultimately dissipated to heat in the resistor.
Rotating Wire Loop
Example 6.12
A
axis in a
its
Fig. 6.16(a).
page.
At an
uniform time-invariant magnetic
The vector B
instant
t
=
0,
field
perpendicular to the plane of drawing and
is
the loop
the plane of drawing.
lies in
co
of flux density B, as depicted in
The
is
directed out of the
resistance of the loop
is
R.
due to induced currents can be neglected. Calculate (a) the induced emf
the loop and (b) the instantaneous and time-average mechanical power of loop rotation.
The magnetic
in
Field
rectangular wire loop of edge lengths a and b rotates with a constant angular velocity
about
CO
Magnetic
in a Static
field
Solution
^ind
(a)
B
Referring to Fig. 6.16(b), the magnetic flux through the loop
®
=B
<t>
C
where 9
is
=B
— abB cos 9,
abn
(6.86)
the angle between the plane of the loop at an instant
perpendicular to the vector B.
©
S
•
is
From
t
and the plane
the definition of angular velocity,
n
T3
1-5
II
3
9o
+ cot
(6.87)
And
Since
co
—
const, the solution for 9
is
=
9(t)
a
where
i
9q
=
0 (9
=
0 for
=
t
0).
(6.88)
,
Thus,
4>(t)
(a)
= abB coscot,
(6.89)
which, substituted in Eq. (6.34), leads to the following expression for the induced
motional emf
in the loop:
— = coabBsmcot.
e jnd (r)
(6.90)
df
(b)
The current
intensity in the loop
amounts
find©
The magnetic
Figure 6.16
A
rotating
uniform
in a
wire loop
time-invariant magnetic
field
field
due
—
to
ej nd
coabB
R
R
to this current
is
neglected in the computation of the
an instant
t
= 0 and
Example 6.12.
/;
in
From
Fig. 4.38
on the loop
is
and Eq.
(4.181), the instantaneous torque of
magnetic forces acting
given by
Tm (t) - m ^abB sin 9 =
coa
2
b2 B 2
sin
i
cot.
(6.92)
~R
(b)
cross-sectional view at an
arbitrary instant
emf
Eq. (6.90).
(elementary ac
generator): (a) top view at
(6.91)
sinoV.
for
Eq. (4.180) tells us that the direction of this torque is such that it opposes the rotation of
the loop (Lenz’s law), i.e., it is opposite to the direction of an externally applied mechanical torque, T mec h, that rotates the loop. In order to sustain the rotation at a constant rate,
R
Tmech must be exactly equal in magnitude to T m (Newton’s second law
Hence, the instantaneous mechanical power used to rotate the loop is
P mech (0 —
co
— Tm CO
T’jnechO)
287
Electromagnetic Induction due to Motion
Section 6.6
angular form).
in
2 2
a b2 B2
sin
R
2
(6.93)
cot.
The same result can also be obtained from energy conservation, as the mechanical
power of loop rotation equals the electric power of the circuit, i.e., the power of Joule’s
losses dissipated in the loop [as in Eq. (6.85)]. This yields
—
Anech(0
—
P](f)
Ri-
j
co
—
2 2
a b 2 B2
sin
(6.94)
cot.
~R
Given that the time-average value of the function
r
1
T
T
T=
2 n/co
is
-cos2 cot
1
.
.
L
where
T
1
Sm2 <**=7 [
JQ
1
2
sin
cot is
T
T
/
~ df = 2 f [Jof
f
.
dt
~
.
power of loop
co
2
\
’
emf and
(6.95)
2
)
the period of time-harmonic variation of the
loop, the time-average mechanical
.
COS20Jtdt
Jo
current in the
rotation equals
a2 b 2 B 2
(6.96)
(^mech) ave
2.
Note that the system in Fig. 6.16 represents a rotational version of the generator
based on translational motion in a static magnetic field in Fig. 6.15. It illustrates the basic
principle of an alternating current (ac) generator, where the loop is mechanically rotated
in a static magnetic field at a constant rate and the emf and current are induced in the
loop of a sinusoidal (time-harmonic) waveform. The angular frequency of this waveform
equals the angular velocity
intensity
=
c.
dc Line Current
Infinite
from
that the current in the straight wire conductor
7,
and that the contour moves away from the wire
in Fig. 6.17.
x
of the loop rotation.
Moving Contour near an
Example 6.13
Assume
co
At
t
=
0,
E indl
Fig. 6.12 is time-invariant, with
at a
constant velocity
v,
as
shown
the distance of the closer parallel side of the contour from the wire
Determine the emf induced
is
in the contour.
Solution The magnetic field produced by the current in the wire is time-invariant, and
the system in Fig. 6.17 represents a motional induction version of the same geometry with
transformer induction in Fig. 6.12. The magnetic field is nonuniform (magnetic flux density
changes in space), and that is why the magnetic flux through the contour is time-varying. As
the contour moves uniformly away from the wire, the distance of its closer parallel side from
the wire increases linearly in the course of time, and is given by
= c + vt.
x(t)
The magnetic
(6.97)
around the wire has the same spatial distribution as in
Fig. 6.12, so that the magnetic flux through the contour has the same form as in Eq. (6.64),
with i(t) substituted by I and c by x(t):
flux density vector
Mo lb
<H0 =
,
x
(6.34), the
d4>
x
motional emf induced
reference direction (Fig. 6.17),
£md(0
+a
In
2n
From Eq.
MO lb
—
—
c
,
In
2n
+ a + vt
c + vt
(6.98)
given with respect to the clockwise
in the contour,
is
dO dx
dO
/xolabv
dx dr
dx
2n
iiolabv
1
x(x
+ a)
2 n{c
+ vt){c + a + vt)'
(6.99)
The emf
in the
contour can also be computed using Eq.
electric field intensity vector Ej n d
E ind2
7
due
to motion, given
by Eq.
(6.70).
Note
(6.68), is
that the induced
perpendicular to the
b
a
Figure 6.17 Evaluation of
the emf
in a
rectangular
contour moving in the
magnetic field due to an
infinitely
long wire with a
steady current; for
Example
6.1 3.
288
Chapter 6
Slowly Time-Varying Electromagnetic Field
pair of contour edges of length a
and hence there
,
Its
magnitude along the
and
left
£.ndi
=
vfli
right
=
^
2.71
emf
in the
contour
e ind
which
is
the
—
same
is
(f
Jc
in Fig. 6.17
and
Eind2
x
equals
= VB 2 =
-
^
2:t(x
(6.100)
+ a)
are the corresponding magnetic flux densities. Finally, the total
obtained as
E; nc
dl
•
|
-
F) ncn£>
Conducting
shows a system
for
—
£ind2^
=
(
x
V
(6.1 01
+ aj
:
\x
2n
)
Eq. (6.99).
result as in
Example 6.14
Fig. 6.18
B2
where B[ and
respectively,
edges
no emf in these edges. On the other hand,
two contour edges, which are parallel to it.
is
Ej ni] does not change along each of the remaining
Fluid
Flow
in a Static
measurement of
Magnetic
Field
fluid velocity that consists of a parallel-plate
capacitor situated in a uniform time-invariant magnetic
the capacitor plates and the voltage between the plates
A liquid flows between
measured by a voltmeter. The
field.
is
The magnetic field lines are parallel
The conductivity and permeability of the liquid are a and p. o, respectively.
The capacitor plate area is S and the separation between plates is d. Fringing effects in the
capacitor can be neglected. The flux density of the applied magnetic field is B. The internal
velocity of the fluid can be considered to be uniform.
to the plates.
resistance of the voltmeter
is
negligible.
Show
is
Ry, while the resistance of the interconnecting conductors
that the velocity of the fluid
is
linearly proportional to the voltage
V
indicated by the voltmeter and find the proportionality constant.
Solution
We
have a motion (flow) of the conducting liquid (with an unknown velocity v)
magnetic field, and therefore an electric field due to motion is induced, given by
Vectors v and B are mutually orthogonal (Fig. 6.18), and we can write
in the static
Eq. (6.68).
E ind =
vB.
This field forces the charge carriers in the liquid to
(6.102)
move
perpendicularly to the direction
of the liquid flow, so that positive and negative excess charges are accumulated on the lower
and upper capacitor
These charges produce a Coulomb
plates, respectively.
to excess charge), which
field
E9
(field
due
perpendicular to the plates and practically uniform in the space
is
between them.
Since the voltmeter
current flowing through
is
not ideal,
its
i.e., its
terminals.
By
current continues with the current of the
/v
internal resistance
is
not
infinite,
same
= /liquid = JS,
a steady
is,
(6.103)
Figure 6.18 Measurement of
based on
motional electromagnetic
induction; for Example 6.14.
is
intensity through the conducting liquid, that
/V
fluid velocity
there
the continuity equation for steady currents, this
V
Section 6.7
Total Electromagnetic Induction
with J standing for the current density in the liquid. This current density, in turn, can be
expressed in terms of the total electric
J
field in the liquid,
where the scalar form of the equation
Ei n d, and E q adopted in Fig. 6.18.
The voltage
terminals
(M and
—
= ctE = cr(E g + Ejnd)
is
J
>
given by Eq. (6.16), as follows:
=
CT(£j nd
- Eq ),
(6.104)
obtained for the reference directions of vectors
that the voltmeter indicates equals the potential difference
between
if
its
N):
V=VM -Vv,
which,
J,
(6.105)
evaluated using Eq. (6.22) along a (straight) path through the liquid, can also be
expressed as
-Vn = Eq d
Vu
(6.106)
(the interconnecting conductors in the voltmeter circuit are ideal,
Finally,
Ohm’s law
RySo (vB —
V = Ryly = RyJS = Ry So (Ejnd — Eq)
from which the solution
see that, indeed, v
—V
(6.1
07)
for the velocity of liquid flow turns out to be
v
We
and thus equipotential).
gives
is
=
oSRy +d
oSdRyB
(6.108)
V.
where the proportionality constant
and the conductivity of the fluid.
linearly proportional to V,
depends on the parameters of the system
in Fig. 6.18
Problems 6.18-6.28; Conceptual Questions (on Companion Website): 6.17-6.22;
:
MATLAB Exercises (on Companion Website).
TOTAL ELECTROMAGNETIC INDUCTION
6.7
Consider now the most general case of electromagnetic induction - that of a moving
conductor in a time-varying magnetic field. This is the case where both sources of
and conductor motion,
Hence, the induced electromotive force in a contour that is
moved and/or deformed in a magnetic field that itself varies with time is the sum
of the transformer emf, Eq. (6.37), and the motional emf, Eq. (6.70). We thus write
the induced electric field, namely, the magnetic field change
act simultaneously.
=-
eind
V*
""
-dS
'
+ ^(v
^
V
transformer emf
and
call e m &
^
x B)
N"
'
-dl,
"
'
(
6 109 )
.
^
motional emf
here the total (transformer plus motional) or complex (combined) emf
in the contour.
The motional emf term of Eq. (6.109) can be transformed as described by
Eqs. (6.71)-(6.75) and Fig. 6.14, so that the total emf in the contour can be expressed
as e ind
= — d<i>/dr
[same as
in
e ind
This last expression
Eq.
— $
(6.34)], or
Ej n d
on the right-hand
dl
—
B
dS.
d t fs
(6.1
1
0)
side of the equation represents the total
derivative of the magnetic flux through the contour with respect to time,
where
total induction
289
.
Chapter 6
Slowly Time-Varying Electromagnetic Field
the change in flux with time
due to a change
These two parts of the
partly
is
due
change
to a
in the shape, orientation,
partly
in the
magnetic
field
and
and/or position of the contour.
change correspond to the transformer and motional
appears, however, that Eq. (6.34) is the most general
form of Faraday’s law of electromagnetic induction, which includes both mechanisms by which the magnetic flux through a contour could change. These two
mechanisms are the magnetic field variation and contour motion, and except for
them, there are no other possibilities that may result in an induced emf in the
emf terms
flux
of Eq. (6.109).
8
It
contour.
Moving Contour near
Example 6.15
Refer to
Fig. 6.17
a Time-Varying Line Current
and assume that the current
time-varying, with intensity
wire conductor
in the infinite
Find the emf induced
i(t).
in the
is
slowly
moving contour.
Solution We now have a motion of the contour in a time-varying magnetic field, produced
by the current in the wire, i.e., a combination of systems in Figs. 6.12 and 6.17. Therefore,
the emf is induced in the contour due to combined (transformer plus motional) induction.
Combining Eqs. (6.64) and (6.98), the magnetic flux through the contour is
<*>(0
From Eq.
=
m(t)b
2n
c
,
In
+ a + vt
c + vt
(6.110) [or Eq. (6.34)], the total (combined)
Cjnd(0
d<t>
—
yu-o
+ a + vt
c + vt
—
b
c
i
~dT
2tz
—
note that the
emf;
it
Eq. (6.65)
Eq. (6.99)
is
/zo i(t)abv
2n(c
dr
+
vt)(c
+a+
(
6 112 )
.
vt)
motional emf
in this expression represents the
transformer part of the total
and becomes the same as in
The second term represents the motional part
case of a stationary contour and becomes the same as
in the case of a steady current in the wire
it
becomes zero
in the case of a
Example 6.16
Assume
term
the contour
case of a stationary contour.
in the
of the total emf;
in
first
becomes zero
emf in
di
transformer emf
We
(6.111)
in
the
steady current in the wire.
Rotating Loop
in a
Time-Harmonic Magnetic
Field
is a low-frequency time-harmonic magnetic field
Rosinwr, and obtain the emf induced in the rotating loop. Identify
the parts of the emf corresponding to the transformer and motional induction.
that the applied field in Fig. 6.16
with flux density B(t)
Solution
=
we
The magnetic
Obviously,
in Fig. 6.16.
are
now adding
flux in
<h(f)
Hence, the emf induced
in
e in d(0
=
abB(t) cos 0
the contour
=
a transformer induction
component
to the
system
Eq. (6.86) becomes
dO
at
=
abB(t) cos cot.
(6.1
1
3)
(6.1
1
4)
is
dB
—ab—- cos cot + coabB{t) sin a>t
df
v
transformer emf
,
motional emf
8
Note that the division of the induced emf between the transformer and motional parts depends on the
chosen frame of reference. The particular division in Eq. (6.109) is given for the stationary frame of
reference, attached to the field B, with respect to which the contour moves at the velocity v. In other
words,
it
however,
is
is
given as measured by a stationary observer (so-called laboratory observer).
The
total emf,
unique and the same for any chosen frame of reference and any observer, including the one
moving with the contour.
291
Total Electromagnetic Induction
Section 6.7
For the given time-variation B(t), the terms corresponding to the transformer and motional
induction appear to be
= — COdbBo COS
^ind(transformer)
respectively,
=
e ind (r)
We
and the
total
—coabBo (cos
and
cot
(6.1 15)
cot,
emf
2
cot
—
sin
2
=
cot)
—coabBo coslcot
emf (and
see that the frequency of the induced
frequency of the applied magnetic
=
coabBocos(2cot
+ n).
current) in the contour
is
(6.1
1
6)
twice the
field.
Stationary Loop
Example 6.17
— (OClbBo sin
Cind(motional)
in a
Rotating Magnetic Field
R
A rectangular loop of resistance R
produced by two mutually
perpendicular large coils with low-frequency time-harmonic currents. The field due to
each coil can be considered to be uniform. The currents in the coils are of equal amplitudes and 90° out of phase, so that the magnetic flux densities they produce are given as
Bi(t) = Bo cos cot and Bi{t) = So sinruf, respectively, and shown in Fig. 6.19(a). The sides of
the loop are a and b long. Neglecting the magnetic field due to induced current in the loop,
find the time-average torque of magnetic forces on the loop.
Solution From
situated in the magnetic field
is
magnitude of the resultant magnetic
Fig. 6.19(b), the
B(f)
at
=
Bi(r)
flux density vector,
+ 82(f),
(6.117)
an arbitrary instant of time equals
|B(f)|
i.e., it is
=
^B\{t)
+ B\{t) = 7s 2 (cos 2 cot T sin 2 cot) =
that the tip of the vector B(f) traces a circle
Tm
Such a vector belongs to a class of so-called circularly
mark the angle between vectors B(f) and Bi(f) at
(b)
constant with respect to time. This
of radius
Bq
(6.118)
Bq,
in the course of time.
means
polarized time-harmonic vectors. Let 6
time
t,
Fig. 6.19(b).
The tangent
of this angle
is
Figure 6.19
=
tan 0(f)
and hence the rate
at a
which it changes
#2(0
Bq
Bi{t)
Bq cos cot
in
time
is
sin cot
=
tan cot
and currents
= cot [see Eq.
0(f)
At
in the coils.
t
=
0,
B2
(6.119)
,
given by Eq. (6.87). This means that
constant angular velocity, equal to the angular frequency
densities
and
at
=
0 and
B=
co
B rotates
of the individual magnetic flux
Bi, which implies that 0(0)
=
0
(6.88)].
and vector B changes in time (its magnitude is constant, but
its direction changes), this is a system based on transformer induction. On the other hand,
for the generation of emf it is irrelevant whether B rotates around a stationary loop or a
loop rotates (at the same rate) in a static B. This latter case is exactly the system based on
As
the contour
is
motional induction in
Fig. 6.19,
stationary
Fig. 6.16.
Exploiting this equivalency, 9 the flux through the loop in
emf, current, and instantaneous torque of magnetic forces on the loop are given by
Eqs. (6.89), (6.90), (6.91), and (6.92), respectively, with
the time-average torque
coa
direction of this torque
which
is
in
(6.95),
is
the
same
2
b 2 Bl
(
2R
6 120 )
.
as the direction of the field rotation [Fig. 6.19(b)],
accordance with Lenz’s law. Namely, the induced current in the loop and the
associated magnetic
9
B substituted by Bq. Using Eq.
is
(Lm ) ave
The
moment of the
loop produce a torque (T m ) that tends to rotate the loop
The possibility to approach the problem of a stationary contour in a rotating magnetic field
induction case
is
the reason for which
we
analyze
B2
it
in the section
devoted to
as a motional
total induction.
A
rectangular
wire loop exposed to two
time-harmonic magnetic
equal amplitudes
and 90° out of phase (a),
fields of
which superposed to each
other represent a rotating
magnetic field
Example 6.1 7.
(b); for
292
Chapter 6
Slowly Time-Varying Electromagnetic Field
along with the applied rotating
decrease the angle
which
0],
field
in
is
contour (caused by the increase
r~
Assume
same
that the loop
generated the emf
in 6) that
Rotating Loop
Example 6.18
Wo
B
to rotate n closer to
[i.e.,
in Fig. 6.19(b)
and thus
opposition to the change in the magnetic flux through the
in a
the
first
place.
Rotating Field - Asynchronous Motor
exposed to the rotating magnetic
direction with an angular velocity
in
<
coo (too
field
from
Fig. 6.19 also rotates in the
as indicated in Fig. 6.20. This device
to),
represents an elementary asynchronous motor. Calculate (a) the time-average power of
Joule’s losses dissipated in the loop, (b) the time-average torque of magnetic forces
(0
loop,
and
on the
motor.
(c) the efficiency of the
Solution
PB
(a) This
in
field
asynchronous motor in the
form of a rectangular wire
loop rotating with an
co o in
angular frequency
> wo (top view
for
Example
co,
at
t
on
total
(mixed) induction - the magnetic
field
called the asynchronous
It is
field.
The
changes (rotates)
motor because the
relative rate of rotation of the
with respect to the rotating part of the motor (the loop in our case), called the rotor,
equals
Aco
= co — coq,
(6.121)
a
which
rotating magnetic field of
co
a system based
loop does not rotate in synchronism with the
Figure 6.20 Elementary
angular velocity
is
time and the loop moves (rotates).
where
—
referred to as the slipping angular velocity of the asynchronous motor.
is
Consequently,
this
system can be replaced by either an equivalent system with a station-
ary loop and a magnetic field rotating with a velocity Aco (transformer induction case, as
0);
in Fig. 6.19) or
6.1 8.
magnetic
field
(6.121), the
an equivalent system with a loop rotating with a velocity Aco
(motional induction case, as in
magnetic flux through the loop
0(f)
Eq. (6.94) then
tells
=
From
Fig. 6.16).
in a static
Eqs. (6.89), (6.118), and
in Fig. 6.20 is
ab |B| cos Acot
— abBo cos(a> — coo)t.
us that the time-average
(6.122)
power of
Joule’s losses in the loop can be
where
=
written as
(Fj)ave
(b)
By means
= k(co - coor,
a2 b2 B2
(6.123)
2R
of Eq. (6.92), the time-average torque of magnetic forces on the loop
(Tm )ave
where the
k
coefficient k
is
1
k(a>
—
is
(6.1
coq),
that in Eq. (6.123). This torque has the
same
24)
direction as the
slipping velocity of the motor.
(c)
The
rate of the loop rotation in Fig. 6.20
is coo,
so that Eq. (6.93) gives the following
expression for the time-average mechanical power used to rotate the loop:
(P mech)ave
The
efficiency of the
motor
is
n
where we neglect the
Example 6.19
=
Tm)ave
(
too ==
k too(to
—
too).
(6.1
25)
given by
=
(P mech)ave
toO
(6.126)
’
(Pmech)ave
T"
(P]) ave
losses in the stationary part of the
Charge Flow due to
a
to
motor
(the stator).
Magnetic Flux Change
Consider a wire contour of resistance R situated in a magnetic field, as shown in Fig. 6.21. If
this field is changed and/or the contour is moved in the field during an arbitrary interval of
time so that the corresponding net change of the magnetic flux through the contour is A0,
find the total
charge flow
Q in
the contour during this process.
Section 6.7
Since the magnetic flux of the contour,
Solution
emf
which we can
process, an
for
is
induced
in the
varies in time during the considered
$>,
contour (due to the total induction
d4>
is
i
in the
general case),
write
fiind
where
= -—r
dr
and
e ind
= Ri,
the intensity of current in the contour (Fig. 6.21).
(6.127)
Combining these two equations
= -Ri dr.
d<J>
d<2
=
dr,
i
We
we
(6.128)
Figure 6.21 Evaluation of
the charge flow
(3.4),
the charge that flows through the contour during an elementary time dr
is
resulting in
= -RdQ.
d<l>
tion,
e ind
Q
gives
From Eq.
293
Total Electromagnetic Induction
(6.129)
On
then integrate both sides of the above equation.
AO, from
thus obtain the total change of flux,
its
the left-hand side of the equa-
in
a wire
contour as a consequence of
a change of the magnetic
flux through the contour;
for
Example
6.1 9.
starting value (4>i) to the ending
value (O 2 ) in the process,
dO =
4>2
4>i
= AO.
(6.130)
Q
equals the total charge flow, Q,
1
On
the right-hand side of the equation, the integral of d
and hence
(6.131)
charge flow
due
where the reference directions of the charge flow and the magnetic
by the right-hand
flux are interconnected
to
in
a wire contour
a magnetic
flux
change
rule, as indicated in Fig. 6.21.
