Download Module 3 - University of Illinois Urbana

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
3-1
Module 3
Maxwell’s Equations
In Differential Form
Faraday’s law and Ampere’s Circuital Law
Gauss’ Laws and the Continuity Equation
Curl and Divergence
3-2
Instructional Objectives
8. Determine if a given time-varying electric/magnetic
field satisfies Maxwell’s curl equations, and if so
find the corresponding magnetic/electric field, and
any required condition, if the field is incompletely
specified
9. Find the electric/magnetic field due to onedimensional static charge/current distribution using
Maxwell’s divergence/curl equation for the
electric/magnetic field
10. Establish the physical realizability of a static electric
field by using Maxwell’s curl equation for the static
case, and of a magnetic field by using the Maxwell’s
divergence equation for the magnetic field
3-3
Faraday’s Law and
Ampère’s Circuital Law
(FEME, Secs. 3.1, 3.2; EEE6E, Sec. 3.1)
3-4
Maxwell’s Equations in Differential Form
Why differential form?
Because for integral forms to be useful, an a priori
knowledge of the behavior of the field to be
computed is necessary.
The problem is similar to the following:
If
1
0 y(x) dx  2, what is y(x)?
There is no unique solution to this.
3-5
However, if, e.g., y(x) = Cx, then we can find y(x),
since then
1
x 2 1
0 Cx dx  2 or C  2 0  2 or C  4
 
 y(x)  4x.
On the other hand, suppose we have the following
problem:
dy
If
 2, what is y?
dx
Then y(x) = 2x + C.
Thus the solution is unique to within a constant.
3-6
FARADAY’S LAW
First consider the special case
E  Ex (z,t) a x and H  H y (z, t) a y
and apply the integral form to the rectangular path
shown, in the limit that the rectangle shrinks to a
point.
y
z
(x, z)
x
z (x, z + z)
S
C
(x + x, z) (x + x, z + z)
x
3-7
d
C E d l   dt S B dS
 Ex z 
z

E 

Lim
x z z
x 0
z 0

d
 By  x z
x   Ex z x  
x, z
dt

  Ex z x
x z
 Lim
x 0
z 0
By
Ex

z
t
 B 

dt
d
y
x, z

x z
x z
3-8
General Case
E  Ex (x, y, z,t)a x  Ey (x, y, z,t)a y  Ez (x, y, z, t)a z
H  H x (x, y, z,t)a x  H y (x, y, z,t)a y  Hz (x, y, z, t)a z
 Ez  E y
 Bx
–
–
y
z
t
 By
 E x  Ez
–
–
z
x
t
 Ey  Ex
 Bz
–
–
x
y
t
Lateral space
derivatives of the
components of E
Time derivatives of
the components of B
3-9
Combining into a single differential equation,
ax
ay
az

x

y

B
–
z
t
Ex
Ey
Ez
B
E–
t
Differential form
of Faraday’s Law



  ax
 ay
 az
x
y
z
B
Del Cross E or Curl of E = –
t
3-10
AMPÈRE’S CIRCUITAL LAW
Consider the general case first. Then noting that
d
C E • dl  – dt S B • dS
  E  –  (B)
we obtain from analogy,
t
d
C H • dl  S J • dS  dt S D • dS
  H  J   (D)
t
3-11
D
HJ
t
Thus
Special case:
E  Ex (z,t)a x , H  H y (z,t)a y
ax ay az
0
0
0
Hy

D
J 
z
t
0
 Hy
 Dx
–
 Jx 
z
t
Differential form
of Ampère’s
circuital law
3-12
 Hy
 Dx
 – Jx –
z
t


Ex. For E  E0 cos 6 ×10 t  kz a y
8
in free space    0 ,   0 , J = 0 ,
find the value(s) of k such that E satisfies both
of Maxwell’s curl equations.
Noting that E  Ey (z,t)a y , we have from
B
E–
,
t
3-13
ay
az
B
 –  E  – 0
t
0

z
0
Ey
0
ax
 Bx  Ey

t
z
 
8


E
cos
6


10
t

kz


0

z
 kE0 sin  6 108 t  kz 
kE0
8
Bx  
cos
6


10
t  kz 

8
6 10
3-14
Thus,
kE0
8
B
cos
6


10
t  kz  ax

8
6 10
B
B
H

0 4 107
kE0
8

cos
6


10
t  kz  ax

2
240
Then, noting that H  H x (z,t)a x , we have from
D
H
,
t
3-15
ax
ay
az
D
 ×H  0
t
0

z
Hx
0
0
 Dy  H x

t
z
2
k E0
8

sin
6


10
t  kz 

2
240
3-16
2
k E0
8
Dy 
cos  6  10 t  kz 
3
8
1440  10
k 2 E0
8
D
cos  6  10 t  kz  a y
3
8
1440  10
D
D
E
 9
0 10 36
k 2 E0
8

