College of Engineering
Fayoum University
Electromagnetic
Fields
Dr. Tarek Said
Electrical Engineering
tms02@fayoum.edu.eg
Recommended Textbook
Content:
 MAGNETIC MATERIALS.
 ELECTROMAGNETIC ENERGY.
 TIME VARYING FIELDS.
 INTRODUCTION TO ELECTROMAGNETIC WAVE
 PROPAGATION IN FREE SPACE.
Grading Policy
Exams
 Quizzes (20 point)
 Midterm Exam (40 point)
 Final Exam (90 point)
Students can miss a midterm exam without
penalty only in serious unpredictable situations
like death in the immediate family, or a serious
accident, or a serious disease that leaves you
incapacitated. Experience shows that most
students that miss a midterm exam for one
reason or another often end up failing the final
exam, and therefore fail the course.
Classroom Etiquette!
•
•
•
•
•
Do arrive on time!
Do be present in Body and Mind!
Do stay AWAKE and be attentive!
Do stay for the whole class!
Always be respectful of yourself, and your
classmates.
• Do put cell phones away and on vibrate…
• Do be prepared for class!
• Do show effort, try your best, ask for help!
Electromagnetic Fields
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What is Electromagnetic Wave?
• By the very nature of the word, electromagnetics implies
having to do with a phenomenon involving both electric and
magnetic fields and furthermore coupled. This is indeed the
case when the situation is dynamic, that is, time-varying,
because time-varying electric and magnetic fields are
interdependent, with one field producing the other.
• In other words, a time-varying electric field or a time-varying
magnetic field cannot exist alone; the two fields coexist in time
and space, with the space-variation of one field governed by
the time-variation of the second field. This is the essence of
Faraday’s law and Ampere’s circuital law, the first two of the
four Maxwell’s equations resulting in wave propagation.
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What is Electromagnetic Wave?
Electric field
Produced by the presence of
electrically charged particles,
and gives rise to the electric
force.
Magnetic field
Produced by the motion of
electric charges, or electric
current, and gives rise to the
magnetic force associated
with magnets.
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• Electromagnetic waves travel VERY FAST
– around 300,000 kilometres per second
(the speed of light).
At this speed they can
go around the world 8
times in one second.
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Electromagnetic Spectrum
Radio Spectrum
Optical Spectrum
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Electromagnetic Spectrum
• Frequencies is called the electromagnetic spectrum.
• Different parts interact with matter in different ways.
• The ones humans can see are called visible light, a
small part of the whole spectrum.
• Antenna of a radio detects radio waves.
• A radio picks up radio waves
through an antenna and converts it
to sound waves.
Each radio station in an area
broadcasts at a different
frequency.
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Electromagnetic Spectrum
 Microwaves are radio waves with higher frequency
than radio waves & shorter wavelength.


Cell phones and satellites use microwaves between
1 cm & 20 cm for communication.
In microwave ovens, a
vibrating electric field causes
water molecules to rotate
billions of times per second
causing friction, which in
turn transfers energy in the
form of heat to the food.
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Electromagnetic Spectrum
• X- RAYS have shorter wavelength and higher
frequency than UV-rays
• Carry a great amount of energy
• Can penetrate most matter.
• Bones and teeth absorb x-rays. (The
light part of an x-ray image indicates a
place where the x-ray was absorbed)
• Too much exposure can cause cancer
(lead vest at dentist protects organs
from unnecessary exposure)
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Electrostatic Fields
o Only when the fields are not changing with time, that is, for
the static case, they are independent; a static electric field or a
static magnetic field can exist alone.
o Thus, in the entire frequency spectrum, except for dc, all
electrical phenomena are governed by interdependent electric
and magnetic fields, or electromagnetic fields.
Statics
Dynamics
dc
Frequency, f
Light
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Quasistatic Approximation
However, at low frequencies, an approximation,
known as the “quasistatic approximation,” can be
made in which the time-varying fields in a physical
structure are approximated to have the same spatial
variations as the static fields in the structure obtained
by setting the source frequency equal to zero.
As the frequency becomes higher and higher, this
approximation violates the actual situation more and
more, and it becomes increasingly necessary to
consider the wave solution.
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Quasistatic Approximation
Statics
Quasistatics
Dynamics
dc
Statics: f = 0;
Frequency, f
Light
  t  0 ; dc
Dynamics: No restriction; complete Maxwell’s equations;
Electromagnetic waves
Quasistatics: low-frequency approximation of dynamics;
d << λ (d is the dimension of the circuit or device).
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Power Plant Frequency (f) is @ 60 Hz
6
Wavelength (l) is 5 10 m
( Over 3,100 Miles)
Consumer
Home
Power
Plant
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Magnetostatics
Fayoum University
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Electricity => Magnetism
• In 1820 Oersted discovered that a steady
current produces a magnetic field while
teaching a physics class.
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Magnetism => Electricity ?
Eleven years later, and at
the same time, Mike
Faraday in London and
Joe Henry in New York
discovered that a timevarying magnetic field
would produce an electric
current!
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Maxwell’s Equations for Magnetostatics
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Maxwell’s Equations for Magnetostatics
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Maxwell’s Equations for Magnetostatics
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Maxwell’s Equations for Magnetostatics
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The Integral Form of Magnetostatics
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The Integral Form of Magnetostatics
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Scalars and Vectors