Example 6.20
Fluxmeter Based on a Charge-Flow Measurement
B
A fluxmeter consists of a small coil (magnetic sonde) connected to a ballistic galvanometer,
is S and the number of wire turns is N.
and the galvanometer is R. The coil is placed in a uniform
time-invariant magnetic field such that the magnetic field lines are perpendicular to the flat
surface spanned over the coil cross section (Fig. 6.22). The coil is then removed from the
field, and the charge flow indicated by the galvanometer is Q. What is the flux density of the
magnetic field?
as in Fig. 6.22.
The
The
cross-sectional area of the coil
total resistance of the coil
Solution The magnetic flux of the
coil while
<Di
it is
in the field
= NBS,
whereas $2 = 0 after the coil is removed from the
through the galvanometer is given by
Q=
A4>
4>2
=
IT
is
(6.132)
field.
- $1
=
R
Using Eq.
(6.131), the charge flow
galvanometer
(BG); for Example 6.20.
a ballistic
<J>i
~R
which yields the following expression for the magnetic
=
NBS
(6.133)
~1T'
flux density of the field:
(6.134)
NS
We see that B is linearly proportional to
Q, where the proportionality constant is determined
by the parameters (R, N, and S) of the fluxmeter in Fig. 6.22.
Thus, by directly measuring the charge flow through its terminals, a ballistic galvanometer connected to a magnetic sonde can be used to indirectly measure the flux density of an
unknown magnetic field. In some cases, like in this example, the measurement is based on
motional electromagnetic induction (field is time-constant and a coil is either removed from
the field or brought into
it).
In other applications, like in the apparatus for
magnetization curves in Fig. 5.19, a coil
is
Figure 6.22 Fluxmeter
consisting of a small coil
stationary in a field that
is
measurement of
from
either established
and
)
294
Chapter 6
Slowly Time-Varying Electromagnetic Field
zero to
its final
measurement
value
is
B
Prove that the magnetic
<1>
= const
Solution
A
R=
(
B
to zero, so that the
Magnetic Flux through a Superconducting Contour
Example 6.21
resistance,
be measured) or reduced from some value
(to
based on transformer electromagnetic induction.
through a superconducting wire contour cannot be changed.
flux
superconducting wire has zero
0,
resistivity
and zero
[see Eq. (3.23)]
total
so that Eqs. (6.127) give
— =0
d<t>
through a
(R
=
(6.135)
0),
dr
superconducting contour
that
is, <t>
=
const, which completes our proof.
Having
in
mind
that
denotes the
<t>
can be explained as follows.
(external or primary magnetic field)
current
is
induced
in the
total existing flux
through the contour,
is
changed and/or the contour
contour whose magnetic
field
(secondary
<t>
moved
its
in the field, a
completely cancels
extreme form), such
through the contour remains constant.
Problems'. 6.29-6.33; Conceptual Questions (on
MATLAB
is
field)
the change of the magnetic flux through the contour (Lenz’s law in
that the total flux
this result
the magnetic field in which a superconducting contour resides
If
Exercises (on
Companion
Companion
Website): 6.23-6.25;
Website).
EDDY CURRENTS
6.8
Whenever
induced in a conducting medium, electric current is
same time-dependence as the field (we assume that the
medium is linear in terms of its conductivity). An example is a conducting wire loop
in a time-varying magnetic field (Fig. 6.8), where the intensity of the induced current in the wire is given by Eq. (6.38). This section is devoted to studying volume
induced currents in solid conducting bodies, where many current contours are established throughout the volume of the body as a result of the induced electric field.
These currents are perpendicular to the magnetic flux in the body and, since they
flow like “eddies” (in water), we call them eddy currents. The eddy current density
vector, Jeddy is related to the electric field intensity vector, E, through Ohm’s law
electric field
is
also established, with the
>
in local
form:
density of eddy currents
(6.136)
where a
is
The vector E represents the actual (meais composed of the induced
excess charge, E [see Eq. (6.16)]. The vec-
the conductivity of the material.
surable) electric field in the material, which, in general,
electric field, Ej ncj,
and the
field
due to
tor Jeddy in Eq. (6.136), therefore,
material, the corresponding
component
cr
f/
the actual (total) current density vector
components of which are a Ej n d and a E
f/
.
Note
in
the
that the
is
usually identified as the induced current density vector
However,
as the accumulation of excess charge in the majority of
Ej n d alone
in the material.
is
practical systems with an induced electric field
is
also a result of electromagnetic
components of the eddy current density vector can be said to be
induced by the same cause that induces the field Ej n d and electromotive force in
the system, i.e., to represent the induced current in the material. This cause, on the
induction, both
other hand, can be related to either transformer or motional induction, as well as to
total
(combined) induction.
Section 6.8
As a consequence of eddy currents,
according to Joule’s law, and this
is
electric
power is
lost to
295
Eddy Currents
heat in the material,
the principle of induction heating. In so-called
induction furnaces, for instance, eddy currents are created on purpose to produce
local heating in metal pieces
and high enough temperatures to melt the metal.
In ac machines and transformers, however, the power loss due to eddy currents
induced in ferromagnetic 10 cores
sity
of the
power of
is
undesirable.
From Eq.
(3.31), the
Joule’s losses at a point in the material
is
volume den-
proportional to the
square of the density of eddy currents, /e ddy> at that point. Using Eq. (3.32), the total
instantaneous power of Joule’s (ohmic) losses in the entire body is obtained as
r I
Pj=
2
dv,
Jv
(6.137)
o
By
due
to
eddy currents
where v denotes the volume of the body.
Another important consequence of eddy currents
produce.
joule's (ohmic) losses
is
the magnetic field that they
Lenz’s law, this field (secondary magnetic field) opposes the change in
the primary magnetic flux inside the body, which caused the eddy currents in the
place.
first
While the secondary magnetic
circuits is practically
often
this
the
rial
is
field
due to induced currents
in thin-wire
always negligible with respect to the primary magnetic
field,
not the case with volume eddy currents in solid bodies. The larger
volume of the body and areas of eddy current contours (eddies) in the matethe larger the induced emf along the contours and the current intensities, as
well as the magnetic field they produce. This effect
is also usually not desirable.
For instance, the magnetic field due to eddy currents in ferromagnetic cores of ac
machines and transformers tends to cancel the primary magnetic flux in the core
and thus considerably reduces the efficiency of the device. As an illustration, consider a large core of a rectangular parallelepipedal shape in a uniform time-varying
primary magnetic field B, as depicted in Fig. 6.23(a). The secondary (induced) mag-
(a)
aj
:
netic field, Bi n d,
is
the strongest at the center of the cross section of the core,
i.e.,
center of all eddy current contours, where all the fields due to these contours
add up. Hence, the resultant (primary plus secondary) magnetic flux density is not
uniformly distributed over the core cross section; it is the smallest at the center and
at the
the largest near the core surface. In other words, practically only the “skin” region
below the surface of the core
carries the
the operation of the device. This
magnetic
phenomenon
is
flux
and
is
effectively used for
referred to as the skin effect in
ferromagnetic cores.
Note that the skin effect in current conductors at high frequencies is also a
consequence of induced (eddy) currents and their magnetic field. To show this, consider a cylindrical conductor carrying a time-harmonic (ac) current of density J,
Fig. 6.23(b). Using the analogy with the dc case in Fig. 4.15, the lines of the magnetic field B due to this current are circles centered at the conductor axis. This field
induces an electric field Ej ncj, which is axial in the conductor (field lines are parallel to the conductor axis) and can be qualitatively analyzed by applying Faraday’s
law of electromagnetic induction, Eq. (6.35), to the rectangular contour C shown in
Fig. 6.23(b). The direction of vectors Ej nC and J e ddy is determined by Lenz’s law (i.e.,
by the minus sign in Faraday’s law). It is such that the secondary magnetic field, Bj ncj,
opposes the primary field B. Hence, the eddy current density vector tends to cancel
j
10
Ferromagnetic materials are electrically conducting, with large conductivities
iron).
(e.g.,
ape
=
10
MS/m for
(b)
Figure 6.23 Illustration of
the skin effect
in a
parallelepipedal
ferromagnetic core with a
time-varying magnetic
(a)
and
field
in a cylindrical
conductor with a
time-harmonic current
(b).
296
Chapter 6
Slowly Time-Varying Electromagnetic Field
the current density J inside the conductor volume, while adding to
near the conductor surface. Therefore, the magnitude of the
vector
is
its
magnitude
total current density
small at the conductor axis and increases towards the conductor surface.
The higher the frequency
(i.e.,
law) the larger the induced
skin effect.
At very high
the faster the time rate of change d/d f in Faraday’s
emf
in the
contour
C
frequencies, the current
and the more pronounced the
is
restricted to a very thin layer
(“skin”) near the conductor surface, 11 practically on the surface
itself, and can be
considered therefore as a surface current and described using the surface current
density vector, J s [see Eqs. (3.12) and (3.13)].
Examples
in this section
include evaluation of eddy current distributions and
Joule’s losses in several characteristic systems based
on each of the three types
of electromagnetic induction (transformer, motional, and total induction). In
cases,
we
skin effect in
body would require
on numerical field-computation techniques.
the distribution of eddy currents in the
analysis based
Example 6.22
A
Eddy Currents
in a
is
a
much more complex
Thin Conducting Disk
thin conducting disk of radius a, thickness 8 (8
lio
all
magnetic field due to eddy currents and the associated
conducting bodies. Taking this field into account in the evaluation of
shall neglect the
positioned inside an infinitely long
air-filled
low-frequency time-harmonic current of intensity
a),
conductivity a, and permeability
shown
solenoid, as
/(f)
=
/q
cos
cot
in Fig. 6.24(a).
A
flows through the winding.
The number of wire turns per unit of the solenoid length is N'. (a) Determine the distribution
of eddy currents induced in the disk, neglecting the magnetic field that they produce, (b) Find
the time-average power of Joule’s losses dissipated in the disk, (c) Evaluate the magnetic
field due to eddy currents at the disk center.
Solution
(a)
Eddy
currents are induced in the disk due to transformer induction in this structure.
induced electric
field inside
the solenoid
Using Eq. (6.136) and referring to
is
given by the
first
The
expression in Eqs. (6.51).
Fig. 6.24(b), the distribution of
eddy currents
in the
Figure 6.24 (a) Thin
conducting disk inside an
infinitely
long solenoid with a
low-frequency time-harmonic
current and (b) evaluation of
in the disk and
magnetic field at the disk
center; for Example 6.22.
eddy currents
their
11
For example, we shall see
(b)
in a later
chapter that the thickness of the layer that carries most of
the current in copper conductors at frequencies higher than about
millimeter.
I
MHz
is
less
than a fraction of a
=
0
0
Section 6.8
disk
is
described by the following expression for the induced current density:
7eddy (C 0
—
E\n& (f
>
0
—
liQcrN'r
di
2
dr
cofioaN'Ior
=
sin cur
where r is the radial distance from the solenoid
due to excess charge is practically zero.
(b)
From Eq.
(6.137), the instantaneous
(0
<
axis
power of Joule’s
27rrdr<5
= J/r=
losses in the disk
=
is
sin
3
2
r
cot
dr
Jo
.
sin
?
(6.139)
cur,
with dv being the volume of an elementary hollow disk of radius
[Fig. 6.24(b)],
electric field
4
no{con.pN'Ip) a 8
ness (height) 5
(6.138)
a),
ra
na(a)/xoN' Ip) 8
dv
2
<
and we assume that the
2
PjU)
r
2
Having
mind Eq.
in
(6.95), the
r,
width
and
dr,
time-average of
this
thick-
power
amounts to
na (cofipN'Ip) 2 a 4 8
>
(6.140)
(7 j)ave
16
(c)
As
8
a,
the elementary hollow disk of volume dv in Fig. 6.24(b) can be replaced by an
equivalent circular current contour (wire) of radius r and current intensity
d/eddy (r,
0 =
./eddy O',
0 S dr
(6.141)
which the current of density J e ddy flows is a
The magnetic flux density
the center of the disk is obtained using Eq. (4.19) for z = 0 and r
(cross section of the hollow disk through
small rectangle of side lengths 8 and dr, and surface area 8 dr).
due to
this
current at
substituting
a:
MO d/eddy( r 0
>
d/?i n d(r
By virtue of the
is
,
(6.142)
t)
2r
superposition principle, the resultant magnetic field due to eddy currents
given by
—
a
co/aiaN'IpS
/^ind(0
[ dB m<xir
Jr =
—
,
t)
Comparing the amplitude
primary
field B(f)
=
(iqN'Iq cos
f
sin cot
4
Jo
dr
=
——
co/iiaN'IpaS
i !
sin cot.
4
Bj n do of the field B; n d(f) to the amplitude
cot
[Eq. (6.48)] inside the solenoid,
BindO
_
~
we
(6.143)
Bo of the
see that
co/dpaaS
(6.1
4
44)
Hence, the magnetic field due to eddy currents in the disk is negligible with respect to
the magnetic field due to primary currents in the solenoid only if
nfunaaS
where /
tion
is
=
«
(6.145)
i
frequency of the currents, and whether or not this condidepends on the numerical values of the parameters of the structure in
As an example, for / = 60 Hz (power frequency), a — 58 MS/m (copper),
co/(2it) is the
satisfied
Fig. 6.24(a).
and 8 — a/20, we obtain that only for disks with quite small radii (a
5 cm), Bj n do
Bo,
whereas eddy currents in larger disks produce magnetic fields that cannot be neglected
with respect to primary fields (at least at the disk center, where the secondary magnetic
maximum).
Note that the magnetic field Bj n d in Eq. (6.143) is evaluated based on the distribution of eddy currents given by Eq. (6.138). However, in the case when the condition in
Eq. (6.145) is not satisfied, this evaluation is not accurate enough and provides only qualfield is
itative results,
because the starting expression for the induced electric
field in
Eqs. (6.51)
Eddy Currents
297
298
Chapter 6
Slowly Time-Varying Electromagnetic Field
is
obtained taking into account only the primary magnetic
on the right-hand
field
side of
Faraday’s law of electromagnetic induction.
Example 6.23
Eddy Currents
in a
Thin Ferromagnetic Plate
A thin conducting ferromagnetic plate of length b, width a
in a
uniform low-frequency time-harmonic magnetic
The
field lines are
and thickness d(d
<5C
of flux density B(t)
a)
=
is
situated
Bq cos cot.
perpendicular to the plate cross section, as shown in Fig. 6.25(a). The
conductivity of the plate
by eddy currents, find
total
,
field
is
a. Neglecting the
end
effects
(a) the distribution of these currents
and the magnetic field produced
throughout the plate and (b) the
time-average power of Joule’s losses associated with them.
Solution
(a)
This
By
is
another example of a structure with eddy currents due to transformer induction.
neglecting the end effects (since d <& a),
we assume
that the current streamlines in
the plate are straight and parallel to the plate surfaces, as indicated in Fig. 6.25(b).
From
Eq. (6.136), the same is true for the lines of the total electric field intensity vector, E,
in the plate. By the same token, the magnitude E of this vector does not depend on
the coordinate y in Fig. 6.25(b). Applying Faraday’s law of electromagnetic induction,
Eq. (6.37), to the rectangular contour C shown in Fig. 6.25(b), we get
2E(x,
t) l
=—
2x1,
—-<x<^-,
2
df
(6.146)
2
which yields
E{x,t)
dB
= —x -—=
coBqx sin cot,
(6.147)
dr
where we neglect the magnetic
these currents is given by
field
•^eddyCr
Figure 6.25 (a) Thin
conducting ferromagnetic plate
in a uniform low-frequency
time-harmonic magnetic field,
(b) distribution of
eddy
currents in the plate, (c)
N
insulated thin plates forming a
core of an ac machine or
transformer,
and
(d) a core
with the same dimensions
made
of a single piece of
material; for
Example 6.23.
0
due
to
eddy currents
in
= oE(x, t) = cooB^x sin cot.
the plate.
The density of
(6.148)
2
Section 6.8
(b)
The
total instantaneous
the plate
power of
volume
v of
is
Jldd^X ^
rd/2
’
Pj(0
Joule’s losses dissipated throughout the
299
Eddy Currents
-l
abdx
— a>,2 oabBQ9 sin"9 tot
2
x dx
=
to
2
oabd 3 Bl
/
Jx=-d/
dv
-
sin
2
art.
12
(6.149)
where dv represents the volume of a differentially thin slab of thickness dx used
volume integration. Finally, averaging in time [Eq. (6.95)] results in
to
2
aabd3 Bq
(6.150)
(^*j)ave
Note
24
machines and transformers are made of mutu-
that ferromagnetic cores of ac
ally insulated
in the
stacked thin plates, as portrayed in Fig. 6.25(c), rather than of a single
shown
With this, the areas of eddy current contours
and so are the electromotive forces and current
intensities along the contours. Consequently, both undesirable effects of eddy currents
(Joule’s losses and secondary magnetic field) are reduced significantly as well. In specific, the time-average power of Joule’s losses in the laminated core in Fig. 6.25(c) can be
obtained as N times the power given in Eq. (6.150) for a single thin plate, where N is the
piece of material,
in Fig. 6.25(d).
in the core are considerably reduced,
number
of insulated thin plates.
homogeneous core
losses in the
On
the other hand, the time-average
in Fig. 6.25(d)
thin-plate expression in Eq. (6.150) for the plate thickness Nd. Hence,
(^j)avel oc
AW3
and
power
of Joule’s
can roughly be estimated using the same
(Ej) a vc2 o< (Nd)
3
,
we can
write
(6.151)
ohmic
respectively, for these
laminated core
is
two powers.
We
see that the reduction of Joule’s losses in the
estimated to be as large as by roughly a factor of TV 2 as compared to
losses in a
laminated
homogeneous core of an ac machine or
core
vs.
transformer
homogeneous core with the same dimensions. Also, the skin effect caused by the
magnetic field due to eddy currents is much more pronounced in the core in Fig. 6.25(d),
where the resultant magnetic flux is “pushed” to the “skin” region near the surface
the
of the core only, so that the entire interior of the core
is
practically flux-free. In the
core in Fig. 6.25(c), on the other side, the “skin” regions are formed in each of the thin
insulated plates, so that, although not entirely uniform over the cross section of the
core, the resultant flux
is
much more
core and the ferromagnetic material
densely distributed throughout the volume of the
is
much more
effectively
used for the machine or
transformer operation.
Note
also that,
from Eq.
(6.150),
2
(Pj)ave CX
/
(6.152)
CT,
dependence of ohmic
due
which means that the power loss due to eddy currents in the core increases very rapidly
with an increase in the operating frequency of the device (f) and also that it can be
reduced by using core materials that have low conductivity (u). That is why ferrites
(which have high permeability but low conductivity) are used instead of ferromagnetics in
some
applications at high frequencies
(e.g.,
for the cores of high-frequency
transformers or multi turn loop antennas).
Example 6.24
A
Eddy Currents
in a
Rotating Strip
very long, thin conducting strip of length
about
its
/,
axis at a constant angular velocity
width
to
in a
a, and thickness 8 (5 <sc a <SC /) rotates
uniform time-invariant magnetic field
shown in Fig. 6.26(a). At an instant t = 0, the strip is perpendicular
The conductivity of the strip is a, and permeability ptQ. Determine (a) the
eddy currents and (b) the instantaneous power of Joule’s losses in the strip.
of flux density B, as
to the vector B.
distribution of
Neglect the end effects and the magnetic
field
produced by eddy
currents.
to
losses
eddy currents on
frequency and conductivity
300
Chapter 6
Slowly Time-Varying Electromagnetic Field
Solution
This
(a)
a structure with
is
induced electric
field
®B
due to currents
shown
the strip,
/
=
Ejnd
o-, /i
0
where 6
a
t
v x
is
in
in Fig.
The vector
At a point P,
perpendicular to the cross section of
the strip.
Ej n<j
6 26(b).
the velocity of which
is
Note
between the
is
=
v
~\ <x<
B = vB sin 9{— z) = —coxB sin cot z,
the angle
[see Eq. (6.88)].
Ej n d
(a)
= cot
eddy currents generated because of motional induction. The
given by Eq. (6.68), where we neglect the magnetic
field in the strip is
cox,
(6.153)
\'
and the reference horizontal plane
strip
at
time
that, for the position of the strip in Fig. 6.26(b), the direction of
> 0 and into the page for x < 0 (point P'). The density of
given with respect to the reference direction out of the page in
out of the page for x
>s
eddy currents
in the strip,
Fig. 6.26(b), is
/eddy
where we neglect the end
near the ends of the
lates
= crFind = ojcjxB sin cot,
(6.1
54)
i.e., the electric field due to excess charge that accumuand causes the current streamlines to bend and close into
effects,
strip
themselves near the ends.
(b)
The
total
instantaneous power of Joule’s losses in the strip comes out to be
f
—
Pj(t)
a/2
^eddv
/
Jx=-a/2
where dv
rotating
in a
and
at
dx
co
77
oI8B 2
sin
a
f
2
cot
a
is
7
x2 dx
/
=
volume of an elementary
2
oa i l8B 2
—
sin
2
cot,
(6.155)
12
J -a/2
the
co
strip of length
/,
thickness
8,
and
in a strip
uniform
top view at
t
=
(b) cross-sectional
an arbitrary time
r;
Rotating Cylinder
Example 6.25
time-invariant magnetic
field: (a)
18
7
2
=
width dr.
Figure 6.26 Evaluation of
eddy currents
=
dv
0
view
for
Example 6.24.
in a
A very long conducting cylinder of length
angular velocity
density
B and
coo
about
field
radius
a,
and conductivity a rotates
while being exposed to a rotating magnetic
its axis,
at a constant
field
of flux
(w > wo), as shown in Fig. 6.27(a). Neglect the end effects
due to eddy currents, and calculate the total instantaneous power of
angular frequency
and the magnetic
/,
Rotating Magnetic Field
co
Joule’s losses in the cylinder.
Solution
This
we can use
is
a system based
by a cylinder rotating
Rotating
Figure 6.27
(a)
cylinder
rotating magnetic
field
and
in a
(b) equivalent system
with the cylinder rotating
magnetic
Example 6.25.
static
field; for
in
a
on
total
(transformer plus motional) induction. However,
the concept of slipping velocity given in Eq. (6.121) and replace this system
in a static
magnetic
field
with the rate
Aw = w — wq
in the
opposite
Problems
direction, as depicted in Fig. 6.27(b). In the equivalent system,
induction only, the induced electric field
point
P
found
is
is
based on motional
Eq. (6.153). At a
in a similar fashion as in
in Fig. 6.27(b),
=
Ej n d
= vBsin6(—x) = — AcorB sind z
v x (By)
= — AtorB sin(Acot + 9q)z,
where
which
301
r is
<
<
a,
the radial distance from the cylinder axis (r
=
0
r
between the vector
—n <
|r|),
v
<
jt ,
(6.156)
= A cor
is
the velocity of
0
and the x-axis, which equals 6q at
t = 0 and is given by Eq.
(6.88) with A co as rotation rate. Note the opposite directions
of Ei n d at points P and P' in the cylinder cross section, i.e., for 0 positive and negative,
respectively, which tells us how the streamlines of eddy currents (J e ddy) close throughout
the point, and 0
is
the angle
r
the cylinder.
