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Fundamentals of Electromagnetics:
A Two-Week, 8-Day, Intensive Course for
Training Faculty in Electrical-, Electronics-,
Communication-, and Computer- Related
Engineering Departments
by
Nannapaneni Narayana Rao
Edward C. Jordan Professor Emeritus
of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, India
Amrita Viswa Vidya Peetham, Coimbatore
August 11, 12, 13, 14, 18, 19, 20, and 21, 2008
2-1
Module 2
Maxwell’s Equations
In Integral Form
The line integral
The surface integral
Faraday’s law
Ampere’s circuital law
Gauss’ Laws
The Law of Conservation of Charge
2-2
Instructional Objectives
5. Apply Faraday's law in integral form to find the
electromotive force induced around a closed loop,
fixed or revolving, for a given magnetic field
distribution
6. Apply Ampere's circuital law in integral form to find
the magnetomotive force induced around a closed
path, for a given current and/or electric field
distribution
7. Apply Gauss' law for the electric field in integral
form, Ampere's circuital law in integral form, the
law of conservation of charge, and symmetry
considerations, to find the line integral of the
magnetic field intensity around a closed path, given
an arrangement of point charges connected by wires
carrying currents
2-3
The Line Integral
(FEME, Sec. 2.1; EEE6E, Sec. 2.1)
2-4
The Line Integral
Work done in carrying a charge from A to B in an
electric field:
B
E2
A
E1 a
2
a1 l2
l1
n
WAB   dWj
j1
2-5
dWj   qE j cos a j  l j 
 qE j l j cos a j
 qE j l j
n
WAB  q
E
j
• l j
j 1
VAB
WAB


q
n

E j • l j (Voltage between
A and B)
j 1
2-6
In the limit n   ,
B
VAB  A E • dl
= Line integral of E
from A to B.

E  d l = Line integral of E
C
around the closed
path C.
2-7
A
If
R
C
C
L
B
 ARBLA E

C
E dl = 0
B
then  E d l
A
is independent of
the path from A to B
(conservative field)
d l ARB E d l  BLA E d l
 ARB E d l  ALB E d l  0
ARB E
d l = ALB E d l
2-8
Ex. For F  yza x  zxa y  xya z , find

(1,2,3)
F
d l along the straight line paths,
(0,0,0)
from (0, 0, 0) to (1, 0, 0), from (1, 0, 0) to
(1, 2, 0) and then from (1, 2, 0) to (1, 2, 3).
z
(0, 0, 0)
(1, 0, 0)
x
(1, 2, 3)
y
(1, 2, 0)
2-9
From (0, 0, 0) to (1, 0, 0),
y  z  0 ; dy  dz  0
F0 ,
(1,0,0)
(0,0,0) F • dl  0
From (1, 0, 0) to (1, 2, 0),
x  1, z  0 ; dx  dz  0
F  ya z
dl  dx a x  dy a y  dz a z  dy a y
F • dl  0,
(1,2,0)
(1,0,0) F • dl  0
2-10
From (1, 2, 0) to (1, 2, 3),
x  1, y  2 ; dx  dy 0
F  2za x  za y  2a z , dl  dz a z
F • dl  2 dz ,

(1,2,3)
(1,2,3)
(1,2,0) 2 dz  6
(0,0,0) F • dl  0  0  6  6
2-11
In fact, F d l   yzax  zxa y  xyaz 
 dx a
x
 dy a y  dz az 
 yz dx  zx dy  xy dz
 d  xyz 
1,2,3

0,0,0 
F d l 
1,2,3
 0,0,0 
d  xyz    xyz 
1,2,3
 0,0,0 
 1 2  3  0  0  0
 6, independent of the path.
2-12
The Surface Integral
(FEME, Sec. 2.2; EEE6E, Sec. 2.2)
2-13
The Surface Integral
Flux of a vector crossing a surface:
B
a nn
Flux = (B)(S)
S
B
S
B
an
a
S
an
Flux = 0
Flux  (B cos a ) S
 B S cos a
 B • S a n
 B • S
2-14
Normal
anj
aj
Bj j
B
Flux 
Sj
S