Scalar refers to a quantity whose value may be represented by
a single (positive or negative) real number.
Some examples include distance, temperature, mass, density,
pressure, volume, and time.
A vector quantity has both a magnitude and a direction in
space. We especially concerned with two- and threedimensional spaces only.
Displacement, velocity, acceleration, and force are examples of
vectors.
• Scalar notation:
• Vector notation:
A or A
(italic or plain)
→
A or A
(bold or plain with arrow)
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Rectangular Coordinate System
Differential surface units:
dx  dy
dy  dz
dx  dz
Differential volume unit :
dx  dy  dz
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Vector Components and Unit Vectors
R PQ ?
r  xa x  ya y  za z
a x , a y , a z : unit vectors
R PQ  rQ  rP
 (2a x  2a y  a z )  (1a x  2a y  3a z )
 a x  4a y  2a z
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Vector Components and Unit Vectors
For any vector B, B  Bxa x  By a y + B:z a z
B  Bx2  By2  Bz2  B
aB 
Magnitude of B
B
B

2
2
2
B
Bx  By  Bz
Unit vector in the direction of B
Example
Given points M(–1,2,1) and N(3,–3,0), find RMN and aMN.
R MN  (3a x  3a y  0a z )  (1a x  2a y  1a z )  4a x  5a y  a z
a MN
4a x  5a y  1a z
R MN
 0.617a x  0.772a y  0.154a z