The
total instantaneous
power of Joule’s
losses dissipated in the cylinder
is
obtained by
integration:
Pj(t)= f crEfnd
Jv
dv= f f
aEfnd (r,d)lrd6dr
Jr=0Jd=—7x
'fl
a
=
(A co)
i
2
oB 2
f
l
/
71
As dr
JO
,
f
7
=
sm 2 6 dQ
,
n (co — coq ) 2 a a 4 IB 2
(6.157)
,
4
J-7T
where dS is an elementary patch in the cylinder cross section [Fig. 6.27(b)], the sides of which
are dr and r d 0 long, and dv = / dS. We see that the dissipated power is constant with
respect to time, which is a consequence of the cylindrical (rotational) symmetry of this
problem.
Problems 6.34-6.45; Conceptual Questions (on Companion Website): 6.26-6.30;
:
MATLAB Exercises (on Companion Website).
Problems
6.1.
Induced
electric field of a circular current loop.
Assuming
in Fig. 4.6
is
electric field
above a square current
vector at a point
N
placed at a height a above
induced electric field
intensity vector at an arbitrary point along the
the vertex representing the junction of sides 1
z-axis (point P).
coordinates of the point thus being x
Induced
and
i(t),
electric
find the
field
of a
triangular
and 2 of the square contour
cur-
Fig. 4.39, calculate the electric field intensity
vector at the point
P induced by
time-varying current of intensity
6.5.
a slowly
i(t)
in
the
loop.
Magnetic field of an EMI source (square contour). Consider the square current contour
described in Example 6.2, and find the magnetic field intensity vector at the point
Fig. 6.2(a).
Compare
M
in
the result with that for
z
in Fig. 6.2(c), the
= y = a /2
= — a, induced by the pulse current i(t) in
Fig. 6.2(b).
rent loop. For the triangular current loop in
6.3.
Induced
contour. Determine the electric field intensity
not steady but slowly time-varying,
with intensity
6.2.
6.4.
that the current in the circular loop
Magnetic field of a current contour of complex
shape. For the wire contour with semicircular
and linear parts carrying a low-frequency timeharmonic current from Example 6.4, compute
the magnetic field intensity vector at the point
O in Fig. 6.4.
6.6.
Induced
electric
field
of
a
semicircular-
rectangular loop. Find the induced electric field
intensity vector at the point
O
in Fig. 4.40,
the induced electric field intensity vector in
assuming that a slowly time-varying current
Fig. 6.2(d).
of intensity
i(t)
flows along the loop.
2
302
6.7.
Chapter 6
Slowly Time-Varying Electromagnetic Field
Current contour with circular and straight segments. A current of intensity i(t) = sin(10 8 /) A
(r in s) flows along a wire contour with two
circular (quarter-circle and 3/4-circle) and two
linear parts,
,
shown
in Fig. 6.28,
where a
lengths b and c
(
b c
,
field
(a)
= 3 cm
>
2a), as
is
placed coaxially around
shown
in Fig. 6.29.
it
The magnetic
due to induced currents can be neglected.
What
is
the total induced
Find the emf
the edge
in
emf
MN
in the
loop?
of the loop,
and b
— 9 cm. (a) Verify that this is a lowfrequency current, and compute (b) the in-
^indMN» in the following two ways, respectively:
(b) by integrating along the edge the induced
duced electric field intensity vector and (c) the
magnetic field intensity vector at the point O.
electric field intensity vector
in the
due to the current
solenoid winding [use the relationship
Eq. (4.43) to solve the integral] and (c) by
showing that Cj n dMN equals the induced emf in
the triangle
in Fig. 6.29, and then comin
AOMN
Figure 6.28 Wire
contour with a
puting this latter
quarter-circle,
from
emf
as a part of the total
emf
(a).
and two
3/4-circle,
linear parts carrying a
N
P
low-frequency
time-harmonic current;
Problem 6.7.
Figure 6.29
for
Rectangular wire loop
placed coaxially around
6.8.
Induced electric field at the axis of a circular
segment. Repeat Example 6.3 but for the field
the solenoid
for
in Fig. 6.9;
Problem 6.12.
point at an arbitrary location (defined by the
coordinate z) along the z-axis (normal to the
plane of drawing) in Fig. 6.3(a).
6.9.
Induced
electric field
composed of two
made of a
i(t).
lengths,
Determine the induced
parts with the
is
same con-
and S 2 (Si ^ S2), and different
determined by angles a and 2n — a
respectively, as
shown
in Fig. 6.30.
an arbitrary
electric field intensity vector at
point along the z-axis
assum-
S\
carries a slowly time-varying
current of intensity
6.6 but
ductivity, a, but with different cross-sectional
areas,
it
Repeat Example
ing that the wire loop around the solenoid
Example 4.4,
semicircle and a straight line from
Solenoid and a loop of wire with nonuniform
cross section.
above or below a semicir-
cular loop. Consider the wire contour
and assume that
6.13.
M
in Fig. 4.9(a).
from current distribution and total
electric field. In a domain v in free space, there
is a slowly time-varying distribution of volume
currents and charges. We know the current
6.10. Voltage
density vector,
J, at
every point of
v, as
Figure 6.30
Structure
Fig.
well
a loop of wire of
as the (total) electric field intensity vector (due
to currents
and charges
outside
Find the voltage between any two
points,
6.11.
it.
in v), E, at
M and N, outside
A
in a quasistatic
is
placed
quasistatic electromagnetic field, for
magnetic vector potential, A,
tric field
What
6.12.
is
in
a
which the
known
at
every
are (a) the total elec-
inside the wire
between the ends,
and
(b) the voltage
M and N, of the wire?
Example
6.14.
6.1 3.
Complex wire assembly inside a solenoid. Let
the wire loop composed of two semicircular parts
with conductivities a\ and 02 from
Example
6.6
6.5,
a rectangular wire loop of edge
have radius a/
and be placed
coaxially inside an air-filled solenoid, and let
two additional linear pieces of wire, with con<73 and 04, be attached to it, at points
M and N, as depicted in Fig. 6.31. Linear wire
segments are positioned radially with respect
to the solenoid axis, and their ends (points
ductivities
Rectangular wire loop around a solenoid. Consider the solenoid described in
and assume that
solenoid; for
Problem
straight metallic wire
point of space.
nonuniform cross
section around the
every point
v.
Voltage along a straight wire
field.
in
6.10(a) but with
303
Problems
P and Q)
are very close to one another (the
gap between them
Compute
is
much
smaller than
the voltage between points
a).
P and Q.
Figure 6.31 Wire
assembly of four parts
with different
conductivities placed
inside an air-filled
solenoid (the gap
between points
Q
is
P
and
very small); for
Problem 6.14.
6.15.
Emf in a rectangular loop due to a two- wire line
current.
A very long lossless thin two-wire line
in air, with distance
equal to 4a,
is
between axes of conductors
generator of low-frequency time -harmonic current intensity ig (t) — 7go cos cot, while the other
rectanguend of the line is short-circuited.
A
lar
(b)
(a)
fed at one end by an ideal current
wire loop of side lengths a and b
in the plane of the line,
such that
its
is
Figure 6.32 Electromagnetic induction
in a
rectangular
loop due to a time-varying current of a thin two-wire
(a)
loop between
of the line; for
line
line:
conductors and (b) loop on a side
Problem
6.1 5.
placed
two sides
are parallel to the line and the distance of one
of the sides from one of the line conductors
a.
Neglecting end and propagation
effects,
Figure 6.33
is
Magnetically coupled
i.e.,
computing the magnetic field of the line as if
it were infinitely long and assuming that the
line current
is
the
same
small circular
concentric coplanar
in every cross section,
loops
free
wound uniformly and densely along
respectively.
core,
Large square and small circular concentric
coplanar loops. Fig. 6.33 shows two concentric
current
space; for
Problem 6.16.
/o
wire loops lying in the same plane, in free
space, a large square loop of side length a
and
a small circular one of radius b (b
The
a).
wise, direction.
The square loop
is
R.
Determine the induced current
circular loop, neglecting
its
in the
own magnetic field.
Electromagnetic induction in a nonlinear mag-
of
i(t) = Iq sin cot,
where
= 10 6 rad/s. The secondary
N2 = 4 wire turns encircling the
intensity
and
has only
co
H
carries a low-
= Iq sin cot, and the resistance of the circular
loop
the entire
fed by a low-frequency time-harmonic
= 0.1 A
coil
frequency time-harmonic current of intensity
i(t)
is
primary winding, and is open-circuited. The
idealized hysteresis loop of the core material is
that in Fig. 6.13(c), with B m = 0.5 T and
m —
170 A/m. Sketch roughly the voltage waveform across the secondary coil terminals within
one period of time-harmonic variation of the
current in the primary circuit (T = 2 n/co), that
is, for 0 < cot < 2tz
loops are oriented in the same, counterclock-
6.17.
in
magnetic field due to induced
current, find the emf induced in the loop for situations in (a) Fig. 6.32(a) and (b) Fig. 6.32(b),
as well as the
6.16.
and
large square
.
A
the form of a thin toroidal ferromagnetic core
Rotating rod in a uniform magnetic field.
conducting rod rotates uniformly with angular velocity w about an axis that splits it onto
with two windings, like the one in Fig. 5.19.
two unequal parts of lengths
The length and
area of the
in a
respectively.
flux density B, as
netic circuit. Consider a magnetic circuit in
core are
/
=
The primary
50
cross-sectional
cm and
coil,
5=1
with N\
=
cm 2
,
850 turns of wire
6.18.
l\ and I2
( l\ / h)
uniform time-invariant magnetic field of
of rotation
is
shown
in Fig. 6.34.
The
axis
perpendicular to the rod, and
F
304
Chapter 6
vector
total
emf
B
is
Slowly Time-Varying Electromagnetic Field
parallel to the axis, (a) Find the
induced emf
if
(b)
/i
=
What
in the rod.
(c)
I2 ,
I2
—
and
0,
is
linearly sliding bar
Consider the system with
a metallic bar moving in a uniform static
magnetic field described in Example 6.11
l\
—
magnetic
in a
the total
(d)
motor - with a
6.20. Electric
0,
respectively?
and assume that an ideal voltage
(Fig. 6.15),
B
generator of time-constant emf £ is added in
series with the resistor (of resistance R), as
i
shown
h
h
field.
R
parameters £,
Figure 6.34 Conducting rod
itive
uniformly rotating about an
from Example
excentric axis in a uniform static
magnetic
6.19.
field; for
Problem
sis
6.1 8.
(b)
the disk surface.
is
A
B
,
with
B
defined by taking leads off
field
1
to
(a) the total
induced emf
=
as follows, (a) For the
0), find the current in the cir-
and mechanical force acting on the bar.
For the bar sliding (uniformly), sketch the
I
Fme ch)- (d) What are the ranges of
Fm ech in which the system in Fig. 6.36
function of
induced
values of
operates as a motor (Fmec h < 0 - the main
mode of operation of the system) and as a gen-
in the disk
Coulomb
and perform the analy-
on the bar (Fme ch)> for —00 < Fme ch < 00,
and sketch this dependence (note that v is a
with respect to the reference direction from the
disk center to the rim, (b) the
6.11),
force
currents in the disk and thickness of the rod,
compute
given and pos-
of the algebraic intensity of the mechanical
and 2,
the rim and the
due
all
on the velocity of the bar (v),
and negative (movement
away from and toward the voltage generator),
respectively, (c) Determine the mechanical
power of the bar movement (P m ech) in terms
center of the disk (via the rod), respectively.
Neglecting the magnetic
B be
for v both positive
perpendicular to
pair of terminals,
,
new system
of this
dependence of
w
about its axis together with an attached axial
copper rod, between the poles of a large permanent magnet producing a uniform magnetic
field of flux density
a and
cuit (7)
6.35
a that rotates at a constant angular velocity
,
(disregard the concrete numerical values
bar at rest (v
shows Faraday’s
wheel, consisting of a copper disk of radius
Faraday’s wheel. Fig.
Let the values of the system
in Fig. 6.36.
\
erator (Pmec h
electric
vector (due to excess charge) at
an arbitrary point of the disk, and (c) the voltage (V = V 12 ) across the open terminals of the
field intensity
>
0 ), respectively? (e)
Compute
Fmech for
which the mechanical power of the
motor
m ech
(|
I
) is
maximum,
operation of the motor,
its
(f) If
for the safe
current has to be
smaller in magnitude than Imax (so that the
motor does not burn out), what is the corre-
wheel.
(g) Establish and
power balance for the system, in
both motor and generator mode of operation.
sponding range of velocity v?
discuss the
£
Figure 6.36 Electric motor - with a linearly
magnetic field
dc voltage generator
for Problem 6.20.
sliding metallic bar in a static
(as in Fig. 6.1 5),
added
in
the
and
circuit;
a
Figure 6.35 Faraday's wheel (a copper
between the
permanent magnet),
with open terminals; for Problem 6.19.
disk uniformly rotating
poles of a large
6.21.
Computation
For the
problem,
for a sliding-bar electric motor,
electric
let
7
motor from the previous
V, R = 2 £2, a = 1 m,
=5
305
Problems
'f’mech
the
= 2.5
power
N, and v
=
10 m/s. Find
B and
R
that the voltage generator {£,
intensity
motor - with a rotating bar
netic field.
A metallic bar of length a
one end to a
in a
is
that
mag-
attached
vertical metallic rod,
the
R
rotates with
w
t
=
shown in Fig. 6.38(a). At
and conductor lie in
it,
as
0,
the loop
same plane (plane of drawing). Neglecting
the magnetic field due to induced currents in
emf
the loop, determine (a) the induced
in
the loop and (b) the instantaneous mechani-
A
T mec h
is
an instant
axis, it
reference direction of vector
rectangular wire loop of edge
about its axis
parallel to the conductor and at a dis-
tance c from
about
can rotate so that its other
end slides without friction along a circular horizontal metallic rail (of radius a ), as portrayed in
Fig. 6.37.
voltage generator of time-invariant
emf £ and internal resistance R is connected
between the rail and the rod (axis), and the
whole system is situated in a uniform static
magnetic field, whose field lines are perpendicular to the plane of the rail and flux density is B.
The algebraic intensity of an externally applied
mechanical torque on the rail is Tme c h, for the
at its
which, as an
A
a constant angular velocity
delivers to the rest of the structure.
6.22. Electric
I.
lengths a and b and resistance
)
power of loop rotation. [To compute the
magnetic flux through the loop, adopt the intecal
gration surface consisting of a cylindrical part
of radius
r\
and a
flat
part of width
indicated in Fig. 6.38(b), where
—
and
r\, as
can
be found using the cosine rule. See also the flux
computation in Fig. 7.9(c).]
r\
r2
in Fig- 6.37.
The losses in the bar, rod, and rail, and the magnetic field due to the current in the circuit (/)
can be neglected, (a) What are I and rmec h for
the bar at rest? (b) If the bar rotates uniformly,
find its angular velocity (w). (c) Sketch the
dependence of the mechanical power of the bar
rotation (P me ch) on Tmec h, for -oo < Tmec h <
oo, and mark the ranges of system operation
as a
motor and
Compute
all
as a generator, respectively, (d)
relevant powers in the system, and
discuss the overall
power balance.
(b)
(a)
Figure 6.38 Rectangular loop rotating
due
in
the magnetic
field
to an infinitely long wire with a steady current: (a) side
view at an instant t = 0 and (b) top (cross-sectional) view
at an arbitrary instant t, with a suggested integration surface
(in
two
6.24.
Figure 6.37 Electric motor -with a uniformly
forming an electric circuit with
dc voltage generator; for Problem 6.22.
field,
6.23.
Rotating loop near an
An
infinitely
uated in
infinite
dc line current.
long straight wire conductor
air carries a time-invariant
bounded by the
loop; for Problem 6.23.
System for measurement of fluid velocity
with an ideal voltmeter. If in the system
for measurement of fluid velocity based on
motional electromagnetic induction described
in Example 6.14 the voltmeter is ideal, i.e.,
the intensity of the current flowing through
its terminals is so low that it can be assumed
to be zero, express (a) the electric field due to
excess charge in the region between the capacitor plates and (b) the voltage indicated by the
voltmeter in terms of the fluid velocity, v, and
other parameters of the system.
rotating metallic bar in a static magnetic
a
parts)
sit-
current of
6.25.
Thevenin generator for fluid flow in a magnetic field. Consider the Thevenin equivalent generator for the system with motional
306
Chapter 6
Slowly Time-Varying Electromagnetic Field
electromagnetic induction from Example 6.14 that replaces the rest of the system with respect
to the voltmeter in Fig. 6.18, as indicated in
Fig.
6.39.
nal
resistance
Show
(a)
of
that the
emf and intercomputed
generator,
this
as the open-circuit voltage of the circuit (in
Fig. 6.18)
it
represents and input resistance of
the circuit (with
the excitations shut down),
all
= vBd and Rj =
equal £j
d/(crS), respectively,
Obtain the expression for the velocity of
the fluid in Eq. (6.108) using the generator
from (a).
(b)
Figure 6.40 System for measurement of conducting
Rj
1
+
V
£t
Figure 6.39
Thevenin equivalent
M*i-
inside a cylindrical capacitor with steady current; for
Problem 6.26.
generator at
N
terminals
(M and N)
nonuniform, given by v(x) = vo[l — (2x/d) 2 ],
—d / 2 < x < d/ 2, where vo is a constant, as well
of the voltmeter in
Fig. 6.1 8; for
-/v
as that a variable resistor (rheostat)
Problem 6.25.
nected to the capacitor plates
6.26. Fluid flow
dc current.
fluid
flow velocity using motional electromagnetic induction
is
con-
in place of the
voltmeter in Fig. 6.18, as shown in Fig. 6.41. (a)
Find the parameters of the Thevenin equiv-
through a cylindrical capacitor with
A nonmagnetic liquid of conductiva very
alent generator - to replace the rest of the
long coaxial cable of conductor radii a and
structure with respect to the rheostat, as in
ity
a flows between the conductors of
b (a
<
depicted in Fig. 6.40. Both the
b), as
Fig. 6.39. (b) If the resistance of the rheostat
R
inner conductor and the inner surface of the
,
what
is
the voltage across
is
it?
outer conductor of the cable are insulated by
At one end,
a thin layer of perfect dielectric.
the cable
is
fed by an ideal current genera-
tor of time-constant current intensity / while
g
,
the other end
is
short-circuited.
cal capacitor of length
cable such that
its
is
/
A
cylindri-
placed inside the
electrodes (thin cylindrical
and
plates), with radii a
6, are tightly
pressed
against the respective (insulated) cable conductors.
A
voltmeter,
whose
internal resistance,
Figure 6.41 Structure in Fig. 6.18 but
with a nonuniform fluid velocity,
including the resistance of the interconnecting
conductors,
tor.
The
uniform,
is
/?y,
is
connected to the capaci-
is
v.
intensity vector, (c) the current density vec-
and (d) the Coulomb
electric field intensity
vector (due to excess charge) -
between the electrodes of the
itor, as
in
the region
cylindrical capac-
well as (e) the voltage indicated
by the
voltmeter.
6.27.
and
a variable resistor
to the capacitor plates; for
connected
Problem 6.27.
Calculate (a) the magnetic flux
density vector, (b) the induced electric field
tor,
v(jc),
velocity of the fluid, considered to be
Nonunif'orm fluid flow and motional induction.
Consider the structure from Example 6.14,
and assume that the velocity of the fluid is
i.
6.28.
Measurement of
ity.
fluid velocity
For the structure described
problem,
let
S -
0.5
m2
,
d
=
and conductivin the
previous
10 cm, and
B =
When the resistance of the rheostat is
set to R = Ro — 40 mfi, its voltage is V = Vq =
30 mV, whereas V = 2Vo if R = 4 Rq. Based
0.1 T.
on these data, compute the central velocity
(for x — 0), vo, and conductivity, a, of the fluid,
using the Thevenin equivalent generator (from
the previous problem).
m
307
Problems
6.29.
Moving contour
in
area
A
rectangular contour of side
magnetic field.
lengths a and b moves along the x-axis with
a constant velocity v in a time-harmonic magnetic field of angular frequency co, which can
be considered to be low, and flux density
B(x, t) = Bq cos kx cos cot, where Bo and k are
constants, and vector B is normal to the plane
of the contour, as
shown
At
in Fig. 6.42.
t
=
0,
the center of the contour coincides with the
coordinate origin (x
= 0). Find the emf induced
in the contour.
What
0
b
—
®B(x,f)
5=1
length / = 20 cm, and
two windings each having N = 100 wire turns
and resistance R = 10 £2. The first winding is
connected via a switch K to a dc voltage generator of emf £ = 9 V and internal resistance
R\ =20 Q. The second winding is terminated
in a ballistic galvanometer, whose resistance
= 5 £2. The switch K is first open, and
is
there is no residual magnetization in the core.
The switch is then closed, and the charge flow
indicated by the galvanometer is Q = 600 /zC.
a nonuniform dynamic
is
cm 2 mean
,
the relative permeability of the core?
V
I
X
a
Figure 6.42 Rectangular
contour moving in a nonuniform
low-frequency time-harmonic
magnetic
6.30.
Rotating loop near an
Problem 6.29.
infinite ac line current.
Repeat Problem 6.23 but
for a low-frequency
time-harmonic
of
Iq
cos
cot
current
intensity
i(t)
Fig. 6.38(a),
where
Small loop in the magnetic
ing large loop.
Assume
=
co
time-constant, of intensity
I,
w.
mary and secondary
pair of
axis of
its sides.
symmetry
At an
6.34.
Problem 6.16 is
and that this loop
that
instant
t
is
=
gap of a mag-
in the
from Example
assume that the
5.14,
cross section of the core in Fig. 5.30(a)
parallel to a
0, it is in
Thin conducting disk
circuit
co
circle of radius a (S
the
completely
same plane with the small circular loop (as in
The magnetic field due to induced
radius
a,
filled
=
7ra
2
),
that the air
is
a
gap
is
with a thin conducting disk of
thickness
Iq,
meability /zo, like the
conductivity a, and per-
one
in Fig. 6.24,
N
and
that
currents in the circular loop can be neglected.
the current in the coil (with
Find (a) the induced emf in the circular loop
and (b) the instantaneous and time-average
low-frequency time-harmonic, with intensity
torque of magnetic forces acting on
meability of the ferromagnetic portion of the
Two
field of the rotating large
the
(oq
i(t)
it.
rotating loops. If in the previous prob-
same
( coo
<
circuit are
determine
/
The length and
and
/z r ,
is
relative per-
respectively. Neglecting
the magnetic field due to eddy currents in the
loop also rotates in
disk and core, find Iq such that the time-average
power of Joule’s losses dissipated in the disk is
in the
direction with an angular velocity
<w),
= Iq cos cot.
turns of wire)
magnetic
lem the small loop situated
(a) the
maximum, and
time-average
power of Joule’s (ohmic) losses in the small
loop and (b) the time-average mechanical
power of its rotation.
6.33.
circuits,
netic circuit. For the simple linear magnetic
Fig. 6.33).
6.32.
coils, and a
galvanometer (BG) in the prirespectively; for Problem 6.33.
ballistic
of a rotat-
field
uniformly rotates with an angular velocity
its
Figure 6.43 Linear ferromagnetic core with two
voltage generator and
that the current in
the large square loop from
about
=
flowing through the infinitely long wire
conductor in
6.31.
field; for
6.35.
find that
maximum
Eddy currents in Faraday’s wheel,
the time-average
power of
power.