In the limit n   ,
Flux,  =
n
  j
j1
n
B j
j1
• S j
S B • dS
= Surface integral of B over S.
2-15
S B • dS = Surface integral of B
over the closed surface S.
D2.4
(a)
A  x  ax  a y 
z
x  0, a n   a x
A  0, A • dS  0
2
 A • dS  0
 A • dS  0
2
x
y
2-16
(b)
z
2
x  2, an   a x
A  2  ax  a y 
d S   dy dz a x
A • dS   2 dy dz
2
2
x
2
2
 A • dS   y 0 z 0 2 dy dz   8
 A • dS  8
y
2-17
z
(c)
2
y  0, an   a y
A  x  ax  a y 
d S   dz dx a y
y
2
x
A d S   x dx dz
2
2
 A • dS   x 0 z 0 x dx dz   4
 A • dS
4
2-18
(d) From (c),
A • dS   x dx dz
2– x
2
 A • dS   x0 z 0 x dx dz
2
  0 x(2 – x) dx
4

3
4
 A • dS  3
z
2
x+z=2
y
x
2
2-19
Faraday’s Law
(FEME, Sec. 2.3; EEE6E, Sec. 2.3)
2-20
Faraday’s Law
d
C E • dl  – dt S B • dS
B
S
C
dS
2-21
C E • dl = Voltage around C, also known as
electromotive force (emf) around C
(but not really a force),
 V m  m, or V.
S B • dS = Magnetic flux crossing S,
2
2
Wb
m

m

 , or Wb.
d
– S B • dS = Time rate of decrease of
dt
magnetic flux crossing S,
Wb s, or V.
2-22
Important Considerations
(1) Right-hand screw (R.H.S.) Rule.
The magnetic flux crossing the
surface S is to be evaluated toward
that side of S a right-hand screw
advances as it is turned in the sense of C.
C
2-23
(2) Any surface S bounded by C.
The surface S can be any surface bounded by
C. For example:
z
R
z
R
C
O
C
Q
O
y
P
x
x
Q
y
P
This means that, for a given C, the values of
magnetic flux crossing all possible surfaces
bounded by it is the same, or the magnetic flux
bounded by C is unique.
2-24
(3) Imaginary contour C versus loop of wire.
There is an emf induced around C in either
case by the setting up of an electric field. A
loop of wire will result in a current flowing in
the wire.
(4) Lenz’s Law.
States that the sense of the induced emf is
such that any current it produces, if the closed
path were a loop of wire, tends to oppose the
change in the magnetic flux that produces it.
2-25
Thus the magnetic flux produced by the
induced current and that is bounded by C must
be such that it opposes the change in the
magnetic flux producing the induced emf.
(5) N-turn coil.
For an N-turn coil, the induced emf is N times
that induced in one turn, since the surface
bounded by one turn is bounded N times by
the N-turn coil. Thus
d
emf  – N
dt
where  is the magnetic flux linked by one turn

D2.5 B  B0 sin t ax  cos t a y
B
S

d S = B0 sin t
z
d
C E d l   dt  B0 sin t 
  B0 cos t V
1
C
1
x
y
2-27

B0
0
 dec.

2
3
t
 inc.
–B0
emf
B0
0
–B0
emf < 0

2
3
emf > 0
Lenz’s law is verified.
t
2-28
(b)
S B • dS
z
1
1
1
 B0 sin t – B0 cos t
2
2
1