2
2
2
R MN
4  (5)  (1)
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The Dot Product
Given two vectors A and B, the dot product, or scalar product, is
defines as the product of the magnitude of A, the magnitude of B,
and the cosine of the smaller angle between them:
A  B  A B cos  AB
The dot product is a scalar, and it obeys the commutative law:
AB  B A
For any vector
A  Axa x  Ay a y + Az a z and B  Bxa x  By a y + Bz,a z
A  B  Ax Bx  Ay By + Az Bz
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The Cross Product
Given two vectors A and B, the magnitude of the cross product, or
vector product, written as AB, is defines as the product of the
magnitude of A, the magnitude of B, and the sine of the smaller angle
between them.
The direction of AB is perpendicular to the plane containing A and B
and is in the direction of advance of a right-handed screw as A is
turned into B.
ax  a y  az
A  B  a N A B sin  AB
a y  az  ax
az  ax  a y
The cross product is a vector, and it is not
commutative:
(B  A)  (A  B)
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The Cylindrical Coordinate System
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The Cylindrical Coordinate System
Differential surface units:
d   dz
 d  dz
d    d
Differential volume unit :
d    d  dz
Relation between the rectangular and
the cylindrical coordinate systems
x    cos 
y    sin 
  x2  y 2
1 y
  tan
zz
zz
x
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The Cylindrical Coordinate System
az
az
?
A  Axa x  Ay a y + Az a z  A  A a   A a + Aza z
a
ay
a
A  A  a 
 ( Axa x  Ay a y + Az a z )  a 
 Axa x  a   Aya y  a  + Az a z  a 
 Ax cos   Ay sin 
ax
A  A  a
 ( Axa x  Ay a y + Az a z )  a
Dot products of unit vectors in cylindrical
 Axa x  a  Aya y  a + Az a z  a
and rectangular coordinate systems
  Ax sin   Ay cos 
Az  A  a z
 ( Axa x  Ay a y + Az a z )  a z
 Axa x  a z  Ay a y  a z + Az a z  a z
 Az
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The Spherical Coordinate System
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The Spherical Coordinate System
Differential surface units:
dr  rd
dr  r sin  d
rd  r sin  d
Differential volume unit :
dr  rd  r sin  d
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The Spherical Coordinate System
Relation between the rectangular and the spherical coordinate systems
x  r sin  cos 
r  x2  y 2  z 2 , r  0
y  r sin  sin 
  cos 1
z  r cos
  tan 1
z
x y z
2
2
y
x
Dot products of unit vectors in spherical and
rectangular coordinate systems
2
, 0    180
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Charge
proton: q = 1.602  10-19 [C]
electron: q = -1.602  10-19 [C]
1 [C] = (1 / 1.602 x10-19) protons = 6.242 x 1018 protons
e
p
Atom
Ben Franklin
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Charge Density
1) Volume charge density v [C/m3]
a) Uniform (homogeneous) volume charge density
v
Q
v 
[C/m3 ]
V
V
Q  v  x, y, z  V
Q
b) Non-uniform (inhomogeneous) volume charge density
Q   v  x, y, z  dV
V
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Charge Density (cont.)
2) Surface charge density s [C/m2]
S
s (x,y,z)
Q
Uniform
s 
Q
S
dQ  s  x, y, z  dS
[C/m2 ]
Non-uniform
Q    s  x, y, z  dS
S
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Charge Density (cont.)
3) Line charge density l [C/m]
++++ + +
+
+
+
+
+
+
+ +
l
l (x,y,z)
Q
Uniform
Q
l 
[C/m]
l
Non-uniform
dQ  l  x, y, z  dl
Q   l  x, y, z  dl
C
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Example
z
v = 10 [C/m3]
a
y
x
Find: Q
Q   v  x, y, z  dV
V
  10 dV
Q  10 V
4 3
=10   a 
3

V
 10  dV
V
40 3
Q   a [C]
3
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Example
z
v = 2r [C/m3], r < a
a
y
Find: Q
x
Q   v  x, y, z  dV
V
  2r dV
V
2  a

2
2
r
r
sin  dr d d


 
0 0 0
Q
2

 sin  d
2
2
r
r
dr



0
0
0
 d
a
1 
  2  2   a 4 
2 
Q  2 a 4 [C]
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Example: Find the Equivalent Surface Charge Density
for a Thin Slab of Charge
y
v  2 xyz C/m3  ,
0 zh
h
x
V
seq  x, y  xyh2 C/m2 
z
Q 

V

v dV  
S
eq
s

h
0
v dz dS 

 seq dS
S
 x, y   0 v  x, y, z  dz
h
S


h
0
2 xyzdz  xyh 2 C/m 2 
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Current
Current is the flow of charge:
the unit is the Ampere (Amp)
1 Amp = 1 [C/s]
Sign convention:
a positive current
flowing one way is equivalent to a negative
current flowing the other way.
Ampere
flow rate is 1 C per second
+
+
+
+
velocity
1 [A]
or
-1 [A]
Note: The arrow is called the reference direction arrow.
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Current (cont.)
Mathematical definition of current
reference direction arrow
I
Q
Q = amount of charge (positive or negative) that crosses the plane in the
direction of the reference arrow in time t.
Q
I
t
More generally,
dQ
i t  
dt
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Current Density Vector J
The magnitude of the current-density vector tells us the current that is
crossing a small surface perpendicular to the current-density vector.
+
+
+
+
+
+
+
+
+
v
+
+
+
S
I = the current crossing the surface S in the direction of the
velocity vector.
I
J
[A/m 2 ]
S
 I 
J 
 vˆ
 S 
[A/m 2 ]
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Current Density Vector (cont.)
v
Q
S
+
+
+
L
+
I
reference plane
L = distance traveled by charges in time t.
I
Q / t  Q / t  L  Q  L 
J