(a)
Compute
Joule’s losses
due
in
Faraday’s wheel
described in Problem 6.19. (b)
What is the mag-
to induced (eddy) currents
Charge flow through the secondary coil on
a magnetic core. Fig. 6.43 shows a thin lin-
netic field that these currents
ear ferromagnetic core with cross-sectional
center of the wheel?
produce
at the
.
308
6.36.
Chapter 6
Eddy
der.
Slowly Time-Varying Electromagnetic Field
currents in an infinite conducting cylin-
An infinitely long solenoid with N'
wire per unit of
length
its
is
with a rectangular cross section, placed coax-
around the solenoid and centrally with
its length. The inner and outer radii
of the toroid are b and c (a < b < c), and its
turns of
wound about a
ially
con-
respect to
ducting ferromagnetic (infinitely long) cylinder
of radius
permeability
a,
a The winding
and conductivity
/x,
height
harmonic current of intensity i(t) = Iq sinaV,
and the medium outside the solenoid is air.
Determine the time-average per-unit(a)
length power of ohmic losses due to eddy
currents induced in the cylinder, neglecting the
magnetic field they produce, and then (b) find
this magnetic field at the axis of the cylinder
(use the procedure from Example 4.14).
6.37.
Hollow disk
field.
A
a,
intensity
<
and permeability
/xo is
i(t)
mately melts
=
Iq
it)
power of Joule’s
b), conductivity
field,
TV turns of wire
cos
cot.
The channel
is
como
metal, namely, induction heating, that
and
ulti-
(b) the total time-average
losses dissipated to heat in the
metal.
situated in a uniform
slowly time-varying magnetic
The solenoid has
and permeability /xo- Neglecting the end effects
(i.e., computing the induced electric field of a
solenoid as if it were infinitely long) and the
magnetic field due to eddy currents, find (a)
the distribution of eddy currents in the channel
(these currents produce local heating in the
thin hollow conducting disk of radii
a and b, thickness 8 (8 <£ a
h.
pletely filled with a metal of conductivity
a triangular-pulse magnetic
in
is
with a low-frequency time-harmonic current of
carries a low-frequency time-
such that
the field lines are perpendicular to the disk, as
shown
H{t),
is
in Fig. 6.44.
The
intensity of this field,
a periodic alternating triangular-pulse
time function of amplitude
sketched
in Fig. 6.13(d).
netic field
Hm
and period T,
Neglecting the mag-
produced by eddy currents,
find (a)
the distribution of these currents throughout
the disk and (b) the total instantaneous and
time-average powers of Joule’s losses asso-
and then (c) compute the
due to eddy currents at the disk
ciated with them,
magnetic
field
center (point O).
Figure 6.45 Induction furnace: eddy currents
accompanying the induced
electric field of a very
long
solenoid with a low-frequency time-harmonic current
in the winding heat and melt metal in a toroidal
channel of rectangular cross section placed around
the solenoid; for Problem 6.38.
6.39.
Eddy
currents in a thin conducting spherical
shell.
Consider the solenoid described in
6.22, and assume that a thin conduct-
Example
Figure 6.44 Thin hollow conducting
ing spherical shell of radius b (b
disk in a uniform slowly time-varying
magnetic
is
field,
sketched
whose
<$(<$«; b ), conductivity
intensity, H(t),
in Fig. 6.1 3(d); for
is
Problem 6.37.
placed inside
shows an induc-
solenoid, with circular cross section of radius
that they produce.
nel
(/
» a),
and a toroidal chan-
(carrying a metal piece that
is
heated),
thickness
fiQ
such that the sphere center
on the solenoid
to
/
a),
axis.
average power of Joule’s losses
tion furnace consisting of a very long air-filled
a and length
<
and permeability
Determine the timein the shell, due
eddy currents, neglecting the magnetic field
lies
6.38. Induction furnace. Fig. 6.45
it,
or,
6.40.
Loss power in a laminated ferromagnetic core.
laminated conducting ferromagnetic core in
A
309
Problems
the form of a packet of
N insulated stacked thin
d (d
a),
and conductivity a is placed in a uniform slowly
time-varying magnetic field, such that the field
plates of length b, width a, thickness
lines
are perpendicular to the packet cross
shown
section, as
sity
Figure 6.46
Continuously
of this
in Fig. 6.25(c).
field, B(t), is
The
inhomogeneous
flux den-
conducting
situated
a periodic alternating
triangular-pulse time function of amplitude
Bm
and period
T, like the function
Fig. 6.13(d).
End effects and magnetic field due
in
strip
two
mutually orthogonal
time-harmonic
sketched in
magnetic
fields of
to
equal amplitudes
and 90° out of
total
phase; for
eddy currents can be neglected, (a) Find the
time-average power of Joule’s losses in
the core, (b) How large has to be N for a given
Problem 6.42.
total thickness of the packet, c, so that the total
loss
power does not exceed
a given value
field,
PI
of flux density
one of the
on a rotating strip in a magnetic
For the conducting strip rotating in a
uniform time-constant magnetic field from
Example 6.24, find the instantaneous torque of
magnetic forces on the strip - (a) by integrating
torques of magnetic forces on elementary strips
(with eddy currents) of width dx in Fig. 6.26(b)
(see Example 4.22) and (b) from energy con-
6 . 41 . Torque
What
is
Example
(Fig. 6.47).
At
t
= 0,
Neglecting the end effects and the magnetic
field.
servation (see
B
strips is parallel to the field lines.
field
due to eddy currents
in the conductor,
induced electric field vector at an
arbitrary point of each of the two strips, (b) the
instantaneous power of Joule’s losses in each of
the strips, (c) the total instantaneous torque of
find (a) the
magnetic forces on the conductor, and (d) the
time-average torque on the conductor.
6.12), respectively, (c)
the time-average torque on the strip?
6 . 42 . Inhomogeneous strip in two orthogonal mag-
shows a very long, thin
nonmagnetic conducting strip of length /, width
a, and thickness 8 (8 <$; a
/) lying in the
section of a
xy-plane of a Cartesian coordinate system.
conductor
The conductivity
of the strip varies with the
composed
x coordinate, and
is
netic fields. Fig. 6.46
-a/2
stant.
<x< a/2,
The
given by a(x)
—
Aoqx2 /a 1
Figure 6.47 Cross
strip
two
rotating in a uniform
where ao is a positive conis exposed to two uniform
low-frequency time-harmonic magnetic
of
crossed strips
,
time-invariant
magnetic field; for
Problem 6.43.
fields
= Bocoscvtx and
82(0 = Bosincatz, respectively. Neglecting the
end effects and the magnetic field due to eddy
currents, compute (a) the time-average power
of Joule’s losses dissipated in the strip and (b)
the time-average torque of magnetic forces
acting on the strip.
of flux density vectors Bi(t)
6 43 .
.
Eddy
currents in two crossed rotating strips.
Fig. 6.47
shows a cross section of
conductor, of length
meability
/a 0
,
/,
a very long
conductivity a, and per-
consisting of
two crossed
right angle) thin strips, each of width a
thickness
8
(8
<$C
a «;
formly rotates in
angular velocity
co,
air
in
(at a
and
/). The conductor uniabout its axis, with an
a uniform magnetostatic
6.44.
Continuously inhomogeneous rotating cylinder. Repeat Example 6.25 but for a continuously inhomogeneous cylinder whose conductivity is given by the following function of the
radial coordinate r [Fig. 6.27(b)]: a(r) = a^r/a,
0
6.45.
<
r
<
a,
where ao
is
a positive constant.
Eddy currents in a rotating cylindrical shell. An
infinitely
long thin conducting cylindrical shell
of radius a, thickness 8 (8 <£ a), conductivity
a, and permeability
juo is
formly about
by an externally applied
its
axis
rotated in air uni-
mechanical torque, T[nech per unit length of
the shell and against the torque of forces of a
uniform time-invariant magnetic field of flux
,
310
Chapter 6
Slowly Time-Varying Electromagnetic Field
density 5, in which the shell resides, as
in Fig. 6.48.
to
Neglecting the magnetic
eddy currents
in the shell, find (a) the
lar velocity of rotation of the shell (a>)
the per-unit-length mechanical
Figure 6.48
Rotating cylindrical
conducting
in a
shell
uniform
time-invariant
magnetic field; for
Problem 6.45.
rotate the shell (P'
mech ).
shown
field
due
angu-
and (b)
power used
to
7
W
Inductance and Magnetic
Energy
\
Introduction:
n
this chapter,
we
An
introduce and study the con-
equally important concept of energy in
I cepts of self- and mutual inductance. In general,
magnetic
inductance can be interpreted as a measure of trans-
just as configurations of
former electromagnetic induction in a system of
conducting contours (circuits) with slowly timevarying currents in a linear magnetic medium.
Briefly, self-inductance is a measure of the magnetic
flux and induced emf in a single isolated contour (or
in one of the contours in a system) due to its own
current. Similarly, a current in one contour causes
magnetic flux through another contour and induced
emf in it, and mutual inductance is used to characterize this coupling between the contours. Some
conductor configurations, called inductors, are specially designed to have a desired (large) inductance
(they can have many turns of wire and can be loaded
with magnetic cores). Along with the resistor and
capacitor, the inductor represents another fundamental element in circuit theory and a basic building
block for ac electric circuits. The inductance of an
tric
inductor
is
dual to the capacitance of a capacitor.
fields is also discussed.
We
shall see that,
charged bodies store elec-
energy, configurations of current-carrying con-
ductors store magnetic energy. Using circuit-theory
terminology, inductors (magnetically coupled or
uncoupled) in a circuit contain magnetic energy, in
a manner analogous to capacitors as electric energy
“containers.” We shall introduce magnetic energy
density as well. In the case of conductors in the presence of linear magnetic materials, magnetic energy
will be related to self- and mutual inductances of the
conductors. For systems that contain ferromagnetic
materials with pronounced nonlinearity and hysteresis behavior,
on the other
side, special attention
be paid to establishing a clear physical understanding and precise mathematical characterization
of the energy balance in the system, including
will
so-called hysteresis losses in the material.
The material of this chapter represents a culmination of our investigations of steady and slowly
311
312
Chapter 7
Inductance and Magnetic Energy
time-varying magnetic fields and electromagnetic
On the other hand, it par-
conductors and magnetic energy density in the
field will bediscussed next. Finally, the concept of
to a large extent, the electrostatic analysis of
self-inductance will be revisited from the energy
induction (Chapters 4-6).
allels,
previous work will be referenced and used here, in
Examples will include inductance and
energy computation for a large variety of theoretically and practically important electromagnetic
structures with slowly time-varying and time-
both the theoretical narrative and examples.
invariant currents and fields, ranging from various
capacitors and other systems of charged conduct-
standpoint.
ing bodies, including electric energy considerations
(Chapter
We
2).
Therefore, a substantial portion of the
and mutual inductance in two separate sections, and then analyze magnetically coupled circuits based on both
concepts. Magnetic energy of current-carrying
contours and coils with cores of different shapes
and material compositions to several types of trans-
shall first study self-
mission lines and circuits with magnetically coupled
inductors.
SELF-INDUCTANCE
7.1
Consider a stationary conducting wire contour (loop), C, in a
linear,
homogeneous
or inhomogeneous, magnetic medium, and assume that a slowly time-varying current of intensity
is
i
shown
established in the contour, as
in Fig. 7.1. This current
produces a magnetic field whose flux density vector, B, at any point of space and
any instant of time is linearly proportional to i. The magnetic field, as well as an
induced electric field (see Section 6.1), of intensity Ej nci, exist both around the contour and inside the wire itself. The magnetic flux through a surface S bounded by
C is given by Eq. (4.95) and is also linearly proportional to i. The proportionality
1
Figure 7.1 Current contour
(loop) in a linear magnetic
medium.
constant,
self-inductance (unit:
H)
(7.1)
is
termed the self-inductance or
inductance of the contour.
just
More
precisely, the
inductance defined by Eq. (7.1) is the so-called external inductance, since it takes
into account only the flux O of the magnetic field that exists outside the conductor of the loop.
There
inside the conductor.
is
We
also an internal inductance of the loop,
due to the
flux
shall introduce the concept of internal inductance in terms
of the magnetic energy stored inside the conductor in a later section. Because the
surrounding
medium
magnetically linear,
is
L depends
only on the
medium perme-
ability and on the shape and dimensions of the contour, and not on the current
intensity
tions of
It
i.
C
.
always positive, provided, of course, that the reference orienta-
is
and S
(i.e.,
the reference directions of
i
and O) are interconnected by
the right-hand rule, as in Fig. 7.1.
The induced emf along
portional to di/dt.
It is
the loop
electromagnetic induction, and
emf due
to self-induction
C
is
given by Eq. (6.33) and is linearly proO by means of Faraday’s law of
we can
^ind
—
write
|
dO
(7.2)
dt
This
emf
is
the circuit
1
referred to as the
(it
Since the current
along the contour.
is
is
i
interrelated with the flux
emf due
to self-induction (or self-induced
caused by the magnetic
slowly time-varying,
its
intensity
field
(i) is
due
emf)
in
to the current in the circuit
only a function of time and does not change
Section 7.1
313
Self-Inductance
and not due to currents in other circuits). The unit for the self-inductance
henry (H). From Eqs. (7.1) and (7.2), which represent two equivalent definitions of L, H = Wb/A = V- s/A. One henry is a very large unit. Typical values of
self-inductances in practice are on the order of mH, ,u.H, and nH.
itself,
is
HISTORICAL ASIDE
Joseph
1878), an
sicist
Henry (1797American phy-
and
one
of
greatest scientists
the
and
in-
ventors in the area of
electricity
magne-
and
tism ever, was a professor
of natural philosophy at
Princeton and the
first
secretary of the Smith-
sonian Institution.
From
1819 to 1822, he attended
the Albany Academy,
where he was appointed professor of mathematics and natural philosophy in 1826. Although
his teaching duties were extremely heavy, Henry
soon became the foremost American scientist of
his time.
He
discovered electromagnetic induc-
tion independently of Faraday
fact,
Henry had performed
(1791-1867). In
the key experiments
ahead of
that led to the discovery of induction
Faraday, in 1830, but Faraday published his dis-
covery
first,
in 1831.
On
the other side,
Henry
is
fully credited for the discovery of self-induction.
It
was
his idea to
wind many layers of insulated
wire on an iron core and thus obtain electromagnets of unmatched power. In a demonstration at
Yale University in 1831, his electromagnet lifted
more than a ton of iron (previous electromagnets
were capable of lifting only a couple of kilograms).
Experimenting with electromagnets, he observed
a large spark that was generated whenever the
circuit was broken, and thus discovered selfinduction (in 1831).
He
realized that, in general,
a time-varying current in a circuit not only induces
electromotive force in another circuit (mutual
induction), but also in itself (self-induction).
He
also defined the associated property of a circuit,
its
self-inductance,
as
a
measure of
to “self-induce” electromotive force.
its
ability
He found
that “coiling” of the wire greatly enhances the
self-inductance of the circuit.
The same
year, he
demonstrated his electric motor based on a continuous oscillating motion of a straight electromagnet
with two coils at its ends. The electromagnet
rocked back and forth on a horizontal axis with
its ends being alternately attracted and repelled by
two vertical permanent magnets and its polarity
being reversed automatically in the same rhythm
by alternately connecting the coils to electrochemical cells (sources). Although still an experimental
laboratory device, Henry’s motor was much closer
to a mechanically useful practical machine than
Faraday’s 1821 motor. It was just one year later, in
1832, that William Sturgeon (1783-1850) invented
the commutator and the first motor with continuous rotary motion, which was a rotary analogue of
Henry’s oscillating motor and essentially a “prototype” of our modern dc motors. In another
stunning demonstration to his students at the
Albany Academy, in 1831, Henry strung a mile
of wire all around the inside of the lecture hall to
connect an electromagnet to a battery. The magnet
was placed close to one end of a pivot mounted
steel bar, whose other end, in turn, was next to
a bell. After the circuit was closed and a current
“sent” from the battery to the coil of the electromagnet, a steel bar swung toward the magnet,
striking the bell on the other end. Breaking the
made
connection, next,
the electromagnet lose
force and release the bar, which
to strike again. This
was then
was the world’s
its
free
first electri-
cal relay (electromechanical switch). In addition,
by connecting and disconnecting the battery to
the circuit in a particular pattern, the steel bar,
made
to ring the same
and this is nothing
else but telegraphy. Although, obviously, Henry
invented the telegraph in 1831, it is Samuel Morse
a mile away, could be
series of signals
(1791-1872)
on the
who
is
bell,
credited for this invention
[Henry did not patent any of his devices, and actually helped Morse to put his telegraph model to
practical use and transmit the first telegraph message using the Morse code (invented in 1838) from
314
Chapter 7
Inductance and Magnetic Energy
Baltimore to Washington, D.C. on
In 1832,
May
24, 1844],
Henry became professor of natural
phi-
losophy (physics) at Princeton University (then
College of
New
Institution
was established
Jersey). After the Smithsonian
1846,
in
Henry was
second president (elected in 1868) of the National
Academy of Sciences, and he held both positions
until his death. In his honor, we use henry (H) as
the unit for inductance.
Brady-Handy Photograph
(Portrait: Library
of Congress,
Collection)
named its first secretary (director). He was also the
6
•
Figure 7.2 Circuit-theory
representation of an
inductor and equivalent
controlled voltage
generator.
Note that, while the emf definition of self-inductance in Eq. (7.2) does not make
any sense for steady currents, the flux definition in Eq. (7.1) can be used in practically the same way under both dynamic and static conditions. Namely, a dc current
I that is assumed to flow in the contour produces the flux O through the contour in
the same way a (slowly) time-varying current does, and the inductance obtained as
L = <t>/7 is equal to that obtained, for the same contour, using either Eq. (7.1) or
(7.2) and time-varying current. On the other hand, the flux O due to the current I is
time-invariant, and thus no emf is generated in the contour.
Every conducting loop or circuit has some inductance (self-inductance), usually
as an undesirable side effect, which can often be neglected. In practical applications, however, we frequently design and use conductors that are specially arranged
and shaped (such as a conducting wire wound as a coil), and sometimes loaded with
magnetic cores, to supply a certain (large) amount of inductance. Such a device, with
its inductance L as its basic property, is called an inductor. Just as a capacitor can
store electric energy, an inductor can store magnetic energy, as we shall see in a later
section. Resistors, capacitors, and inductors are basic circuit elements that are combined, together with voltage and current generators, to form arbitrary RLC circuits.
Fig. 7.2 shows the circuit-theory representation of an inductor. When a timeis
varying current of intensity i flows through the inductor terminals, the emf
induced in the inductor, given by Eq. (7.2), where L is the inductance of the inductor. With respect to its terminals, the inductor can now be replaced by an equivalent
ideal voltage generator
whose emf equals e m
<\
(this is a voltage
by the time derivative of a current), as indicated
emf
of this
is
the
same
in Fig. 7.2.
generator controlled
The reference
as the reference direction of the current
voltage v across the inductor terminals in Fig. 7.2
V
element law for an inductor
—
—
Cj n d
is
i
direction
(see Fig. 7.1).
The
hence
di
(7.3)
L,
dr'
This
is
the element law (current-voltage characteristic) for an inductor.
It tells
us
change of i in time, with L as the
proportionality constant. Its form is just opposite to the element law for a capacitor,
Eq. (3.45), and we say that an inductor and a capacitor are dual elements. These two
laws, the element law for a resistor (Ohm’s law), Eq. (3.72), Kirchhoff’s current and
voltage laws, Eqs. (3.42) and (1.92), and element laws for an ideal voltage generator,
Eq. (3.1 16), and an ideal current generator, Eq. (3.127), represent a full set of basic
that v
is
linearly proportional to the rate of
equations of circuit theory for analysis of linear ac circuits.
Note that the model in Fig. 7.2 actually represents an ideal inductor, which does
not include any parasitic effects, such as parasitic capacitances between the adjacent
turns in a coil and Joule’s losses (in wires and magnetic cores). These effects are
present to a greater or lesser extent in
many
practical situations.
The only
all
effect
real inductors, but can be neglected in
modeled by an
ideal inductor
is
the
emf
Section 7.1
due
On
to self-induction, e j nc j.
the other side,
it is
assumed
Self-Inductance
in circuit theory that
the magnetic flux, induced emf, and magnetic energy are concentrated only in the
inductors in a circuit, while the magnetic field and the associated flux, emf, and
energy due to the connecting conductors are assumed to be negligible. In practice,
there is always an induced emf distributed along conductors in the circuit (in the ac
regime), and it depends on the shape and dimensions of the conductors. However,
this emf,
i.e.,
the inductance of the conductors, can, again, be neglected in
practical applications.
We
zero resistance of the conductors have been
many
assumptions of zero capacitance and
recall that similar
made
while introducing circuit-theory
The
representations of a capacitor and a resistor in Figs. 2.15 and 3.9, respectively.
between
the elements in circuit layouts) as if they were all ideal short-circuits, where the
shape, dimensions, and material properties of the interconnects are assumed to be
completely irrelevant for the operation (and analysis and design) of the circuit.
circuit-theory model, therefore, deals with connecting conductors (lines
Finally, inductors filled
Fig. 6.13) are
with magnetically nonlinear materials
(permeability) of the material depend
H
whereas
is always proportional to
depends on the current intensity,
is
the coil of
on the applied magnetic
field intensity,
H,
Consequently, the inductance of the inductor
i.
L=
which
(e.g.,
nonlinear circuit elements. In such cases, the magnetic properties
L(i),
(7.4)
nonlinear inductor
analogous to the relations in Eqs. (2.114) and (3.74) for a nonlinear
capacitor and a nonlinear resistor, respectively.
Example
7.1
Inductance of a Very Long Air-Filled Solenoid
An air-filled solenoidal coil has N =
200 turns of wire. The length of the solenoid is / = 20 cm
and the surface area of its cross section is S = 4 cm 2 The solenoid can be considered as very
long, so that the end effects can be neglected. Under these circumstances, find the inductance
.
of the
coil.
Solution Let us assume a slowly time-varying current of intensity i in the coil, find the
magnetic flux through all of its turns, and use Eq. (7.1) for the inductance. By neglecting the
end effects, we also assume that the magnetic field produced by the current i is uniform inside
the entire solenoid (as for an infinitely long solenoid). This field is given by Eq. (6.48), which
leads to the following expressions for the magnetic flux through a surface spanned over a
single turn of the coil and the total flux through the coil:
—
^single turn
I^qHS
where the reference direction for
<t>
respect to the reference direction for
>
is
i.
<
t
>
=
adopted
^VOgjugie turn
in
=
7~
'»
(7-5)
accordance to the right-hand rule with
The inductance of the
coil
is
hence
(7.6)
l
i
that
is,
L=
100
/u,H for the
given numerical data.
cross section (provided that the solenoid
Example 7.2
Coil
It is
the
same
for
any shape of the solenoid
very long).
on an Inhomogeneous Thick Toroidal Core
A coil with N turns of wire
made
is
is
wound over
a thick toroidal core of rectangular cross section
of two ferromagnetic layers of permeabilities g\ and
/z 2
,
as portrayed in Fig. 7.3.