cos
6


10
t  kz  a y

2
4
3-17
Comparing with the original given E, we have
k 2 E0
E0 
4 2
k   2
E  E0 cos  6  108 t  2 z  a y
Sinusoidal traveling waves in free space, propagating in the
z directions with velocity, 3  108 ( c) m s.
3-18
Gauss’ Laws and
the Continuity Equation
(FEME, Secs. 3.4, 3.5, 3.6; EEE6E, Sec. 3.2)
3-19
GAUSS’ LAW FOR THE ELECTRIC FIELD
S D • dS  V  dv
z
(x, y, z)
x
 Dx xx  y  z   Dx x  y  z
  Dy 

z  x   Dy   z  x
y y
y
  Dz z  z  x  y   Dz z  x  y
  x  y z
z
y
y
x
3-20
 D 

x x x

  Dx x y  z

  Dy 
  Dy  Δ z Δ x
y +Δy
y
Lim
x 0
y 0
z 0
 Lim
x  0
y  0
z  0
  Dz z z   Dz z   x y
 x y  z
  x y  z
 x y  z
3-21
 Dx  Dy  Dz



x
y
z
Longitudinal derivatives
of the components of D
•D 
Divergence of D = 
Ex. Given that
0 for – a  x  a
  
0 otherwise
Find D everywhere.
3-22
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
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•
•
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•
•
•
•
•
•
•
•
•
•
•
•
•
•
•

0
x=–a x=0 x=a
Noting that  = (x) and hence D = D(x), we set


 0 and
 0, so that
y
z
 Dx  Dy  Dz  Dx
• D 



x
y
z
x
3-23
Thus, • D =  gives
 Dx
  (x)
x
which also means that D has only an xcomponent. Proceeding further, we have
x
Dx  –  x  dx  C
where C is the constant of integration.
Evaluating the integral graphically, we have the
following:
3-24

–a
0
a
x
x
–   ( x) dx
2 0 a
–a
0
0
a
x
From symmetry considerations, the fields on
the two sides of the charge distribution must
be equal in magnitude and opposite in
direction. Hence,
C = –  0a
3-25
Dx
0 a
–a
a
– 0a
 0 a a x for x  a

D   0 x a x for  a  x  a
  a a for x  a
x
 0
x
3-26
GAUSS’ LAW FOR THE MAGNETIC FIELD
S D • dS = V  dv
• D  
From analogy
S B • dS = 0 = V 0
dv
•B0
•B0
Solenoidal property of magnetic field lines. Provides test for
physical realizability of a given vector field as a magnetic
field.
3-27
LAW OF CONSERVATION OF CHARGE
d  dv  0
J
•
dS

S
dt V

 • J  t ( )  0

• J 

0
t
Continuity
Equation
3-28
SUMMARY
B
E–
t
D
HJ
t
•D
(1)
(2)
(3)
•B0
(4)

•J
0
t
(5)
(4) is, however, not independent of (1), and (3) can
be derived from (2) with the aid of (5).
3-29
Curl and Divergence
(FEME, Secs. 3.3, 3.6; EEE6E, Sec. 3.3)
3-30
Maxwell’s Equations in Differential Form
B
×E = 
t
D
×H = J 
t
 D
 B
ax
ay
az

Curl  × Α 
x

y

z
Ax
Ay
Az
Divergence  A= Ax  A y  Az
x
y
z
3-31
Basic definition of curl
Lim  C A d l 

 an
×A =
S  0  S 

 max
 × A is the maximum value of circulation of A per
unit area in the limit that the area shrinks to the point.
Direction of  ×A is the direction of the normal
vector to the area in the limit that the area shrinks
to the point, and in the right-hand sense.
3-32
Curl Meter
is a device to probe the field for studying the curl of the
field. It responds to the circulation of the field.
3-33
3-34
a
 2x
for 0  x 
 v0 a az
2
v
2x 
a
 v0  2 
 az for  x  a
a 
2
 
ax
ay
az

×v 
x

y

z
0
vz
0
   × v y
vz

ay
x
a

negative
for
0

x


2

 positive for a  x  a

2
 2v0
  a a y

 2v0 a y
 a
3-35
Basic definition of divergence
A

 A  v  0
Lim
dS
v
is the outward flux of A per unit volume in the limit that
the volume shrinks to the point.
Divergence meter
is a device to probe the field for studying the divergence
of the field. It responds to the closed surface integral of
the vector field.
3-36
x
Example:
At the point (1, 1, 0)
(a)
 x  1
2
ax
Divergence zero
(b)
1
 y  1 ay
y
z
x
1
1
Divergence positive
y
z
x
(c) x a
y
y
1
1
Divergence negative
y
z
1
3-37
Two Useful Theorems:
Stokes’ theorem

C
A d l =   × A dS
S
Divergence theorem

S
A dS =
 
V
A useful identity
 ×A  
A  dv
3-38
ax
ay
az


×Α 
x y

z
Ax
Ay
Az



 ×A =
  × A x    × A y   × A z
x
y
z



x y z




0
x y z
Ax
Ay
Az
The End