S
S
S L
 V  t 
or
J  v v
Hence
so
J  J vˆ   v v  vˆ
J  v v
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Current Crossing Surface
I   J  nˆ  S
Integrating over the surface,
I   J  nˆ dS
S
Note: The direction of the unit normal
vector determines whether the current
is measured going in or out.
n̂
J
S
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Example
J  xˆ  3x 2 y   yˆ  3z 3 y 2   zˆ  3xy  [A/m2 ]
z
(0,1,0)
(1,0,0)
y
x
S
Find the current I crossing the surface S in the upward direction.
I   J  nˆ dS   J  zˆ dS   3xy dS
S
S
S
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Example
y
I   3xy dS
S

 

0
1
1

0
y( x)

0
y=1-x

3 xy dy  dx


1
x
1
y x
3 2
 2 xy 
0
dx
1
3
2
   x 1  x   dx

20 
I  0.125 [A]
1
3 1 4 2 3 1 2 
  x  x  x 
2 4
3
2 0
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Electric Flux
E
A
area A
We define the electric flux ,
of the electric field E,
through the surface A, as:
 = E .A
 = E A cos ()
Where:
A is a vector normal to the surface
(magnitude A, and direction normal to the surface).
 is the angle between E and A
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Electric Flux
You can think of the flux through some surface as a measure of
the number of field lines which pass through that surface.
Flux depends on the strength of E, on the surface area, and on
the relative orientation of the field and surface.
E
E
A
Normal to surface,
magnitude A
area A
Here the flux is
=E·A
A 
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In the case of a closed surface
   d   E  dA
The loop means the integral is over a closed surface.

E
dA
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For a closed surface:
The flux is positive for field lines that leave the enclosed
volume
The flux is negative for field lines that enter the enclosed
volume
If a charge is outside a closed surface, the net flux is zero.
As many lines leave the surface, as lines enter it.
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Gauss’s Law
The total flux within
a closed surface …
   E  dA =
… is proportional to
the enclosed charge.
Qenclosed
0
Gauss’s Law is always true, but is only useful for certain
very simple problems with great symmetry.
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Differential Form of Gauss' Law
GAUSS' LAW
Think about a region of space, enclosed by a box.
Divide Gauss' law by the volume of the box:
E || x
Take the limit
of a small box
Work on the left hand side of the equation:
For a general case where E can point in any direction:
GAUSS' LAW
Differential Form
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Applying Gauss’s Law: Electric field produced by a
 E . dA = Q / 0
point charge
E
 E . dA = E  dA = E A
A = 4  r2
E A = E 4  r2 = Q / 0
Q
E
1 Qq
E
4 0 r 2
k = 1 / 4  0
0 = permittivity
0 = 8.85x10-12 C2/Nm2
Coulomb’s Law !
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Applying Gauss’s Law: Infinite sheet of charge
Gauss’s law is useful only when
the electric field is constant on a
given surface
1. Select Gauss surface
In this case a cylindrical
pillbox
2. Calculate the flux of the
electric field through the
Gauss surface
 = 2 EA
3. Equate  = qencl/0
2EA = qencl/0
4. Solve for E
E = qencl / 2 A 0 =  / 2 0
(with  = qencl / A)
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Problem: Sphere of Charge Q
A charge Q is uniformly distributed through a sphere of radius R.
What is the electric field as a function of r?. Find E at r1 and r2.
r1
r2
R
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E(r1)
r1
E(r2)
r2
R
Use symmetry!
This is spherically symmetric.
That means that E(r) is radially
outward, and that all points, at a
given radius (|r|=r), have the same
magnitude of field.
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First find E(r) at a point outside the charged sphere. Apply Gauss’s
law, using as the Gaussian surface the sphere of radius r pictured.
E & dA
What is the enclosed charge? Q
What is the flux through this surface?
r
   E  dA   E dA
 E  dA  EA  E(4 r )
2
R
Gauss:
  Q / o
Q/ 0    E(4 r )
2
Exactly as though all the
charge were at the origin!
(for r>R)
So
1
Qˆ
E(r ) 
2 r
4 o r
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Next find E(r) at a point inside the sphere. Apply Gauss’s law,
using a little sphere of radius r as a Gaussian surface.
What is the enclosed charge?
E(r)
That’s given by volume ratio: Q enc
r
R
Setting
r3
 3Q
R
Again the flux is:  = EA = E(4 r )
2
  Qenc /  o gives
For r < R
E(r) =
(r 3 / R 3 )Q
E=
4 o r 2
Q
4 o R
ˆ
r
r
3
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Problem: Sphere of Charge Q