The
L -
very long solenoid
315
)
Chapter 7
Inductance and Magnetic Energy
c
b
dr
a
:
m2
N
Ml
Figure 7.3 Evaluation of the
inductance of a
ds
with two
ferromagnetic
Example
b
:
wound on
coil
a thick toroidal core
linear
h
:
layers; for
r
6
7.2.
inner and outer radii of the toroid are a and
surface between the layers
Solution
If
b (a
is
< b <
c, its
height
is
h,
and the radius of the boundary
Calculate the inductance of the
c).
a steady current of intensity /
assumed
is
to flow in the coil (Fig. 7.3), the
netic field intensity H(r) inside the core (in both ferromagnetic layers)
The magnetic
<1>
r
=N
flux
through the
c
B(r)dS
=N
Jr=a
r
coil is
is
mag-
given by Eq. (5.83).
then [see Fig. 7.3 and Eqs. (5.84) and (5.92)]
b
r
n\H(r) dS
/
coil.
+
M 2 #(r) dS =
Jb
Ja
N ih
—
—
2
c
/
2n
c\
b
/
(
mi
In
\
a
+ M2
M 2 In —
b
).
)
(7.7)
and the inductance of the
coil
L=
N2 h
— = -r—
I
Example
7.3
mi
2jt
External Inductance
b
(
<t>
\
p.u.l.
In
a
c
+ M2 In
of a Thin
-
(7.8)
b
Two-Wire Line
Find the external inductance per unit length of a thin symmetrical two-wire transmission
in air.
The conductor
Solution
upon
radii are
The two-wire
itself at
a and the distance between their axes
can be considered as an
line
both ends of the
line at infinity. Fig. 7.4
infinitely
is
d (d
line
» a).
long wire loop that closes
shows the cross section of the
line.
The
procedure of evaluation of the line inductance (external self-inductance) per unit length is
analogous (dual) to the procedure of evaluation of the capacitance per unit length of a thin
two-wire line in Example 2.15. We assume that a slowly time-varying current of intensity i
(or a steady current of intensity /)
is
established in the line (Fig. 7.4).
At
a point
plane containing the axes of conductors, the magnetic flux density vectors B) and
currents (of the
same magnitude and opposite
where x
is
B\
+
fi 2
Figure 7.4 Evaluation of the
external inductance per unit
length of a thin two-wire
transmission line
in air
(cross
section of the structure); for
Example
7.3.
to
is
(7.9)
--
the distance from the axis of the
in the
directions) in individual conductors of the line
are collinear, so that the resultant magnetic flux density
B=
M
B 2 due
first
conductor.
Section 7.1
As
the structure
is
infinitely long,
the magnetic flux through the
we
consider only a part of
surface of length
flat
that
l
is
Seif-Inductance
317
that is l long and compute
spanned between the line
it
is the right choice because B is perpendicular to the surand oriented as required by the right-hand rule with respect to the adopted reference
direction of the line current. This flux is given by
conductors (such integration surface
face)
<t>
=
B
Idx
=
1
Mo il
dx
d—a
unit
=
Mo U
In
n
2 7Z
,
In
n
a
d
-
(7.10)
a
dS
[see the integration in Eq. (2.140)].
meter) of the
line
comes out
Hence, the external inductance per unit length (per each
to be
<D
u — LP
(7.11)
where O'
is
the flux per unit length of the
As a numerical example, for d/a =
external inductance amounts to
Example 7.4
Consider an
L=
coaxial cable.
L' - thin two-wire
H/m)
line.
30, L'
=
1.36
\x
H/m.
If
the line
is
100
m long, its total
136 ^H.
External Inductance
air-filled
a
jt
il
p.u.l.
The
the inner radius of the outer conductor
of a Coaxial Cable
is a and
Obtain the expression for the external
radius of the inner conductor of the cable
is
b (b >
a).
inductance per unit length of the cable.
Let us assume a dc current of intensity I flowing through the line conductors, as
The associated magnetic flux density vector, B, is circular with respect to
Solution
shown
in Fig. 7.5.
the cable axis and
its
magnitude, B(r), in the dielectric
is
given by the second expression in
Eq. (4.61), where r is the distance from the axis and a < r < b. Computing the magnetic flux
through a flat surface of length / spanned between the conductors (Fig. 7.5) then yields the
following expression for the external inductance per unit length of the cable:
1
B{r)
ll
dS
=
1
b
f
W>/
TljM-r
where the dielectric can be any nonmagnetic material (m
voltage and capacitance computation in Example 2.10.
As
Mr=
,
=
.
Mo)-
Mo
^
=
0.6
mm
and b
=
3.62
mm
(dielectric
is
(7.12)
a-
The procedure
a numerical example, the inductance per unit length of an
having a
b
,
ln
polyethylene),
is
is
dual to the
RG-11 coaxial cable,
computed to be L' =
359 nH/m.
Comparing the expression for L! in Eq. (7.12) and the expression for the capacitance per
— 2tzsq/ In (b/a) [see Eq. (2.123)], we note that the following
same cable,
unit length of the
C
Figure 7.5 Evaluation of the
external inductance per unit
length of a coaxial cable;
for
Example
7.4.
L' - coaxial cable
line (unit:
.
318
Chapter 7
Inductance and Magnetic Energy
relationship exists between the two parameters:
duality of L'
and C'
L' C'
for air
—
(7.13)
eoMo-
dielectric
Comparing then
the expressions in Eqs. (7.11) and (2.141),
for a thin two-wire line in
same relationship
tivity
exists
matter of
a
we
fact, as
we
also note that the
shall
prove
same
is
true
in a later chapter, the
between L' and C' for any two-conductor transmission
For a transmission
dielectric.
As
air.
and permeability, L'C'
line
—
line with air
having homogeneous linear dielectric of arbitrary permit-
ep..
Using these
relations,
we can now very
easily obtain the
inductance per unit length of transmission lines (with homogeneous linear dielectrics) for
which we already have the capacitance per unit length (found in Chapter 2). For example,
from the expression in Eq. (2.146) for the capacitance per unit length of the transmission line
consisting of a thin wire conductor and a grounded conducting plane in Fig. 2.24(a), we can
directly write the expression for the external inductance per unit length of this line:
,
B
Example
_
MO
—
—
2n
2/?
£qMo
,
a
Superconducting Contour
7.5
(7.14)
In
C
in a
Magnetic
A superconducting planar contour of area S and inductance
netostatic field of flux density B.
The contour
perpendicular to the plane of the contour, as
intensity
flows along the contour.
I\
The
Solution
magnetic
total
flux
contour
inductance L
magnetostatic
Example
7.5.
in a
field; for
is
situated in a uniform
then turned such that
new
in
steady
the
mag-
B
is
steady current of
its
plane becomes
state.
first
state
is
= LIi+BS,
4>i
uniform
is
in the
through the contour
Superconducting contour of
L
positioned such that the vector
in Fig. 7.6. In this state, a
The contour
parallel to B. Find the current I2 in the
Figure 7.6
is first
Field
(7.15)
where the first term is the flux due to the current (/] ) in the contour (self-flux) and the second
term is the flux of the external magnetic field (B), i.e., the flux due to other currents (or
permanent magnets) in the system (mutual flux).
second
In the
state,
0 2 = LI2
since the mutual flux
The
total
is
(7.16)
,
S = 0).
through a superconducting contour cannot be changed [see
zero (B
magnetic
flux
Eq. (6.135)], which means that
0i
Combining Eqs.
(7.
1
5)— (7.
17),
Note
field,
(7.17)
.
the current in the contour in the second state
that the increment in current
amount
is
hence
= /t + ^.
/2
the contour exactly at the
= 02
between the two
(7-18)
states,
A / = BS/L,
is
induced
in
that completely compensates, via the associated magnetic
the change of the mutual flux (from
BS
to zero), maintaining thus a constant total flux
through the contour.
Problems
:
MATLAB
7.2
7.
1-7.9;
Conceptual Questions (on Companion Website): 7.1-7.10;
Exercises (on
Companion Website).
MUTUAL INDUCTANCE
Consider now two stationary conducting wire contours, C) and C2 in a linear
(homogeneous or inhomogeneous) magnetic medium, as shown in Fig. 7.7. Let
the first contour carry a slowly time-varying current of intensity /]. As a result, a
,
Section 7.2
319
Mutual Inductance
magnetic field, of flux density Bi, is produced everywhere, and Bi, which is a function of both the spatial coordinates and time, is linearly proportional to i\. Some of
the lines of Bi pass through the second contour, i.e., through a surface S2 bounded
by C2 These lines constitute the magnetic flux through the second contour due to
the current q, which can be expressed as
.
02 = (
Bi -dS 2
(7.19)
.
Js2
The
flux
02
is
linearly proportional to
q
02
where the proportionality constant L 21
two contours and is obtained as
as well:
=
^21*1,
is
called the
(7-20)
mutual inductance between the
(7.21)
mutual inductance
(unit:
H)
M
is also used to denote mutual inductance. It is a measure
between the contours. Its magnitude depends on the shape,
size, and mutual position of the contours, and on the magnetic properties (permeability) of the medium. The mutual inductance can be both positive and negative,
depending on the adopted reference orientation of each of the contours for their
given mutual position. Namely, if a positive current q in the contour C 1 gives rise
to a positive magnetic flux 02 for the orientation of the surface S 2 that is in accor-
Note
that the
symbol
of magnetic coupling
dance to the right-hand rule with respect to the orientation of the contour C2 the
mutual inductance is positive. Otherwise, it is negative. 2 The mutual inductance is
,
also expressed in henrys.
By
applying Faraday’s law of electromagnetic induction, Eq. (6.34), to the flux
we obtain the induced emf (due to transformer induction) in the
Eq. (7.20),
contour C2
in
:
^ind2
d0 2
—
(7.22)
dt
Eqs. (7.21) and (7.22) should be regarded as equivalent definitions of mutual
They indicate that the mutual inductance between two magnetically
coupled contours (circuits) can be evaluated by assuming a current (q ) to flow in
one contour (primary circuit) and computing or measuring the magnetic flux ( 02 )
or the induced emf (ej n d 2 ) in the other contour (secondary circuit). Again, note
inductance.
that,
while the flux definition of
L 21
in
Eq. (7.21) can be used for both time-varying
Figure 7.7 Two magnetically
coupled conducting contours.
2
In
some
texts,
of contours
is
the information about the sign of the mutual inductance for specific reference orientations
not included in
its
definition,
i.e.,
mutual inductance
is
defined as always being nonnegative.
emf due
to
mutual induction
320
Chapter 7
Inductance and Magnetic Energy
and steady currents, the emf definition
Eq. (7.22) makes sense for time-varying
in
currents only.
Conversely,
if
we assume
that a current
'2
z
flows along the contour
mine the associated magnetic flux Oi through the contour
between the contours is given by
C
1
,
C2
and deter-
the mutual inductance
(7.23)
emf e m ^\ along
with an analogous expression using the induced
the contour C\.
Because of reciprocity (in a linear system, transfer functions remain the same if the
source location and the location at which the response to the source is observed are
interchanged),
reciprocity of
mutual
^12
=
(7.24)
^21-
induction
If the medium around the contours is air (or any nonmagnetic medium), the
induced electric field intensity vector due to the current in the first contour is given
by [see Eq. (6.5) and Fig. 7.7]
so that the induced
®ind2
i
=
emf in
Ejndl
•
MO
—
Ejndl
dli
dz'i
(7.25)
An
the second contour can be written as [see Eq. (6.33)]
d>2
MO
=
47T
dli
ll
•
dq
dl2
R
dq
=
—Li\
dt
(7.26)
”d7
This yields the following expression for the mutual inductance between the contours
Ci and
Neumann
C2
:
formula for mutual
(7.27)
inductance
which
is
known
as the
Neumann 3
formula for mutual inductance.
It
implies the
evaluation of a double line integral along the contours and underscores the fact
that the
mutual inductance
only a property of the geometrical shape and the
is
physical arrangement of coupled contours, as well as of the permeability of the
medium
[if
the contours are situated in a
homogeneous
linear magnetic
medium
of
permeability p, the constant p 0 needs to be replaced by p in Eq. (7.27)]. It is obvious
from Eq. (7.27) that the change of the reference orientation of one of the contours
means the change of the reference direction
changes the polarity (sign) of the mutual inductance L 21 It is also obvious that interchanging the subscripts 1 and 2 in Eq. (7.27)
does not change the algebraic value (which includes both the magnitude and the
in Fig. 7.7 (but
not both of them), which
of one of the vectors
dlj
and
dl 2
,
.
sign) of the
double integral (since the dot product
is
commutative and the two
line
integrations can be performed in an arbitrary order). This proves the identity in
Eq. (7.24). The
Neumann formula
represents a good basis for numerical evaluation
of the mutual inductance of contours of arbitrary shapes, where the involved line
integrals along the contours are
computed using numerical
3
German
Franz Ernst
Neumann
(1798-1895), a
mineralogist, physicist, and mathematician, a profes-
sor of mineralogy and physics at the University of Konigsberg.
electromagnetic induction, and derived,
in
Neumann
1845, the formula for the
equal parallel coaxial polygons of wire, the generalization of which
inductance of arbitrary wire contours (loops), as
integration methods.
we
use
it
today.
is
contributed to the theory of
mutual inductance between two
the Neumann formula for mutual
a
Section 7.2
Mutual Inductance between a Loop and an
Example 7.6
Infinite
Find the mutual inductance between the rectangular loop and the
321
Mutual Inductance
Wire
infinitely
long straight
wire in Fig. 6.12.
Solution
The
infinitely
long wire can be considered as a loop that closes upon
itself at
and the rectangular loop as C2 the current i\ and magnetic
flux <f>2 in Eq. (7.21) are actually the current i carried by the wire in Fig. 6.12 and the flux
through the rectangular contour given by Eq. (6.64), respectively. The mutual inductance
between the contours for their reference orientations given in Fig. 6.12 (upward orientation
for the infinitely long wire and clockwise direction for the rectangular loop) is hence
infinity.
Designating
it
as C\
,
L2 i =
$
Mo
—
=—
b
2
that the
same
result
is
+
(7.28)
.
c
2 Tt
l\
Note
c
In
obtained by applying the emf definition of mutual inductance
Eq. (7.22) to the expression for the induced emf in the rectangular contour in Eq. (6.65).
Note also that computing the mutual inductance as L\ 2 in this case would be prohibitively
in
It would require either finding the induced electric field intensity vector Ei n d 2
an arbitrary point of the infinite wire due to an assumed current i2 in the rectangular
contour, and then integrating it along the wire (which, in fact, is the Neumann formula for
complicated.
at
L 12 )
or finding and integrating the magnetic flux density vector
plane bounded by the loop
Example 7.7
Two
coils are
C
1
B 2 due
to i2 across a half-
.
Mutual Inductance of
wound uniformly and
a thin toroidal core (such as the
Two
on
Coils
a Thin Toroidal Core
densely in two layers, one on top of the other, about
one shown
in Fig. 5.11).
The core
is
made from
a linear
ferromagnetic material of relative permeability Mr = 500. Its length is / = 50 cm and
2
cross-sectional area 5 = 1 cm The number of wire turns is N\ = 400 for the first coil and
.
N2 — 600 for the second one. Compute the mutual inductance of the coils.
Let us adopt the same reference orientations for the two coils (this will give us
mutual inductance of the coils) and assume that the first coil carries a current of
intensity i\. This current produces a magnetic field of the same intensity {H\) everywhere
inside the core. From Eq. (5.53), H\ = N\i\/l so that the total magnetic flux through the
second coil ($ 2 ) is given by [see also Eq. (5.96)]
Solution
a positive
,
^single turn
— pH\S
(m
—
MrMo)
and the mutual inductance between the
4*2
coils
— ^2 ^single turn —
comes out
p-n x
.
(7.29)
to be
n2 s
(7.30)
l
Finally, substituting the
coil
numerical data gives
L 21 =
30
mH.
coil (<t>i).
Magnetic Coupling between a Toroidal
Coil
and
a
Loop
N = 500 turns of wire
and densely wound about a ferromagnetic core. The relative permeability
of the material is Mr = 1000. The inner and outer radii of the toroid are a = 2 cm and
b = 4 cm (thick toroid), and its height is h = 1 cm. A wire loop is placed around the toroid,
as shown in Fig. 7.8. There is a low-frequency time-harmonic current of intensity i = Iq cos cot
An
open-circuited toroidal coil with a rectangular cross section has
that are uniformly
two coupled
thin toroidal core
Note that the same result is obtained by assuming a current i2 to flow in the second
and computing the mutual inductance L 12 from the total magnetic flux through the first
Example 7.8
L 21 -
coils
on a
t
322
Chapter 7
Inductance and Magnetic Energy
Figure 7.8 Magnetic coupling
between
a coil
wound
over a
and
a wire
thick toroidal core
loop encircling the toroid; for
Example
7.8.
flowing along the loop, where /o
terminals of the coil.
= 2A and a> = 7
Let us denote the wire loop as
Solution
no current in the coil
Eqs. (6.62) and (7.22)]
is
(it is
x 10 3 rad/s. Find the voltage between the
circuit 1
and the toroidal
coil as circuit 2.
open-circuited) and the voltage across the coil terminals
v
—
^ind 2
—
dz'i
E21
=
(r'l
There
is
[see
(7.31)
i).
dr
However, the mutual inductance L21
is
extremely
difficult to find
using Eq. (7.21). Namely,
the field Bi due to q, which would have to be integrated through each of the turns of
the toroidal coil in order to compute the flux $2, is not only highly nonuniform but also
impossible to find analytically for a loop of arbitrary, irregular shape.
On
instead.
we can use the identity in Eq. (7.24) and find the inductance L \2
we assume a current of intensity q in the toroidal coil, and no cur-
the other side,
To
this
end,
rent in the loop
( i\
=
the flux
0), find
<Jq
through a surface bounded by the loop, and use
Eq. (7.23). Under these circumstances, there is no field outside the coil [see Example 4.12],
so that the flux <Jq equals the negative of the flux <J> through a cross section of the toroid given
by Eq.
(5.84),
where the minus
comes from
sign (negative)
different reference directions of
the fluxes in Figs. 7.8 and 5.26. Hence,
L12
12
Finally,
=
~~
combining Eqs.
coil terminals in Fig. 7.8
v
— = -— =
h
(7.31), (7.24),
comes out
= L \2 —7“ =
MrMO Nh
r
2n
1
to
and
b
ln-
-693
a
(7.32), the
n H.
(7.32)
time-harmonic voltage between the
be
—cdL\ 2 Iq sincvf
=
9.7 sin(7
x 10 3 f)
V
(
(7.33)
in s).
at
Note
Z21
- transfer impedance
between two circuits (unit: Q)
Eo
=
that the amplitude of the voltage v can
IZ21I/0,
where
Z21
be written as
= 0 L 21 = cuL 12 = — —
In
C
2n
— = —4.85
£2.
(7.34)
a
4
can be measured by a voltmeter connected
)
by directly measuring Vo, we can indirectly measure the
amplitude (or rms value) of the current (Iq) in an arbitrary conductor, without inserting an
ammeter in its circuit. In such cases, the coil in Fig. 7.8 is used as a test transformer, where
the current to be determined is transformed to the voltage that is indicated by the voltmeter
This amplitude (or the corresponding rms value
to the terminals of the coil. Thus,
by means of magnetic coupling and transformer electromagnetic induction. The constant Z21
in Eq. (7.34), expressed in ohms, is said to be the mutual impedance or transfer impedance
4
Note
that
amplitudes
most instruments show rms (root-mean-square) values of measured quantities instead of their
(maximum
values).
u
'
Section 7.2
between the two
the circuits
is,
directions of
Example
i
circuits
= Z21
(Z 12
negative here, as the mutual inductance between
). It is
because of the particular reference orientations of the contours and reference
and v
in Fig. 7.8.
Mutual Inductance
7.9
Two Two-Wire
of
p.u.l.
shows a cross section of a system composed of two
Fig. 7.9(a)
323
Mutual Inductance
The
Lines
infinitely
long thin two-wire
has conductors marked as
1 and T,
and the second one as 2 and 2'. The distances between axes of the first and second conductor
of the first line and each of the conductors of the second line are r\ 2 r\ 2 H'2> and H'2'>
respectively. Find the mutual inductance per unit length between the two lines.
lines
running parallel to each other
in air.
first line
r
,
,
We assume that the first line (line 1-1') carries a current of intensity q
Solution
,
Second
as indicated
Our goal is to find the magnetic flux $2 due to this current through a surface S
bounded by the conductors of the second line (line 2-2'), and to apply Eq. (7.21)
the mutual inductance between the lines.
Let us first find the flux Oa through S due to the conductor 1 alone. The corresponding
in Fig. 7.9(b).
of length
for
/
magnetic flux density vector, Bi
magnitude is [Eq. (4.22)]
circular with respect to the axis of the conductor
is
,
B =
r is
the distance from the
simply as the
flat
which
ri 2
and a
flat
part
this part of integration (note the
integration over the rest of
integrate Bi, however,
r r 12
not convenient to adopt S
more complicated
shown in Fig. 7.9(c), is adopted. It consists of a cylindrical
(strip) whose width is equal to r\ 2 —
The flux through
is
zero, because
Instead, a
Bi
is
tangential to the surface (Bi
analogy with the electric potential computation in
S
(flat
part)
'
J/r=r\2
it is
2'.
2
'
(b)
is
perpendicular to the integration surface (Bi
reduces to the one in Eq. (6.64), and we have
=
its
'
the cylindrical part of the surface
The
To
surface spanned across the conductors 2 and
surface, the cross section of
part of radius
axis.
and
(7.35)
t
where
line
(a)
B t dS =
is
||
r r 12
quite simple to perform, because Bi
dS).
Hence, the
'
dr
l
J r\2
_L
=
,
In
2n
is
now
computation practically
flux
Mo ill
2nr
dS) in
Fig. 1.23).
—
n
r\
(7.36)
2
(c)
Similarly, the flux
<t>b
due
conductor V alone can be found by intethrough a surface conveniently placed with respect
to the current
grating the corresponding magnetic field
i\
in the
However, instead of repeating the above computation for a new position of
the conductor (now conductor 1'), we can use the final result in Eq. (7.36) with just changing
to this conductor.
mutual inductance per unit
length of two parallel
long thin two-wire
infinitely
the notation:
<*>b
=
—
Moil I, r V2'
ln
2n
r V2
lines in air: (a) cross section
(7.37)
•
being opposite to the direction of the current in the conductor
in the
conductor
V
1.
three-dimensional view
showing
superposition, the total magnetic flux through the line 2-2' at the length
/
of the
system amounts to
a part of the
system of length
l
that
only considered, and
^
^ +A =
$2 = 4>a
4>b
The corresponding
of the system with given
geometrical data, (b)
where the minus sign comes from the reference direction of the current q
By
Figure 7.9 Evaluation of the
flux per unit length (for
Moill
—
— In H2'TL'2
~,
,
2n
o\
(7.38)
.
r\ 2 r\' 2 >
integration of the magnetic
field
one meter) of the system
is
due to the conductor
=
(* 2 )^.
1.
= ^,
obtained as
(7-39)
bounded by the
conductors of the
for
Example
line 2-2!)
