Slide Presentations for ECE 329,
Introduction to Electromagnetic Fields,
to supplement “Elements of Engineering
Electromagnetics, Sixth Edition”
by
Nannapaneni Narayana Rao
Edward C. Jordan Professor of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, Urbana, Illinois, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India
1.6
The Magnetic Field
1.6-2
The Magnetic Field
acts to exert force on charge when it is in motion.
Fm  qv  B
B
Fm
v
q
B = Magnetic flux density vector
Alternatively, since charge in motion constitutes
current, magnetic field exerts forces on current
dFm
elements.
B
dFm  I dl  B
I dl
1.6-3
N
N–m
Units of B:
=
2
A m A m
Wb
= 2 =T
m
Sources: Currents;
Time-varying electric field
1.6-4
Ampère’s Law of Force
I1
a21 dl2
I2
a
dl1 12 R
 0 I 2 d l 2 × a 21 
d F1  I1d l1 × 

2
4

R


 I1d l1 × B 2
 0 I1 d l1 × a12 
d F2  I 2 d l 2 × 

2
4

R


 I 2 d l 2 × B1
1.6-5
Magnetic field due to a current element
(Biot-Savart Law)
0 I dl  a R
B
4
R2
Note B  sin 
1
B
R2
I dl

aR
R
P
B
B circular to the axis of the current element
0  Permeability of free space
= 4  10 –7 H m
1.6-6


Ex. I d l  I dx a x  a y A situated at 1,  2, 2  .
Find B at  2,  1, 3 .
0 I d l × a R 0 I d l × R
B

2
4
R
4
R3
R

 since a R  
R

R =  2  1 a x   1  2  a y   3  2  a z
 ax  a y  az
1.6-7
0 I dx  a x  a y  ×  a x  a y  a z 
B=
3
4
3
 
0 I dx

ax  a y 

12 3
1.6-8
Current Distributions
I
(a) Filamentary Current
I (A)
Thin
wire
(b) Surface Current
Surface current density, JS (A/m)
JS
w
I 

JS 
w max
1.6-9
(c) Volume Current
Density, J (A/m2)
area A
J
I 

J   
A max
1.6-10
P1.44
z
a2
P(r, f, z)
2

dz
aR
z – z
z I
r
y
1
x
a1
r
dB 

0 dz a z ×a R
2
2

4 r   z  z   


0 I dz sin  af
2
2

4 r   z  z   


1.6-11
B
z  a2
z   a1
0 I

4 r 2
z  z
cot  
r
dB

d z  sin  af
1   z  z r  
2
0 I  cosec 2 sin  d

af
2




4 r
cosec 
0 I

  cos     af
4 r
0 I

 cos 1  cos  2  af
4 r
2
1
2
1
1.6-12
For infinitely long wire,
a1  – , a2  ,
1  0 ,  2  
0 I
B
af
2r
1.6-13
Magnetic Field Due to an Infinite Plane Sheet of
Uniform Surface Current Density
This can be found by dividing the sheet into
infinitely long strips parallel to the current density
and using superposition, as in the case of finding the
electric field due to an infinite plane sheet of uniform
surface charge density. Instead of going through this
procedure, let us use analogy. To do this, we first
note the following:
1.6-14
(a) Point Charge
P
Current Element
E
P
ar
B
R
I dl
ar
R
Q
E
Q
4 0 R2
aR
0 I dl  a R
B
4R2
1.6-15
(b) Line Charge
z
L0
r
Line Current
z
I
P
r
ar E
r=0
L0
E
ar
2 0 r
P
B ar
r=0
0 I
B
af
2 r
0 I

a z  ar
2 r
1.6-16
Then,
(c)
Sheet Charge
S0
Sheet Current
JS
P
P
an E
S0
E
an
2 0
B
an
B
0
JS  a n
2