B0 sin  t – 
4
2

C E • dl
d  1
 

– 
B0 sin  t –  
dt  2
4 

–
 B0


cos  t –  V
4
2

C
1
x
1 y
2-29
(c)
z
S B • dS
1
 B0 sin t  B0 cos t
C


 2 B0 sin  t  
4

C
E • dl
1 y
1
x
d 
 

 –  2 B0 sin  t   
dt 
4 



 – 2  B0 cos  t   V
4

2-30
Motional emf concept
B
C
l
z
S
x
S
d S = B0ly
= B0l  y0  v0t 
v0 ay
conducting
rails
y
B = B0az
B
dS
conducting bar
y  y 0  v 0t
2-31
d
C E • dl  – dt S B • dS
d
   B0l  y0  v0t  
dt
  B0lv0
This can be interpreted as due to an electric field
F
E   v0 B0 a x
Q
induced in the moving bar, as viewed by an observer
moving with the bar, since
l
v0 B0 l  x0 v0 B0a x • dx a x
l
 x0 E • dl
2-32
where
F  Qv  B
 Qv0 a y  B0 a z
 Qv0 B0 a x
is the magnetic force on a charge Q in the bar.
Hence, the emf is known as motional emf.
2-33
Ampére’s Circuital Law
(FEME, Sec. 2.4; EEE6E, Sec. 2.4)
2-34
Ampére’s Circuital Law
d
C H • dl  S J • dS  dt S D • dS
J, D
S
C
dS
2-35
C H • dl = Magnetomotive force (only by
analogy with electromotive
force),
 A m  m, or A.
S J • dS = Current due to flow of charges
crossing S,
2
2
A
m

m

 , or A.
S D • dS = Displacement flux, or electric
flux, crossing S,
2
2
C
m

m

 , or C.
2-36
d
D • dS = Time rate of increase of

S
dt
displacement flux crossing S,
or, displacement current
crossing S,
C s, or A.
Right-hand screw rule.
Any surface S bounded by C, but the same surface
for both terms on the right side.
2-37
Three cases fo clarify Ampére’s circuital law
(a) Infinitely long, current carrying wire
No displacement flux
S1 J • dS  S2 J • dS  I
C H • dl  I
S1
From 
I
C
S2
To 
2-38
(b) Capacitor circuit (neglect fringing of field
in the capacitor)
S2
C
I(t)
S1
S1 J • dS = I but S2 J • dS = 0
S1 D • dS = 0 but S2 D • dS  0
d
D • dS must be  I

S
dt 2
so that  H • dl is unique.
C
2-39
(c) Finitely long wire
 J • dS  I and
S1
S1 D • dS  0
S2 J • dS  0 and S2 D • dS  0
d
S1 J • dS  dt S1 D • dS must be
d
  J • dS 
D • dS

S2
S
dt 2
S1
Q1
I
C
Q2
S2
2-40
Uniqueness of C H • dl
dS1
S1
dS2
C
S2
potato
rubber band
d
H
•
dl

J
•
dS

C
S1
1
 D • dS1
dt S1
d
H
•
dl

–
J
•
dS
–
C
S2
2
 D • dS 2
dt S2
d
d
  J • dS1   D • dS1  –  J • dS 2 –  D • dS 2
S1
S2
dt S1
dt S2
d
d
D
•
dS


1
 D • dS 2  – S1 J • dS1 – S2 J • dS2
dt S1
dt S2
2-41
d
D
•
dS

–
J
•
dS


SS1 S2
dt SS1 S2
Displacement current emanating from a closed
surface = – (current due to flow of charges
emanating from the same closed surface)
2-42
D2.9 (a) Current flowing from Q2 to Q3.
d
 I  I23 
dt
D
dS  0
S2
 I  I23  2 I  0
Q2
I23  3I A
S2
I23
I
Q1
3I
S1
Q3
S3
2-43
(b) Displacement current emanating from the
spherical surface of radius 0.1 m and centered at Q1.
d
I  3I 
dt
d
dt
D
D
dS  0
S1
d S  4 I A
S1
(c) Displacement current emanating from the spherical
surface of radius 0.1 m and centered at Q3.
d
3I  I23 
dt
d
dt
D
S3
D
dS  0
S3
d S  3I  I23  3I  3I  6 I A
2-44
Interdependence of Time-Varying Electric and
Magnetic Fields
d
C E • dl = – dt S B • dS
d
C H • dl = S J • dS + dt S D • dS
2-45
Hertzian Dipole
I(t)
I(t)
H(t)
E(t)
2-46
Radiation from Hertzian Dipole
2-47
Gauss’ Laws
(FEME, Secs. 2.5, 2.6; EEE6E, Sec. 2.5)
2-48
Gauss’ Law for the Electric Field
 C