7.9.
so that the mutual inductance per unit length of the two lines for the orientations of the lines
given in Fig. 7.9
is
L21
—
(L 2 l)p. u
.i.
—
^21
/
_ ^2 _
q
MO
2n
H2'H'2
j
'
r\ 2 r\' 2
1
over a conveniently chosen
surface
*2
is
(c)
(7.40)
L'2]
- two
two-wire
parallel thin
lines (unit:
H/m)
.
324
Chapter 7
’
Inductance and Magnetic Energy
Depending on the actual mutual position of the lines, L'2l can be both positive (when
> r \2 r Y2') and negative (when r^ryj < ri2 r l'2')- Th e unit for ^21 s H/m.
f\2' r \'2
*
This formula
is
of great importance in assessing (usually undesired) magnetic coupling
and mutual electromagnetic induction between
transmission lines
power
in practical
applications
different combinations of parallel two-wire
coupling between a phone line and nearby
(e.g.,
or between pairs of wire conductors
line
an electronic device).
in
for other types of transmission lines that can be
approximated by wire
It
can be used also
lines in
some con-
rough computation of mutual inductances within multiconductor microstrip
and strip transmission lines that give rise to the so-called inductive “cross talk” between
conductors in printed circuit boards in computers).
siderations
(e.g.,
Problems 7.10-7.15; Conceptual Questions (on Companion Website):
:
MATLAB
73 ANALYSIS OF MAGNETICALLY COUPLED
Having now
in
7.1 1-7.21;
Companion Website).
Exercises (on
hand the concepts of both
self-
CIRCUITS
and mutual inductance,
let
us assume
that slowly time-varying currents exist in both contours (circuits) of Fig. 7.7 at the
same
time, as
shown
in Fig. 7.10.
now caused by both
principle, we have
is
and
i\
z'
2
.
The
total magnetic flux Oj through the first circuit
Using Eqs. (7.1) and (7.23) and the superposition
Ol
with
L
1
= Liz't + Li2i2>
being the self-inductance of the
between the
circuits. Similarly,
— L 12
and L 2
token, the induced
emf
is
in
first circuit
= Lnh + ^2*2
the self-inductance of the second circuit.
each of the circuits
is
=-
d<t>]
~ dT
dz‘i
= ~ L]
composed of both
By
the
same
self-induction
as
~
d /2
Ll2
~dt~
d<J>2
(7.43)
~d7
d/
di\
= -T 21-77 - l 2~T72
T—
d
dt
at
^ind2
is
(7.42)
>
and mutual-induction terms, which can be written
eindl
and L 12 the mutual inductance
the total magnetic flux through the second circuit
$2
where L 21
(7.41)
(7.44)
t
Finally,
from
Fig. 7.10 [see also Figs. 6.6
and
6.1 1(b)],
voltages across the terminals
of the circuits are given by
current-voltage characteristic
V\
=
-<Mndl
—+L
=T —+T —
=L
d/l
/
circuits
V2
=
d/
~eind2
J
with slowly time-varying
currents.
(7.45)
dr
d/ 2
l
,
/
2
21
dr
Figure 7.10 Analysis of two
12-77,
dr
of two magnetically coupled
magnetically coupled circuits
d/ 2
,
,
1
dr
(7.46)
Analysis of Magnetically
Section 7.3
In words, the voltage across the terminals of each of the circuits
nation of time derivatives of currents in both
with L\,
circuits,
is
Coupled
Circuits
325
a linear combi-
L 2 and L \2 = L 21
,
as
linearity (proportionality) constants.
Shown
coupled
in Fig. 7.11
is
the circuit-theory representation of the two magnetically
circuits of Fig. 7.10. This
is
a two-port network
whose current-voltage
char-
given by Eqs. (7.45) and (7.46). It consists of two coupled ideal inductors,
where, in addition to modeling the emf due to self-induction in each inductor, the
acteristic
is
effect of the emf due to mutual induction between the inductors is
The mutual inductance between the inductors is customarily written
also modeled.
as
Figure 7.11 Circuit-theory
representation of
L 12 = ±ky/L\L2,
where k
is
(7.47)
two
coupled inductors.
a positive dimensionless constant called the coefficient of (magnetic)
coupling of the inductors (circuits) and defined as
k
—
0<k<l.
(7.48)
VLiLi’
coefficient of
magnetic
coupling between two circuits
(dimensionless)
We
k
shall see in the next section that
is
always
less
than or eventually equal
to unity. The coefficient k in Fig. 7.11 thus provides the information about the
magnitude of the mutual inductance between the circuits of Fig. 7.10. The sign
of L 12 however, depends on the adopted reference directions of currents
and
and
therefore
cannot
be
given
as
a
single
piece
of
information
(positive
or
2
negative) along with k, independently from the current directions. That is why
we use a so-called two-dot notation to include the information about the sign of
L 12 in the representation in Fig. 7.11, by placing two big dots near the particular
ends of the two inductors. According to this notation (convention), if both currents (z'i and 2 ) enter the inductors at ends marked by a big dot (as in Fig. 7.11),
the mutual inductance, for that particular combination of reference directions of
currents, is positive (note that Z42 > 0 in Fig. 7.10). The same is true if both currents leave the inductors at marked (dotted) ends. Otherwise, if one current enters
and the other leaves the inductor at marked ends, the mutual inductance is negative (note that a change of the reference direction of one of the currents in
Figs. 7.10 and 7.11 would result in Z42 becoming negative). Finally, if k = 0, the
two inductors in Fig. 7.11 become decoupled and independently described by the
current-voltage characteristic in Eq. (7.3) for a single inductor [Eqs. (7.45) and
z'i
,
z'
,
z'
(7.46) with
L \2 =
0].
Note that Eqs.
(7.41)
and
(7.42) can
[<*>]
where [O] and
[z]
are
be represented
=
in
matrix form:
(7.49)
[L][i],
column matrices whose elements are the fluxes and curand [L] is a symmetrical square matrix
rents of the coupled circuits, respectively,
of inductances,
[L]
(Ln
=
L\ and
=
L 22 = L 2 ). Note
L\\
L 12
_L 2 1
L 22_
N x N matrix.
number
(
N
)
L\ L\2
L\2
_
L2 \
(7.50)
also that these equations, as well as Eqs. (7.43)-
(7.46) for the electromotive forces
with an arbitrary
=
and
voltages, can
be generalized to the system
is an
of coupled contours (circuits), in which case [L]
inductance matrix
)
326
Chapter 7
Inductance and Magnetic Energy
Example 7.10
Two Coupled
Inductors Connected
in Series
and
Parallel
L 2 and
Find the equivalent inductance of two coupled inductors of inductances L\ and
coupling coefficient k
if
they are connected
in (a) series, as in Fig. 7.12(a),
and (b)
parallel,
as in Fig. 7.12(b).
Solution
(a)
Voltages
vj
and
their connection
v
=
vj
by Eqs. (7.45) and
V 2 of individual inductors are given
through the inductors
in Fig. 7.12(a)
is
the same,
i\
=
=
12
i,
(7.46).
The current
so that the voltage across
is
+ V2 =
— + L — + T — + Z/2— = (^1 + ^2 + 2L —
dz
dz
L\
dz
inductor
dz
12 )
dr
dr
dz
12
\2
dr
dr
(7.51
.
dr
inductor 2
1
By comparison
with the current-voltage characteristic for a single inductor, Eq. (7.3),
conclude that the total (equivalent) inductance of the connection in series equals
L—
equivalent inductance of two
coupled inductors
L\
+ L 2 + 2L\2,
we
(7.52)
in series
where the mutual inductance between the inductors,
Fig. 7.12(a), is positive (the current enters
both inductors
for
the
situation
given
in
ends marked by a big
at their
dot) and, using Eq. (7.47), amounts to
L12
(b)
=
kyjL\L 2
(7.53)
.
The voltage across the inductors in Fig. 7.12(b) is the same, v\ =V 2 — v. Assuming it to
be known, Eqs. (7.45) and (7.46) provide the following system of equations with d/j /dr
and dz' 2 /dr as unknowns:
dil
L,
~di
Its
solution
,
is
~ L\2
L,2 — T|2
dr
1
Ln
d/2
and
dr
d *l
I
+ L1
,
~ii
d, 2
’
=
dzj
d ~t~ ~di
expressing v in terms of dz/
+
dz'2
~di
^1 ~ ^12
~ L L -L\
2
2
_
and
(b) parallel; for
Example 7.10.
in
=
i\
+
z'2,
(7.56)
x
T 12
dz'
(7.57)
dr
i
(a) series
is i
dr,
x
coupled inductors connected
(7.55)
x
L + L 2 — 2L X2
L L 2 - L 2n
L + L 2 — 2L\2
Figure 7.12 Evaluation of the
(7.54)
'
x
~
L,\ Lj 2
equivalent inductance of two
V
~ii
the terminals of the parallel connection of inductors
dz
By
A
and
z^2
dz'i
As the current through
we have
d
h =
v
+Lu ^i
1
Section 7.3
we conclude
that the
two coupled inductors of
Analysis of Magnetically
Coupled
327
Circuits
can be replaced by a single
Fig. 7.12(b)
equivalent inductor of inductance
L1L2
^12
(7.58)
L\+ L 2 — 2L\ 2
The
coupled inductors
particular (given) placement of big dots in Fig. 7.12(b) tells us that the
inductance
L \2
is
that for
k
in parallel
mutual
negative, and hence Eq. (7.47) yields
Li 2
Note
equivalent inductance of two
=
0 and L\ 2
=
= -kJUL 2
(7.59)
.
Eqs. (7.52) and (7.58) reduce to expressions for
L 2 (that are
0,
equivalent inductances of two ordinary inductors of inductances L\ and
not magnetically coupled together) connected in series and parallel, respectively. Note
also that these expressions (with
L 12 =
0)
have the same form as the correspond-
ing expressions for two connected resistors of resistances R\ and
and
R2
[see Eqs. (3.86)
(3.94)].
Example
Compute the
Example 7.7.
Solution
Coupling Coefficient of
7.1
coefficient of coupling
Using the expression
of the core due to the current
i\
in
Two
Coils
between the
on
coils
a Toroidal
on the
Core
thin toroidal core
from
Eq. (7.29) for the magnetic flux through the cross section
in the first coil, the self-inductance of the first coil
‘I* single
turn
is
pN^S
(7.60)
h
Similarly, the self-inductance of the
between the
coils is given
by Eq.
second coil is L 2 = pN^S/l. The mutual inductance
Hence, their coupling coefficient turns out to be
\Ln\
i
maximum
coupling
is
a
_ NjN2
(7.61
Voltage Transformation by
ings
shown
respectively.
circuit in the
in Fig. 7.13(a).
maximum
coupling
-
coils
a toroidal core
consequence of the magnetic
so that the entire flux passes through every turn of both
Consider a magnetic
)
y/NjNj
flux
due to the current of each of
the coils being concentrated entirely inside the core (flux leakage from the core
Example 7.12
a thin toroidal
core
(7.30).
^L L2
This
L - coil on
is
negligible),
coils.
Two Coupled
Coils
form of a thin toroidal ferromagnetic core with two wind= 1000 and N2 = 500 turns of wire,
The windings have N\
The ferromagnetic material can be considered
to be linear. Losses in the
Figure 7.13
circuit
(a)
Magnetic
with two windings and
(b) equivalent
schematic
diagram with two coupled
inductors; for Example 7.12.
on
.
328
Chapter 7
2
1
2
Inductance and Magnetic Energy
windings and the core can be neglected. The primary winding
=
voltage generator of emf e g
is
connected to an ideal ac
V (r in s). The secondary winding is open-circuited.
100 cos 377f
Find the voltage across the secondary winding.
Solution Fig. 7.13(b) shows the equivalent schematic diagram with two coupled inducwhere L \ 2 is positive for the reference orientations of windings in Fig. 7.13(a). As the
secondary winding is open-circuited, Eqs. (7.45) and (7.46) give
tors,
=
v\
L\-j-
and
v2
= L2
Using Eqs. (7.60) and
(7.30),
vi
coupled
write
L
_
that
is,
numbers of wire
turns.
=
vj
v"2
can say that the circuit
port to
N\
N\N2
_
its
Assume
The voltage across
=
=
eg
in Fig. 7.13
50cos377r
Current Transformation by
equal to the
hence
(rins).
(7.64)
Two Coupled
its
primary
Coils
that the primary winding in the magnetic circuit in Fig. 7.13(a)
secondary winding
From
Solution
is
the equivalent schematic diagram
,
g
=
15 sin 377/
shown
— +L — =0
d/i
mA
(/
is
connected to
in s), while the
in Fig. 7.14
and Eq.
(7.46),
d/7
r
2
(v 2
=0).
(7.65)
at
at
means
/
and determine the current of the secondary winding.
short-circuited,
L 12
that
h_
by
coils
N _
N\N2
L2 _
_
Z^i 2
*2
i.e.,
V
is
is
operates as a transformer of voltage from
an ideal ac current generator of current intensity
coupled
the secondary winding
N2 /N\
Example 7.13
current transformation
(7.63)
N2
secondary port, with the transformation (multiplication) factor being equal to the
wire turns ratio
This
N\
the ratio of the voltages across the primary and secondary windings
ratio of the
We
^
X
L\2
V2
coils
(7.62)
dr
we can
voltage transformation by
= 0).
(*2
\
dr
N
(7.66)
the ratio of the currents in the primary and secondary windings
inverse ratio of the
numbers of wire
is
equal to the negative
turns. Solving for the current in the
secondary winding,
we have
/v
= — —1
*2
We
/'1
N2
=—
/V
N2
/
g
= — 30 sin 377/ mA
(/ins).
(7.67)
see that the circuit in Fig. 7.14 performs a transformation of current between
ports, with the
magnitude transformation factor being the reciprocal of
transformation
in
Eq. (7.64).
——
lJL
+
p
Figure 7.14 Current
1
transformation by two
magnetically coupled
for
Example
coils;
c
—
<N
s
^
2
7.1 3.
—
—
II
l2
'
its
that for the voltage
0
.
Analysis of Magnetically
Section 7.3
Coupling Coefficient of
Example 7.14
Fig. 7.15
Two
shows a cross section of two very long solenoidal
N2
of a linear ferromagnetic material of relative permeability
coils is air-filled.
two
Neglecting the end
329
coil is
/u, r ,
is
the same,
wound on
/,
a core
with
and
made
while the space between the
effects, calculate the coefficient
of coupling between the
coils.
Solution
(Fig. 7.15).
Let us adopt counter-clockwise reference directions of currents in both coils
first that a current, of intensity /1 (steady current), exists only in the first
Assume
(inner) coil, while I2
the
coils positioned coaxially
their length
,
The inner
respectively.
,
Circuits
Coaxial Solenoids
respect to each other. Radii of solenoids are a\ and a 2
numbers of wire turns are Ni and
Coupled
first coil
and
is
= 0.
Under
this
assumption, the magnetic
field is
nonzero only inside
given by [see Eq. (6.48)]
Hi
= Nih
(7.68)
'
l
Therefore, the magnetic fluxes through a single turn of both coils are the same.
single turn
Using
2
—
MrAoNl/lTT^
—
(7.69)
|
Figure 7.15 Cross section
found to be
this expression, the self-inductance of the first coil is
of
two coupled
solenoids; for
Ll
<f>l
=
=
Ni<t> single
turn
I1 t /J.qN^7Z
=
(7.70)
h
~h
7
[note that this can also be obtained by multiplying the inductance in Eq. (7.6) by
the
same expression
(for Single turn), the
i21
4>2
=
=
mutual inductance between the
^2 'h single
turn
1
7T
=
coils
is
/r r ].
Using
obtained as
(7.71)
7
On the other hand, the assumption of a current, of intensity I2
=
,
existing only in the second
(outer) coil (while 7i
0) in Fig. 7.15 gives a nonzero magnetic field everywhere inside the
second coil, of intensity
2 i2
n
H2 =
(7.72)
l
The magnetic flux density in the core is B 2 = ix t h,qH2 while B 2q = ij-qH2 in the air-filled
space between the coils, which leads to the following for the total magnetic flux through the
,
second
coil:
0. 2 =
NAT
2
vn
[B 2 na\2
+ a\- a\)
2 m = MN\-Kh{^a\
1
—
+ Bd 2on(a^2 - af)]
j
f
The self-inductance of the second
h2=
coil is
4>2
k
=
(7.73)
hence
—
-
ixqNItt[(ii x
1 )a\
+ a\]
(7.74)
7
whereas computing <t>i through the first coil and L 12 = <4>i/72 would, of course, give the same
result for the mutual inductance as in Eq. (7.71).
Finally, the coefficient of coupling between the coils, Eq. (7.48), comes out to be
1^121
a/L\L 2
_
Mr
I
V
Mr
Note that
for very large relative permeabilities
(coupling
is
second
[Mrfli
coil
very strong), which
is
-
1
(«2/«l)
/r r (e.g.,
/u, r
=
2
(7.75)
’
1000),
k
is
very close to unity
a consequence of the magnetic flux 4>2 due to 12 in the
being practically entirely concentrated
» a\ - a\ in Eq. (7.73)].
+
in the
core of the
first coil
for
/j, T
3>
1
coaxial
Example 7.14.
h
330
Chapter 7
Inductance and Magnetic Energy
Magnetically Coupled Circuits Containing Two-Wire Lines
Example 7.15
and 2-2') in air, where b = 20 cm
The first line is connected at one end
to an ideal voltage generator of time-harmonic emf e (r) = £ ocos&>r, where £ o = 3 V
g
g
g
and co = 10 6 rad/s, and the other end of the line is short-circuited. The second line is
two
Fig. 7.16(a) depicts
and
c
=
mm. The
5
parallel thin two-wire lines (1-1'
= 0.3 mm.
radii of all wires are a
short-circuited at both ends. Neglecting internal inductances, losses in the wires, capacitive
coupling between the
currents in the
Solution Using Eq.
end
and propagation
effects,
/zH/m
1.565
In
%=—
n
and
a
respectively,
h
M
where we neglect
(r\ 2
=
-
1.125 /zH/m.
(7.76)
a
~
,
=
In 2
In
Cz
2c and r \2
=
=
ryy
CY
it
nH/m.
n-7-7
-i
/
277
-
-7
(7.77)
71
We neglect the end effects and the propagation effects, so that the total self-inductances
and mutual inductance of the magnetically coupled circuits formed by the lines are obtained
by multiplying the per-unit-length inductances by the corresponding lengths of the lines, L\ =
L\3b = 939 nH, L 2 = L'2 b = 225 nH, and L 12 = L'v b = 55.4 nH (the length of the shorter of
the two lines, b is relevant for the mutual inductance between the lines).
Fig. 7.16(b) shows the equivalent schematic diagram of the two coupled circuits (we
neglect the losses in wires and capacitive coupling between the circuits). According to this
,
Figure 7.16 Analysis of
magnetically coupled
containing two thin
lines: (a)
=
ryi
)
,
2ll
two-wire
=
'
MO
MO
—
—r— = —
r
circuits
In
internal inductances of the lines. Eq. (7.40) gives the mutual
inductance per unit length of the lines
Lj ]7 =
(b)
amplitudes of
effects, find the
(7.11), self-inductances per unit length of the lines in Fig. 7.16(a) are
—n — =
=
L\
lines,
lines.
diagram.
structure
geometry and (b) equivalent
schematic diagram with two
dh
L
+ Ln
'~it
The
di2
La
£p
~it
dii
~ir
d
+i2 h =
ir
0
(7.78)
.
solution of these two equations for the time derivatives of currents in the circuits
coupled inductors; for
Example 7.1 5.
L2
di\
dt
which
L\
L-2 ~~
%
7
dM
£go cos cot
and
dr
L\
L 12
—
L,2
L\2
£go cos cot
is
(7.79)
,
then integrated with respect to time to obtain the solution for the currents.
is
Integration in time of time-harmonic quantities results in an additional factor
1
expressions for amplitudes, so that the amplitudes (peak-values) of the currents
the circuits
in
/co in
the
(and the lines) are, respectively,
=
A) 1
^ 12
co{L\L 2
Example 7.16
p.u.l.
symmetrical thin two-wire line
as
shown
is
h
(
The wire
Solution
Assume
is
in Fig.
of a
=
Two-Wire Line above
positioned
in air
7.17(b)
is
a
PMC
Plane
»
(in the
is cl (cl
for the
» a). Compute the
presence of the plane).
magnetic
i is
established in the line
we can remove the ferromagcurrent conductors. The equivalent
field, Fig. 5.14,
images of the original
thus obtained, which consists of two (magnetically coupled) parallel
thin two-wire lines with the
the upper (original) line
(7.80)
over a ferromagnetic (n
/-to) plane,
axes of both wires with respect
that a slowly time-varying current of intensity
By image theory
0.8 A.
^12)
» a), and the distance between the wire axes
netic plane by introducing positive
system
£ 12 %)
=
radii are a, the height of the
inductance per unit length of the line
[Fig. 7.17(a)].
/02
^
A
to the plane
and
co(L\L 2
Inductance
in Fig. 7.17(a).
A
3.24
is
same current
(/) in air.
The
total
magnetic
flux per unit length of
therefore [see Eq. (7.41)]
<t>'
—
L\i
+
L\ 2 i
,
(7.81)
d
331
Magnetic Energy of Current-Carrying Conductors
Section 7.4
where L) is the self-inductance p.u.l. of the upper line when isolated in free space and L'u is
the mutual inductance p.u.l. of the upper and lower lines in the equivalent system in air. These
inductances are computed using Eqs. (7.11) and (7.40), respectively, so that the inductance
p.u.l. of the original line in Fig. 7.17(a) comes out to be
L'
=
— =L
I
1
+ L'u =
l
^
,
In
7T
d
—
,
h In
——h
2
2
—y/d + 4 h
2
+ 4h 2
(7.82)
i
n
2
a
dy/
ii o
h
Mo
2 ah
Mr
(a)
Obviously, L'12 represents the influence of the ferromagnetic material in the lower half-space
on the
^°°
line inductance.
AO'
Problems 7.16-7.24; Conceptual Questions (on Companion Website): 7.22-7.28;
:
MATLAB Exercises (on Companion Website).
MAGNETIC ENERGY OF CURRENT-CARRYING
CONDUCTORS
7.4
Every system of conducting loops with currents contains a certain amount of energy,
called magnetic energy, in a manner analogous to a system of charged conducting
bodies storing electric energy. In other words, current-carrying (coupled or uncoupled) inductors in an ac or dc circuit store magnetic energy, much like charged
capacitors store electric energy. By the principle of conservation of energy, the magnetic energy of a system of loops (or inductors) with currents equals the work done
to the system in the process of establishing these currents from zero to their final
values. Thus, by simulating this process of “loading” the loops by currents, we can
find general expressions for computing the magnetic energy in terms of the current
intensities and associated magnetic fluxes (final values) of the loops.
Let us consider first a single loop with a slowly time -varying current of intensity i. The current flow is maintained in the circuit by an ideal voltage generator of
emf e g as shown in Fig. 7.18. As the magnetic flux through the loop, O, changes in
time, an emf ej nd is induced in the loop, given by Faraday’s law of electromagnetic
induction, Eq. (6.34). Using Kirchhoff’s voltage law, Eq. (3.119), we can write
(b)
Figure 7.17 Evaluation of
the inductance per unit
length of a two-wire line
above
a
PMC)
plane: (a) original
ferromagnetic (or
system and (b) equivalent
free-space system with
two
magnetically coupled
two-wire
lines; for
Example
7.1 6.
,
eg
=
Ri
-
e ind
(7.83)
,
R is the resistance of the loop. The emf e m & (emf due to self-induction)
opposes the current change in the loop (Lenz’s law), that is, it acts against the emf
e g Therefore, an amount of work must be done in establishing the current in the
loop to overcome this induced emf. This work is done by en external agent, with the
energy transfer to the circuit being modeled by the generator (of emf e g ) in Fig. 7.18.
To investigate the energy balance in the circuit during a differentially short time
interval dr, we multiply both sides of Eq. (7.83) by i d t,
where
.
eg i d t
=
2
Ri d t
+
(-<?j n d/ dr).
(7.84)
By means
of Eq. (3.121), the term on the left-hand side of this equation is the work
done by the emf of the generator during the time interval dr. It equals the energy of
external sources delivered to the circuit during that fraction of time.
From
Joule’s
term on the right-hand side of the equation represents
term in the equation
(including the minus sign) is the work done against the emf ej n d in dr. Eq. (7.84)
thus expresses the principle of conservation of energy for the circuit in Fig. 7.18,
law,
Eq.
(3.77), the first
Joule’s (ohmic) losses in the loop during dr. Finally, the last
Figure 7.18 Wire loop of
inductance
R
L and
resistance
with a slowly time-varying
current of intensity
i,
maintained by an ideal
voltage generator of
emf
eg
.
2
.
332
Chapter 7
Inductance and Magnetic Energy
work done by the generator during dr is partly converted into
and partly used to overcome the induced emf in the loop. While the
lost to heat, the second part is given to the emf Cj n d, i.e., to the flux O of
telling us that the
Joule’s losses
first
part
is
the loop. Ultimately,
it
represents the energy delivered to the magnetic field of the
dr. Using Eq. (6.34), — e; nc dr = dO, so that the
elementary work done by external sources to the magnetic field of the loop can be
current
i
in the
loop during the time
i
written as
dWm = — ejncfidr = rdO.
If
the
medium surrounding
the loop
dWm =
where L
is
i
=
d<t>
the inductance of the loop.
of the loop, expressed in joules
magnetically linear, the use of Eq. (7.1) yields
is
id(Li)
=
Lidi,
(7.86)
The magnetic field, on
and
ble of storing the received energy,
(7.85)
this stored
energy
the other hand,
is
Consequently, the work
(J).
capa-
is
the magnetic energy
dWm
in
Eq. (7.86)
represents an increment of the magnetic energy of the circuit in Fig. 7.18.
When
the current in the loop
is
zero, the loop has
stored in the magnetic field of the loop
work done
value
/,
when
its
current
no energy. The
is
is
obtained by adding up
all
from
elementary works dVTm
n
=
o
total
energy
equals therefore the net
to the magnetic field in changing the loop current
and
W = f dWm = L fidi=L^z
Ji=
Jo
From
i
-
ur
i
— 0 to
its
final
:
(7.87)
2
is also the energy of a linear inductor of
an arbitrary electric circuit. Employing Eq. (7.1), the equivalent
expressions for the energy of an inductor are
the circuit-theory standpoint, this
inductance
L
in
11
1
energy of inductor
II
(unit: J )
.
=
d> 2
—
2L
(7.88)
note the complete analogy (duality) with the corresponding expressions for the
electric
energy (Vke ) of a capacitor
Generally, for a system with
energy of
- 0/
2
2
We
1
11
in
Eq. (2.192).
> 1) magnetically coupled loops (inductors),
N (N
N magnetically
(7.89)
coupled loops
where the magnetic
fluxes through the loops are given
by linear relationships
in
Eqs. (7.41)-(7.42) or (7.49). Using these relationships, the energy of the system can
be expressed only in terms of the currents and self- and mutual inductances of the
loops. For example, for
N = 2,
the magnetic energy of two coupled loops can be
written in the form
magnetic energy of two
coupled inductors
(7.90)
sum of energies of the two
from each other (when L \2 = 0), because L can
be both positive and negative. Note that the expression in Eq. (7.89) is entirely
analogous to the expression for the electric energy of a linear multibody system in
This energy can be both larger and smaller than the
loops
when magnetically
Eq. (2.195).
isolated
\
Magnetic Energy of Current-Carrying Conductors
Section 7.4
333
The magnetic energy can also be expressed in terms of distributions of the curJ, and the magnetic vector potential, A, throughout the volume
rent density vector,
of current-carrying loops. Namely, the magnetic flux <t>£ through the loop Ck of the
along Ck [see Eq. (4.121)], so that
system with
loops equals the circulation of
Eq. (7.89) becomes
A
N
A
ik dlfc
(7.91)
i
i k can be brought inside the integral sign in the line integral
does not change along the wire (currents are slowly time-varying). As
J dvk, Eq. (4.10), we then have
where the current
because
ik
dU =
it
Wm =
A
•
ik dv k
{7.92)
,
with v k standing for the volume of the kth loop. Finally, the sum of N integrals over
volumes of individual loops can be joined together into a single volume integral:
U
W
n
J Adv.
(7.93)
Since any slowly time-varying volume current distribution can be considered as consisting of
an
infinite
number of filamentary current loops, this
is
a general expression
for evaluation of the magnetic energy of a system of current conductors of arbitrary
shapes in a linear medium. In general, the volume integration is performed over
all parts of the system populated by current (v curre nt)- We again note the duality
with the corresponding expression for the electric energy, namely, with the volume
W
e in terms of the charge density (p) and the
(V) for an electrostatic system with a volume charge.
integral expression in Eq. (2.196) for
electric scalar potential
Magnetic Energy of Two Coupled Two-Wire Lines
Example 7.17
Find the time-average magnetic energy of the system of two coupled two-wire lines shown in
Fig. 7.16
and described
Solution
From
Example
7.15.
Eqs. (7.79), instantaneous intensities of currents in the lines have the fol-
= Iqi sin cot and 2 (0 = ~h)2 smart, where the current amplitudes /01 and
are given in Eqs. (7.80). Using Eq. (7.90), the instantaneous magnetic energy contained in
lowing forms:
/02
in
i\(t)
z'
the magnetic field of the two lines
Wm
(t)
1
,
=
-
=
- Li/qj
L\i{{t)
is
+
1
sin
1
-
,
L 2 q{t)
+ Li2*i (0*2(0
2
art
+
1
-
L 2 Iq2 sin2
<*>*
~
L,\2h\hi sin
2
(7.94)
where the self-inductances and mutual inductance of the coupled circuits in Fig. 7.16(b) are
also found in Example 7.15. Eq. (6.95) tells us that the time-average value of the function
sin
2
art is 1/2,
so that the time-average magnetic energy of the system
(Wm
)
ave
=
2
1
Q ui
0l
+
~
L 2 I22 - L 12 /01 /02
=
is
2.43 Hi.
(7.95)
)
This energy can also be obtained using the equivalent inductance, L, seen by the generator in Fig. 7.16.
To
find L,
we
rewrite the
first
magnetic energy
volume currents
y current
equation of Eqs. (7.79)
in the following
in
terms of
'
334
Chapter 7
'
Inductance and Magnetic Energy
form:
d/i
!Al
dh
,
\
(7.96)
/dr
L2
dr
J
Hence,
L=
In other words, this
is
=
e2
925.4 nH.
(7.97)
the inductance of an inductor that, as far as the generator
can replace the two coupled inductors
coupled two-wire
l\ 2
Li
lines
is
the
same
as the energy of the equivalent inductor,
can be used. Since the current of the equivalent inductor
Wm
(r)
is
is
and Eq.
(7.88)
q,
Ei^t),
—
concerned,
Consequently, the energy of the two
in Fig. 7.16(b).
(7.98)
^
which, averaged in time, gives the same result as in Eq. (7.95).
Proof that the
Example 7.18
Maximum
Coupling Coefficient
is
Unity
Prove that the largest possible value for the magnitude of the mutual inductance of two
magnetically coupled circuits equals the geometric mean of the self-inductances of circuits.
Solution
The
we have
inequality that
to prove, IL 12
simple fact that the stored magnetic energy
this,
we start
QE L 2
1
,
it
consequence of the
a
is
field
of the two circuits
L\i\f(x),
(7.99)
with the expression for the energy stored in the magnetic
Eq. (7.90) and write
in
<
I
always positive (or eventually zero). To show
is
in the following form:
2
12
Li f
+ 2^
E \ii/
L\
Wm = -Ui\
1
2
1
i\
2
*
where
=
fix)
ax
2
+ bx +
x
1,
h_
and
E\2
—
b
2
(7.100)
h
From
for
all
the fact that
values of
Wm
>
0,
we conclude
that the quadratic function fix)
x. This, in turn, is satisfied
only
the
if
E 2 L 2n
0
L\ L\
is
minimum
^ 12 ^ 12
E\
+
must be nonnegative
of/, given by
1
- - L
ci +I
El
-
(7.101)
nonnegative. Hence,
/min
—0
>
which concludes our proof.
or
'12
We
IL 12
I
< v/LiL 2
—
»
k <
1,
(7.102)
realize that these inequalities can also be written in terms of
the coefficient of coupling of the two circuits, defined by Eq. (7.48), as k
coefficient cannot be larger than unity
we noted
earlier, in Section 7.3,
<
1
(that the coupling
but without a proof).
Problems: 7.25 and 7.26; Conceptual Questions (on Companion Website): 7.29 and
MATLAB Exercises (on Companion Website).
7.30;
7.5
In
MAGNETIC ENERGY DENSITY
analogy with the concept of electric energy density (Section 2.16), we
shall
now
define and use the magnetic energy density to describe the actual localization and
distribution of the magnetic energy of a system of current-carrying conductors,
the total
consider
amount
first
of which
is
given by Eqs. (7.89) and (7.93). To this end, let us
magnetic field of uniform intensity in a thin toroidal
a simple case of a
0
Section 7.5
Magnetic Energy Density
335
ferromagnetic core with a winding that carries a slowly time-varying current of
intensity i. Let the number of wire turns of the winding be N, the length of the core
/, and its cross section S in area. From Eq. (5.53), the current in the winding can
H
in the core as i = Hl/N. On
be expressed in terms of the magnetic field intensity
the other hand, the magnetic flux through the winding (all of its turns) can be writ= NBS, where B is the magnetic flux density in the core.
ten as 0 =
s n gie turn
Substituting these two expressions in Eq. (7.85) leads to the following expression
for the work of external sources needed for a change d<f> in the magnetic flux of the
i
in the winding):
winding (not counting Joule’s losses
dWrM= idO =
where v
— SI
or,
the volume of the core,
The net work
i.e.,
the volume of the
(7.103)
domain where the
changing the flux from zero to its final value
equivalently, the flux density from zero to the final value B is hence
magnetic
O
is
—
d(NBS) =HdBSl = HdBv,
N
field exists.
in
B
f
dWm =
v
JB =
Dividing
it
by
v,
[ HdB.
(7.104)
Jo
we get the magnetic energy density
(energy per unit volume) of the
core (in J/m 3 ),
wm =
More
precisely, for
—=
v
B
\ HdB.
(7.105)
Jo
arbitrary
an arbitrary magnetic material of the core,
this
is
the energy per
volume of the material spent by external sources in establishing the field, and
not the energy stored in the field per unit volume of the material. Namely, as we
unit
shall see later in this section, in the case of materials that exhibit hysteresis effects
(see Fig. 5.21), this energy can only partially be returned
by the field to the sources
in
H to zero, because of losses occurring during the magnetization (while establishing H and demagnetization (while
reducing H of the material. These losses, appearing as heat, are a consequence of
the reverse process of reducing the field intensity
)
)
microscopic frictions encountered as elementary magnetic domains (see Fig. 5.2)
change their size and rotate in the magnetization-demagnetization process of the
known as hysteresis losses. For materials with no hysteresis losses,
on the other side, Eq. (7.105) gives the energy that is contained in the field per unit
volume of the core and can be obtained from it at any time in its entirety by reducing to zero the current in the coil. While using the term magnetic energy density for
the integral expression in Eq. (7.105), we shall always keep in mind this distinction
in its actual meaning between materials with pronounced hysteresis behavior and
those for which hysteresis effects are not present or can be neglected.
Although derived for a special case of a coil on a thin toroidal core, the result in
Eq. (7.105) can be generalized to an arbitrary magnetic field, which can be visualized
as a collection of elementary flux tubes (toroids) formed by the lines of vector B
[see Fig. 4.26 and the proof of the law of conservation of magnetic flux, Eq. (4.99)].
In general, the magnetic energy of a differentially small cell of volume dv in an
material and are
!
i
j
i
j
arbitrary (nonuniform) magnetic field in an arbitrary (nonlinear) material
is
;
dWm = w m dv,
(7.106)
where the energy density is given by Eq. (7.105). By summing up the energies of all
of the cells, that is, by integrating the energy dWm over the entire domain with the
i
magnetic energy
medium
density,
(unit: J/rrr’J
336
Chapter 7
Inductance and Magnetic Energy
magnetic
field
(volume
v),
we
Wm =
obtain the total magnetic energy of the system:
dv
J
=
id:
//dfljdv.
(7.107)
In the case of a linear magnetic material of permeability
the integral with respect to
5
in the
B—
/jl,
/xH, so that
expression for the magnetic energy density in
Eq. (7.105) can easily be solved,
B
magnetic energy
linear
B
—
= - BH — -nH
2
5d5 =
Wr
density,
M
Jo
medium
2/z
Jo
1
1
2
2
,
2
(7.108)
.
Note
that these expressions are entirely analogous to the corresponding expressions
Eq. (2.199) for the electric energy density w e in a linear dielectric of permittivity
5
e. The expression for the total magnetic energy becomes
in
1
2
B Hdv.
l
(7.109)
For nonlinear magnetic materials, the integral in Eq. (7.105), in general, cannot
be solved analytically (in a closed form). Having in mind a typical initial magnetization curve of a nonlinear ferromagnetic material
that
in
H dB
is
(e.g.,
that in Fig. 5.20),
proportional to the area of a thin strip of “length”
A/m) and “width” d B (width measured
the 5-axis at the “height”
B
in
we note
H (length measured
T) positioned between the curve and
with respect to the
H- axis,
as indicated in Fig. 7.19(a).
This means that the integral in Eq. (7.105) represents the sum of areas of all such
strips as the point P' with abscissa
and ordinate B moves in the integration pro-
H
cess
from the coordinate origin
to
its final
magnetic energy density in the material
magnetization curve and the 5-axis, that
OPQ
in Fig. 7.19(a).
position (P).
We
conclude, thus, that the
proportional to the area between the
is
is,
to the area of the curvilinear triangle
The proportionality constant (w m / area) can be expressed
terms of the ratio of the magnetic
H
field intensity
in
H in A/m that corresponds to a cer-
and the length (e.g., a 1-cm division along the
and the similar ratio (5/length) for the
5-axis. This conclusion, of course, applies also to linear materials, which can always
be considered as a special case of nonlinear ones. In a linear case, the hypotenuse
of the triangle OPQ becomes straight, and the area of the triangle is computed as
5/7/2, Fig. 7.19(b), which is the same result as in Eq. (7.108).
tain physical length along the
axis
may represent
-
axis
a 100- A/m field intensity)
In ferromagnetic materials that exhibit hysteresis effects, the function
value at the point P in Fig. 7.19(a) to zero,
O
5
(b)
Figure 7.19
Correspondence between
the magnetic energy density
given by the integral in
Eq. (7.105) and the area
between the magnetization
curve and the B-ax is for
(a) nonlinear and (b) linear
magnetic materials.
As
in
5
H
B{H)
is
reduced from its
does not go to zero, but to the remanent
not only nonlinear, but also has multiple branches. Thus,
if
is
the electrostatic case, the magnetic energy of a system of current-carrying conductors might alter-
natively be viewed to reside in the system current, and not the magnetic field [see the corresponding
discussion on the two energy localization viewpoints (field-based and charge-based) for an electrostatic
system, in Section 2.16]. With this approach, the magnetic energy density would be evaluated as being
equal to J
A/2
at
points where the current exists in the system [from Eq. (7.93)] and would be zero
we choose
netic energy localization of a system in terms of the magnetic field distribution (energy exists
magwherever
and whenever the
As noted
elsewhere. Although this viewpoint
in
field exists)
is
also “correct”
and has
its
merit,
and use the associated energy density expressions
the discussion of the electric energy localization, the field-based approach
in
is
to describe the
Eq. (7.108).
much
better suited to
modeling electromagnetic wave propagation and the associated energy flow, where both electric and
magnetic fields of a wave exist (and carry energy) independent of the charge and current that produced
them
at
previous instants of time.
Section 7.5
Magnetic Energy Density
given
Figure 7.20 Evaluation of hysteresis losses
in a
ferromagnetic material as the difference between
the energy given to the
to the sources
in
field
and the energy returned
the process of magnetization and
demagnetization of the material.
magnetic flux density B = B r (see Fig. 5.21). As B also decreases, d B is negative.
means that the energy
d B per unit volume of the material given to the magnetic field in this process is negative, that is, the field is returning its energy to the
external sources (e.g., to the generator in Fig. 7.18). However, this returned energy,
H
This
which
is
is
proportional to the area of the curvilinear triangle
the material (curvilinear triangle
the material in the process of
losses),
and the difference
OPR. Consequently,
is
RPQ shown in Fig. 7.20,
smaller than the energy spent by the sources in increasing the magnetic field in
OPQ). The
its
difference in energy
is
lost to
heat in
magnetization and demagnetization (hysteresis
in area in Fig. 7.20
is
the area of the curvilinear triangle
the energy of hysteresis losses per unit
proportional to the area of the triangle
volume of the material
OPR.
Finally, let us consider a full hysteresis cycle in the material.
The density
of
energy spent on changing the magnetic field in this cycle is given by the integral
in Eq. (7.105) with the integration being carried out all around the hysteresis loop.
Dividing the loop (contour Ch) into four characteristic segments, as indicated in
H dB
0 along the first line segment (H > 0 and d B > 0)
and the third segment (H < 0 and dB < 0), while
dB < 0 along the second line
segment (H > 0 and dB < 0) and the fourth one (H < 0 and dB > 0). The areas
between line segments 1 and 3 and the 5-axis are therefore a measure of the energy
density given to the field at a point in the material, while the areas between line
segments 2 and 4 and the 5-axis correspond to the energy density returned to the
sources at the same point. The difference, the area enclosed by the hysteresis loop,
Sh, represents hysteresis losses in the material. In other words, the energy density
Fig. 7.21,
we note
that
>
H
Figure 7.21 Evaluation of hysteresis losses
in a
complete
magnetization-demagnetization hysteresis cycle of the material.
337
338
Chapter 7
Inductance and Magnetic Energy
Figure 7.22 Cases with no hysteresis
material with retracing of the
in
nonlinear
losses:
magnetization curve
initial
the periodic magnetization-demagnetization of the
material,
which
results in zero
material (or material
approximated by
loop area, and linear
whose magnetization curve can be
a linear function).
of hysteresis losses, Wh, in one complete magnetization-demagnetization hysteresis
cycle
is
proportional to the area Sh,
wh =
energy density of hysteresis
=
H'm
H dB a Sh-
<p
(7.110)
JCh
losses (see Fig. 7.21)
If
the field
is
time-harmonic (ac
field),
with a frequency/, the time of one cycle
of the periodic magnetization-demagnetization of the material, that
the field variation,
obtained by dividing the energy
Thus, in the case of a uniform
lost
an ac
within a cycle by T,
i.e.,
field intensity in the material,
time-average power of
hysteresis losses for
is,
the time dur-
r
P circumscribes once the hysteresis loop, equals the period of
T — 1/f. The time-average power of hysteresis losses, (Ph)ave, is
ing which the point
(Ph)a\e CX
by multiplying
we can
it
by/.
write
(7.111)
/
field
where
v
is
the
volume of the material with
losses.
We now recall that hard ferromagnetic materials, having large Sh
used primarily for dc applications, so that / = 0 in Eq. (7.111)
(see Fig. 5.23),
and hysteresis
losses do not represent any problem. Soft ferromagnetic materials have small Sh,
and that is why they are very suitable for ac applications. In the limiting case, if
no hysteresis is present and the initial magnetization curve is retraced, as depicted
= 0 in Eq. (7.111) and the periodic magnetization-demagnetization
in Fig. 7.22,
process is accomplished with no hysteresis losses. Finally, 5h = 0 for linear magnetic
are
materials as well (Fig. 7.22).
In general, total losses in a ferromagnetic material in an ac field are the
sum
of hysteresis losses and Joule’s losses due to eddy currents, given by Eq. (6.137).
We
recall that the
time-average power of eddy-current losses
is
proportional to the
frequency squared [Eq. (6.152)].
Energy Distribution
Example 7.19
Consider the toroidal
of intensity
i
is
coil
from Example
established in the
magnetic energy
in
coil.
in a
5.1
1,
Under
Thick Linear Toroidal Core
and assume
that a slowly time-varying current
these circumstances, find (a) the distribution of
the core and (b) the total energy of the coil.
Magnetic Energy Density
Section 7.5
339
Solution
(a)
The
As
distribution of energy in the core
the material of the core
the core
is
linear,
(b)
use Eq. (7.108).
The magnetic
wm
.
field intensity in
given by Eq. (5.83), and hence
is
w m (r) =
where
described by the magnetic energy density,
is
we
r is the
- ixH
=
(r)
<
(a
8n 2 r2
r
<
(7.
b ),
112 )
distance from the toroid axis.
The energy of the
coil, that
the total magnetic energy stored in the core, can be
is,
obtained by integrating the energy density
wm
/xN2 i2
w m over the volume v of the core
we adopt dv
[Eq. (7.109)].
form of a differentially thin toroid of radius r and thickness d r, the cross section of which is the thin
strip of length h and width dr shown in Fig. 5.26. The volume of this elementary toroid is
thus dv = / d5, where l = 2nr (length of the toroid) and d S = h dr, so that the magnetic
energy comes out to be
Because
a function of the coordinate r only,
is
-f
Wr
,(r)
2nrhdr =
J r=a
The above
/xN 2 i2 h
/j.N
f
4n
dv
result can also
—
dr
be obtained from Eq.
(7.88), as
2 2
i
in the
h
b
In-.
(
47r
a
Wm =
4> total */2,
where
7
.
113 )
<l>total is
N times the flux through the cross section of the core, found in Eq. (5.84).
Energy of a Simple Nonlinear Magnetic Circuit
Example 7.20
Calculate the energy spent for establishing the field in the magnetic circuit
and described
Solution
in
Example
shown
in Fig. 5.30
5.13.
Final (established) values of the magnetic flux density and field intensity in the
ferromagnetic core and the
material of the core
is
air gap,
(
B H)
,
and (Bo, Hq), are found
nonlinear and the operating point
P
in
Example
5.13.