3
D
d
S

r
dv

m
,
or
C
 3

S
V
m


D
r•
V
S
dS
Displacement flux emanating from a closed surface S =
charge contained in the volume bounded by S = charge
enclosed by S.
2-49
Gauss’ Law for the Magnetic Field
S B • dS = 0
B
dS
S
Magnetic flux emanating from a closed
surface S = 0.
2-50
P2.21 Finding displacement emanating from a surface
enclosing charge

2
2
2
(a) r  x, y, z   r0 3  x  y  z

Surface of cube bounded by
x   1, y   1, and z   1

S
D d S   r dv
V

1
1
 
1
x 1 y 1 z 1
 8r0 
1
1
 
r0  3  x2  y2  z2  dx dy dz
1
x 0 y 0 z 0
2
2
2
3

x

y

z

 dx dy dz
 1 1 1
 8r0  3    
 3 3 3
 16 r0
2-51
(b) r  x, y, z   r0  x y z 
Surface of the volume x > 0, y > 0, z > 0, and (x2 + y2 + z2) < 1.

S
D d S   r dv
V

1

1 x2
x 0 y 0

r0

r0

2
1
 
1 x2  y2

z 0
1 x2
x 0 y 0
r0 xyz dx dy dz
xy 1  x2  y2  dx dy
1 x2
 xy x y xy 


dx



x

0
2
2
4  y 0
 2
r0 1 
1
3
3
5
3
1
2
3
2
4
5 
x

x

x

x

x

2
x

x
dx





x

0
4
2


r0
1
r0  x2 x4 x6 
x 3 1 5
    x  x  dx     
4 x 0  2
2 
4  4 4 12 0

r0
48
1
2-52
P2.23
z
dS4
1
dS1
dS2
y

x
dS3
S B • dS

S1 B • dS + S2 B • dS2
 S B • dS3  S B • dS 4
3
4
0
2-53
  B d S1
S1
   B d S2   B d S3   B d S4
S2
 
1
S3




z 0 x 0
S4
B0  yax  xa y 
y 0
0  0
 
1
z 0 x 0
B0 x dx dz
B0 2

2
Absolute value =
B0  2
2
Wb
 dx dz a 
y
The Law of
Conservation of Charge
(FEME, Sec. 2.5; EEE6E, Sec. 2.6)
2-55
Law of Conservation of Charge
d
S J • dS + dt V r dv  0
J
r(t)
V
S
dS
Current due to flow of charges emanating from a closed surface S
= Time rate of decrease of charge enclosed by S.
2-56
Summarizing, we have the following:
Maxwell’s Equations
d
C E d l   dt S B d S
d
C H d l  S J d S  dt S D d S


S
S
(1)
(2)
D dS   r dv
(3)
B dS  0
(4)
V
2-57
Law of Conservation of Charge

d
J dS 
S
dt

r dv  0
(5)
V
(4) is, however, not independent of (1), whereas (3) follows
from (2) with the aid of (5).
2-58
Example: Finding

C
H d l around C.
dS
I1
S
Q(t)

d
H d l  I2 
C
dt

1
D dS  Q
S
2

D dS
I2
C
(Ampére’s Circuital Law)
S
(Gauss’ Law for the electric
field and symmetry
considerations)
2-59
dQ
I2 – I1 
0
dt
(Law of Conservation
of Charge)
dQ
 I1 – I2
dt
d 1 
  H d l  I2   Q 
C
dt  2 
1
 I2   I1  I2 
2
1
  I1  I2 
2
The End