As
the
of the circuit, in Fig. 5.30(b), does
segment of the idealized initial magnetization curve (which could be
we must use Eq. (7.105) for the density of energy spent
to change the field in the core from zero to (B, H). Having in mind Fig. 7.19(a), this density is
proportional to the area of a polygon formed by the curve and the B-axis, from the coordinate
origin to the point P, as shown in Fig. 7.23. This polygon, in turn, is a sum of a triangle and a
trapezoid, which gives the following solution for the integral in Eq. (7.105):
not belong to the
described by an
first
initial
wm =
permeability),
HdB =
jj
1
Hk B k + \i.Hk + H)(B - B k = 272 J/m
point
Bk =
0.4
T and Hk =
1000
(7.1 14)
,
trapezoid
triangle
where
3
)
A/m are the magnetic flux density and field intensity at the
K (“knee” point between the two segments of the curve) in Figs. 7.23 and 5.30(b).
Air
is
a linear
medium, so
that the magnetic energy density in the gap can be obtained
Figure 7.23 Evaluation of
using Eq. (7.108),
the density of energy spent
w m0 = BqHq =
\
Finally, the total
magnetic energy of the
77
kJ/m3
(7.1
.
1
5)
amounts
This energy can be completely returned by the field only
to
if
the
Energy Lost
in
(7.116)
initial
magnetization curve
is
Magnetization and Demagnetization
In a thin toroidal core of cross-sectional area
field is
established of intensity
Hm =
1
5
=
1
cm 2 and
in Fig.
Example 7.20.
retraced in the process of reducing the field intensities in the circuit, as in Fig. 7.22.
Example 7.21
for establishing the field in
the core
circuit
Wm = w m Sl + WmoSo/o = 290 mJ.
!
B
length
kA/m, and then reduced
to
/
=
20 cm, a magnetic
H = 0.
In this process,
5.30; for
.
340
Chapter 7
Inductance and Magnetic Energy
B
shown
the operating point described the path
// a
= 0.001 H/m.
where the
in Fig. 7.24,
Find the net magnetic energy spent
in the
permeability
initial
is
magnetization-demagnetization
of the core.
The magnetic energy spent
Solution
H
ative in the process of reducing
H = Hm is positive, while it is negThe net magnetic energy, VTm spent in the
to establish
to zero.
,
entire magnetization-demagnetization process represents hysteresis losses in the core (see
Fig. 7.20). Its density is
is
in
wm = [
JO
the magnetization-
HdB+
where
H dB =
f
B m = ji a Hm =
1
2
2
Jp
and
Wm
T, so that
Hm =
B m Hm
\
4
=
250 J/m
3
(7.117)
,
AOPR
demagnetization
magnetization
demagnetization of a
ferromagnetic core; for
Fig. 7.24
R
P
Figure 7.24 Evaluation of
the net magnetic energy
spent
OPR in
proportional to the area of the shaded triangle
given by
— w m S/ =
5 mJ.
Example 7.21
Time-Average Power of Hysteresis Losses
Core
in a
Consider the core depicted in Fig. 6.13(a) and (c), and assume that a low-frequency timeharmonic current of intensity i = Iosin(2nft ) is established in the coil, where /o = 0.1 A and
f —l kHz. Compute the time-average power of hysteresis losses in the core.
The energy
Solution
density of hysteresis losses in one magnetization-demagnetization
hysteresis cycle of the material
parallelogram in
Fig. 6.13(c).
is
given by Eq. (7.110), with Sh
The current amplitude
the peak-values of H(t) and B{t) are also the same,
that B(t)
is
same
the
is
Hm =
now
being the area of the
as in
Example
A/m
100
and
6.10, so that
Bm =
0.1
T
[note
not a time-harmonic function, due to the nonlinearity and hysteresis behavior of
the core material]. Thus, computing the parallelogram area,
wh =
<t
we
obtain
HdB = 2B m Hm = 20 J/m 3
(7.118)
.
Jch
Based on Eq.
core
(7.111), the time-average
power of
hysteresis losses in the
volume
v
—
SI of the
is
CPh)ave=MS/ = 0.8W.
(7.119)
Evaluation of Force from Energy for an Electromagnet
Example 7.23
An electromagnet consisting of an iron core in the shape of a horseshoe and a coil is sketched
The
in Fig. 7.25.
cross-sectional area of the core
the coil, the electromagnet
that there are
two
is
capable of
tiny air gaps
density in the core and the gaps
The
lifting a
is 5.
With a steady current established
weight
W
(made
also of iron).
between the core and the weight and
is fi,
in
Assuming
that the magnetic flux
find the lifting force of the electromagnet.
consequence of the magnetic field in
F m on the lower part of
the circuit that is lifting its weight. This force can be determined from the magnetic energy
of the system using the principle of conservation of energy. Let us suppose that F m moves
the weight by an elementary distance dr upward. We recall the energy considerations in
connection with Fig. 7.18. Here, however, the elementary work of external sources i d<t> from
Eq. (7.85) is split to the change in the magnetic energy of the system dWm and the work dW/r
Solution
lifting force
of the electromagnet
the magnetic circuit in Fig. 7.25, namely,
of the force
F m along
the displacement dr.
id<t>=
If
the magnetic flux
the current
magnetic force from energy
i
in
is
it is
maintained
the coil
is
is
a
the magnetic force,
As d\VF = Fm
dr,
we can
dWm + dWF = dWm + Fm dx.
at a
constant value (4>)
varied appropriately, then
_
~
d<t>
in this
dr
<J>=const
write
(7.120)
experiment, which means that
= 0 and
dwm
,
the above equation yields
Section 7.5
341
Magnetic Energy Density
Figure 7.25 Evaluation of
the
lifting
force of an
electromagnet from the
magnetic energy contained
the
When
system
is
the weight
moved upward by
is
air
in
gaps; for Example 7.23.
dx, the only change in magnetic energy of the
the reduction in energy contained in the two air gaps due to their decreased length.
dWm
The energy change
in Eq. (7.121)
thus negative and corresponds to the energy con-
is
tained in the parts of the air gaps that are dx long and that vanish in our experiment. In
other words,
it
equals the negative of the magnetic energy density
by Eq. (7.108) with m
=
mo> times the change in the
(7.122)
MO
'
v
in the air gaps, given
-^
dWm = -^25dx
—=
2mo
—
dv
—
'
wm
volume of the gaps:
'
(25 is the total cross-sectional area of the gaps).
electromagnet is
From Eq.
(7.121), the lifting force of the
dWm _ B^5
dx
As
this
a numerical example, for
electromagnet can
eration of free
fall).
lift
T and 5 == 0.125 m 2 Fm
of m = Fm /g « 10 tons (g =
B=
a weight
(7.123)
mo
1
,
Such powerful electromagnets are used
100 kN, which means that
9.81
m/s 2 - standard
accel-
in cranes for lifting large pieces
of iron.
Magnetic Pressure
Example 7.24
(a)
For the electromagnet of
of the lifting force
result in (a),
Fm on
,
Fig. 7.25,
determine the magnetic pressure, that
the surface of the iron piece that
is
being
compare the maximal values of the magnetic and
is,
lifted, (b)
the pressure
Based on the
electric pressures attained in
practical situations.
Solution
(a)
The force
on the parts of the surface of the iron piece that form the air gaps with the
is, on the surface of area 25 in total. The corresponding pressure is
therefore given by
acts
core in Fig. 7.25, that
Pm
Fm
/-»
rt
Zo
B2
r\
(7.124)
’
2/Xq
and is called the magnetic pressure. This expression is
between a ferromagnetic material (with m
Mo) and
any boundary surface
any other nonmagnetic
valid for
air (or
magnetic pressure
(unit:
Pa)
342
Chapter 7
Inductance and Magnetic Energy
medium), with
B
that the pressure
being the local magnetic flux density in
pm
netic side of the surface.
p
(b)
-
p0
Note
near the surface.
air
actually equals the local magnetic energy density
It
acts
on
the
We
see
nonmag-
from the ferromagnetic material toward the medium with
.
that the expression in Eq. (7.124)
entirely analogous to the expression in
is
Eq. (2.133) for the electric pressure on a metallic surface in air. These two expressions
provide us now with an opportunity to compare the electric and magnetic pressures and
forces.
Combining them, we
To estimate the
get
maximal magnetic and maximal electric pressure attained
maximal permissible electric field intensity, £,
is determined by the dielectric strength of air, Eq. (2.53), and we take therefore £ max =
3 MV/m in this estimation. On the other side, there is no such limit for B. However,
maximal magnetic flux densities that are normally attained in typical magnetic circuits
are on the order of one tesla and it is reasonable to assume that B ma x = 1 T for the
comparison. Hence,
ratio of the
in practical situations,
ratio of
and
we
recall that the
2
Emax
(Pm) max
1
(Pe)max
e 0M0
10 000
maximal magnetic
,
electric pressures
attained in practice
We
(7.126)
.
( Emax
conclude that practically attainable magnetic pressures are several orders of magni-
tude stronger than electric ones. This
is
why magnetic
forces,
the actual workhorses of our industrial world. Practically
all
and not
electric ones, are
devices for electromechani-
energy conversion, such as different types of electric motors and generators, are based
on magnetic forces and their work and power.
cal
Problems: 7.27-7.35; Conceptual Questions (on Companion Website): 7.31-7.33;
MATLAB
7.6
Exercises (on
Companion Website).
INTERNAL AND EXTERNAL INDUCTANCE
IN
TERMS OF
MAGNETIC ENERGY
In this section,
it,
now from
we
revisit the
concept of self-inductance and techniques to compute
We consider, thus, an arbitrary conductor in
the energy standpoint.
a linear magnetic
medium. From Eq.
(or steady) current of intensity
inductance) of the conductor,
in the
magnetic
due
field
to
i,
i
(7.88),
if
we assume
L can be expressed
,
Wm
,
a slowly time-varying
to flow in the conductor, the inductance (selfin
terms of the energy contained
as
L=
inductance from magnetic
2
W
n
(7.127)
energy
This expression can be viewed as the third equivalent definition of self-inductance,
the other two being the flux definition in Eq. (7.1) and emf definition in Eq. (7.2). It
can hence be used, as an alternative general means for computing the inductance of
different structures,
where the magnetic energy of the structure
is
computed using
the integral expression in Eq. (7.109).
Since the energy
Wm
can be written as the
sum
of energies localized inside and
outside the conductor,
VFm
=
Wmi + Wme
,
(7.128)
Section 7.6
L
the inductance
and External Inductance
Internal
in
Terms of Magnetic Energy
can be decomposed accordingly into the internal inductance,
and external inductance,
Le
,
343
Lu
of the conductor:
L=
+ Le
Li
(7.129)
.
In other words,
—
j
2Wmi
^
,
Lq
and
,rs
—
2Wme
jz
(7.130)
,
internal
and
external
inductances
where the internal and external magnetic energies of the conductor are obtained by
integrating the energy density over the conductor interior (volume Vj) and exterior
(volume v e ), respectively, that is,
W
n
More
precisely, v e
is
ml
all
-
H dv
W
and
B Hdv.
n
(7.131)
only that part of the conductor exterior which
the magnetic field (there
In
B
is
no need
occupied by
is
to integrate zero energy density).
of the examples of Section 7.1,
we
only evaluated the external inductance
of conductors, by employing the flux (or emf) definition of self-inductance with only
the external magnetic flux taken into account. Evaluation of Lj from the internal
magnetic flux is also possible, but is physically less clear and mathematically more
complicated than from the internal stored magnetic energy.
At high frequencies, due to skin effect (see Fig. 6.23), the current and magnetic
field in a conductor are confined to a very thin region on the surface of the conductor, which considerably reduces the high-frequency value of the internal inductance
of the conductor from its low-frequency (or dc) value. Therefore, in most highfrequency applications, Lj is negligible as compared to L e and assumption that
L « L e yields very accurate results.
,
Example 7.25
Conductor
Internal Inductance p.u.l. of a Cylindrical
Find the internal inductance per unit length of an infinitely long cylindrical conductor of
radius
a.
The permeability of the conductor
Assume
Solution
tributed over
conductor
i,
that
is
its
is pr.
r
is
not pronounced.
that the conductor carries a current of intensity
cross section (skin effect not pronounced).
<
given by the expression for r
that
i
The magnetic
a in Eq. (4.56) with
/iq
is
uniformly
dis-
flux density in the
substituted by
p and 1 by
is,
B{r)
with
Skin effect
being the distance from the conductor
=
pir
2na 2
axis, as
(7.132)
’
Figure 7.26 Evaluation of
shown in Fig.
7.26.
From
the
first
sion in Eqs. (7.131), the internal magnetic energy per unit length of the conductor
is
expres-
obtained
by integrating the magnetic energy density over the cross section S of the conductor [see
Eqs. (2.206) and (2.207) for the similar integration of the electric energy density],
=
where dS
The
is
I
^^
the low-frequency internal
inductance per unit length
of an infinitely long
conductor of
cylindrical
permeability
P
p
;
for
Example 7.25.
d'
= l£'
f7' 1
33 >
the surface area of an elementary ring of radius r and width dr (Fig. 7.26).
equation of Eqs. (7.130) then gives the following expression for the lowfrequency internal inductance per unit length of the conductor:
first
L =
,
low-frequency internal
2W'.
—
(7.134)
8tt
inductance
cylindrical
p.u.l.
of a
conductor
344
Chapter 7
Inductance and Magnetic Energy
We
note that this inductance
=
conductors, L[
Determine the
Two-Wire Line
low-frequency inductance per unit length of the
assuming that the conductors are nonmagnetic.
total (internal plus external)
line in Fig. 7.4,
Solution
independent of the conductor radius. For nonmagnetic
is
50 nH/m.
Internal Inductance p.u.l. of a Thin
Example 7.26
two-wire
=
po/(8n)
is much larger than the
magnetic energies of individual conductors (for an assumed lowthe line) can be evaluated independently from each other. Therefore,
Since the distance between the conductors of the line
conductor
radii, the internal
frequency current in
the low-frequency internal inductance per unit length of the line can be found as twice that
of a single isolated conductor, Eq. (7.134) with
p = po (conductors
are nonmagnetic), which
yields
=
L[
+
L[ x
L[ 2
=
=
2L'
^=
Using the expression for the external inductance
low-frequency per-unit-length inductance
100
nH/m.
p.u.l.
of the
(7.1 35)
line,
Eq. (7.11),
p.u.l.
total
(7.136)
low-frequency total
inductance
its
is
of a thin
two- wire line
Inductance from Energy for a Coaxial Cable
Example 7.27
Find the total low-frequency inductance per unit length of the coaxial cable from
Example
4.11.
Solution
If
we assume
that a dc current of intensity I
is
Eqs. (4.61)-(4.63).
established in the cable conductors,
from the cable
as in Fig. 4.17, the magnetic flux density at a distance r
We use the energy definition of inductance, Eq.
axis, B(r), is
(7.127).
The
given by
total (internal
plus external) magnetic energy per unit length of the cable can be obtained by integrating
the magnetic energy density over the cross section of the cable (the magnetic field outside
the cable, for r
>
c, is
zero) in the
same manner
as in Eq. (7.133),
^2nrdr.
Jr = 0
As
<
the function B{r) for 0
(4.62),
we break
and from b
to
c.
”
m
r
<
c
is
(7.137)
-^M0
given by three different expressions in Eqs. (4.61) and
the integration with respect to r
up
from 0 to a from a to
into three parts:
.
b,
This yields
Mo l
2
MO I
2
An
16tt
W'
.
b
p0 I
4n(c2
a
2
-b
c
(
2
)
\c2
4
-
3c
c
2
-
b2
(7.138)
b2
b
4
W’
tv:
where W7, W'me and WU 2 are the magnetic energies residing in the inner conductor, dielectric, and outer conductor of the cable per unit of its length, respectively. The total dc (or
low-frequency) inductance per unit length of the cable is hence
,
,,
low-frequency total
inductance
p.u.l.
2
,
Wn
7
of a coaxial
2
Mo[l
2;r
|_4
1
c*
/
4
c
- b 2 \c2 -
c
b*
b
3c
2
-b 2
4
\
)
Mo,
b
+ =-lna
2n
,
p
(7.139)
cable
where
cable.
L'-
and L'e are the internal and external inductance, respectively, per unit length of the
1
Problems
The above
result for L'e
is,
of course, the
same
using the flux definition of inductance. Note that the
magnetic energy
345
which was obtained
component of Lj, that corresponding
as that in Eq. (7.12),
first
inductance per unit length
Eq. (7.134) with \l = hq.
As a numerical example, the internal and external inductances per unit length of a
coaxial cable with a = 1 mm, b = 4 mm, and c = 5
amount to L[ = 66.6 nH/m and
to the
in the inner conductor, equals the internal
of an isolated cylindrical conductor in
air,
mm
L;
= 277.3 nH/m.
Problems’. 7.36-7.38; Conceptual Questions (on
MATLAB Exercises
(on
Companion Website):
7.34;
Companion Website).
Problems
7.1.
Induced emf and voltage of an inductor. A
current of intensity i(t) = Iq e~ f/ r where Iq
and r > 0 are constants, flows through an
c
•
a
—
,
\
i
inductor of inductance L. Calculate (a) the
magnetic flux, (b) the induced emf, and
M2
N
(c) the
i
j
Ml
voltage of the inductor. Specify the reference
directions/orientations for
7.2.
1
0
Inductance of a solenoid with a two-layer
core. If the solenoidal coil described in
Example 7.1 is wound over a ferromagnetic
core with two coaxial layers of relative perme-
=
abilities /x r i
= 500
in Fig. 7.27,
where the cross-sectional area of
the inner layer
tance of the
is
and
Si
/x r 2
= 1 cm 2
,
1000, as
J/z/2
i
quantities.
all
<!
l
i
Figure 7.28 The same toroidal
Fig. 7.3
coil as in
but with ferromagnetic layers
stacked on top of one another; for
Problem
shown
7.3.
layers of permeability
find the induc-
/x
and thickness
a,
analogously to the line with dielectrically
coil.
coated conductors in
Fig. 2.31.
Determine the
external inductance per unit length of the
new
7.6.
Figure 7.27 Solenoid with
a core
composed
of
two
coaxial linear ferromagnetic
layers; for
7.3.
Problem
7.2 but for the two-layer core
7.7.
7.4.
Inductance of a coil in a simple linear magnetic circuit. Find the inductance of the coil in
the simple linear magnetic circuit with an air
gap from Example 5.14.
Planar line with two magnetic layers. For the
planar transmission line in Fig. 3.24, assume
that the two material layers between metallic
strips are lossless (cq
with permeabilities
d2 so
,
7.5.
Two-wire
with a piece-wise homogeneous ferro-
magnetic material composed of two coaxial
layers of relative permeabilities /x r i and /x r 2
and dimensions as in Fig. 2.50.
shown
in Fig. 7.28.
Coaxial cable filled with an inhomogeneous
magnetic material. Compute the external
inductance per unit length of a coaxial cable
filled
7.2.
Toroidal coil with a two-layer core. Repeat
Example
line.
=02 =
/x 1
and
0),
/X 2 -
but magnetic,
Let w ;» d\ +
that fringing effects can be neglected,
with ferromagnetic coatings
over conductors. Let the conductors of the
i.e.,
thin symmetrical two-wire transmission line
the two layers, and not in the air around
in
Fig.
line
7.4
be
coated
by
ferromagnetic
the magnetic field of the line can be con-
sidered to be uniform and localized only in
(also see the magnetic field
computation
in
2
346
Inductance and Magnetic Energy
Chapter 7
Example
Under
11.10).
these circumstances,
(Fig.
7.8.
External inductance
What
of a three-wire
p.u.l.
line.
length of the transmission line consisting of
7.14.
an isolated wire and two short-circuited wires
7.9.
intensity
A superconducting square contour
L
of edge length a and inductance
uated outside any magnetic
field,
7.15.
the contour and (b) the current intensity
7.11.
new
in the
steady state.
A
phone line due to a nearby power
power conductor of a cable car and
the distances of axes of wires of this line
from the power-conductor axis are d\ — 5.5 m
and d2 = 5.9 m, respectively. Both lines can
be considered to be thin. If a time-harmonic
wire line in both cases in Fig. 6.32, in the
current of amplitude (peak-value) Iq
following two ways: (a) from the definition of
and frequency /
mutual inductance in Eq. (7.21) or (7.23) and
(b) using Eq. (7.40), respectively.
power line, find the amplitude of the induced
emf per / = 1 km length of the phone line,
inductance
one shown
in Fig. 5.26,
assuming that the two
7.16.
being a
—
3 cm,
b
—
6 cm, and h
pled
=
i.e.,
lines are in free space.
Equivalent input inductance of structures
with coupled
with
the inner and outer radii and height of the
toroid
= 150 A
= 60 Hz is established in the
neglecting the influence of the earth,
Mutual inductance of two coils on a thick
toroidal core. Repeat Example 7.7 but for
section, as the
coils,
Two
coils.
wound on
inductances L\
—
magnetically cou-
a cardboard core, have
— L2 —
50
\x
H
and coupling
Calculate the equivalent
2 cm, respectively.
coefficient k
Mutual inductance between a toroidal coil
and axial wire. Consider the thick toroidal
inductance between terminals 1 and 2 of the
structure for connections between coils as in
Fig. 7.29(a), (b), and (c), respectively.
coil
with
Fig. 4.18,
a
rectangular
section
cross
and assume that an
infinitely
in
long
7.17.
the coil and the wire, computing
L
and
\
(b)
it
both as (a)
L 2 \.
Mutual inductance between a solenoid and
rectangular loop.
Assume
the solenoid in Fig. 6.29
the total
N=
If
the
number
that the length of
is / (/
of wire turns
a), so that
amounts
to
open.
N'l, as well as that its terminals are
the rectangular loop positioned around
solenoid
at
the
middle of
its
length
Equivalent inductance for a unity coupling
Repeat the previous problem
assuming that the cardboard core is replaced
by a linear ferromagnetic one, with which
L\ — L 2 = 60 mH and k = 1.
Find the mutual inductance between
(z-axis).
0.1.
coefficient.
straight wire runs along the axis of the toroid
7.13.
in a
line.
between a loop and
two-wire line. Find the mutual inductance
between the rectangular loop and the two-
Mutual
a thick toroidal core of a rectangular cross
7.12.
Emf
approximated by a two-wire line, in a vertical
plane, with distance between wire axes h =
5 m. In the same horizontal plane containing the power conductor, running in parallel
to it, there is a two-wire telephone line, and
tion of B, find (a) the magnetic flux through
7.10.
openin the
track form a transmission line that can be
the right-hand rule with respect to the direc-
-
is
compute the induced emf
large loop.
in a
it
flows along the small circular
i(t)
circuited,
is
in it. The contour is then brought
uniform time-invariant magnetic field of
flux density B and positioned so that the vector B is perpendicular to the plane of the contour. Orienting the contour in accordance to
the two-loop system
time-varying current of
loop, while the large square loop
is first sit-
and there
no current
along
If in
in Fig. 6.33 a slowly
Bringing a superconducting contour in a magnetic field.
time-
Mutual inductance between large and small
concentric loops.
described in Example 2.16?
in air
low-frequency
a
of the solenoid.
the external inductance per unit
is
carries
6.29)
harmonic current of intensity i = /o cos cot,
determine the voltage between the terminals
find the p.u.l. ex