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ENE 325
Electromagnetic Fields
and Waves
Lecture 10 Time-Varying Fields and Maxwell’s
Equations
Review


Magnetic boundary conditions
Bn1 = Bn2
H 1  H 2  a n12  K .
and
Inductance and mutual inductance
 self inductance L is defined as the ratio of flux linkage to the
current generating the flux,
Ntotal
henrys or Wb/A.
L
I
N 212
 mutual inductance M ,
M 12 
I1
where M12 = M21.
N
M 21  1 21
I2
Time-Varying fields and Maxwell’s
equations

Concept

The electric field E is produced by the change in the
magnetic field B.

The magnetic field B is produced by the change in
the electric field E.
Faraday’s law
d
emf  
dt
V
where emf = electromotive force that may establish a current
in a suitable closed circuit and d is a voltage that arises
dt
from conductors moving in static or changing magnetic fields.

d is arisen from
dt
1. the change of flux in a closed path
2. the moving closed path in a stationary magnetic field
3. both 1 and 2
d
 For the N number of loops, emf   N
v.
dt
emf in the closed loop is not zero
d
emf   E d L    B d S  0
dt S
a) direction of the induced current
b) emf
Changing flux in a stationary path
(transformer emf)
From
emf   E d L  
d
 B d S,
dt S
apply Stokes’ theorem,
B
dS
 (  E ) d S   
S
S t
B
(  E ) d S  
dS
t
So we have
B
(  E )  
.
t
1st Maxwell’s equation
Transformer

To transform AC voltages and currents between a pair of
windings in magnetic circuits
N1i1  N2i2
With Faraday’s law, we have
d
d
v1  N1
, v2  N 2
.
dt
dt
Since the term d is the same for both voltages, so we get
dt
N
v2  2 v1.
N1
Ex1 Assume B  Boekt a z, prove the 1st Maxwell’s
equation.
z
r
Changing flux in a moving closed
path (1)

a conductor moves in a uniform magnetic field.
yw
   B  d S     Bo a z  dxdya z   Bo yw.
00
d
dy
emf  
 Bo
w  Bo vw.
dt
dt
The sign of emf determines the
direction of the induced current.
Changing flux in a moving closed
path (2)

Examine in a different point of view
z
y
B
va y
w
x
So we get
emf   (v  B) d L
V.
Changing flux in a moving closed
path (3)

Combing both effects yields
emf  
d
B
  E d L  
d S   (v  B) d L.
dt
S t
Ex2 Let B  6cos10 t sin 0.01xa z mT, find
6
a) flux passing through the surface z = 0, 0 < x < 20 m,
and 0 < y < 3 m at t = 1 S.
b) value of closed line integral around the surface specified
above at t = 1 S.
Ex3 A moving conductor is located on the
conducting rail as shown at time t = 0,
z
y
B
0.05
x
a) find emf when the conductor is at rest at x = 0.05 m and
B  0.3sin104 ta z
T.
b) find emf when the conductor is moving with the speed
v  150a x m/s.
Displacement current (1)

The next Maxwell’s equation can be found in terms of timechanging electric field
From a steady magnetic field,
 H  J


  H   J  0.
From the equation of continuity,
rv
 J 
t
rv
this is impossible!
therefore
 0.
t
Displacement current (2)

Another term must be added to make the equation valid.
 H  J  J d .
In a non-conductive medium, J  0.
2nd Maxwell’s equation
Displacement current (3)

We can show the displacement current as
D
Id   J d d S  
d S.
s
s t

The more general Ampere’s circuital law:
 H d L  I  Id
Where is the displacement current
from?
Consider a simple current loop, let emf = Vocost
Ex4 Determine the magnitude of J d for
the following situations:
a)
in the air near the antenna that radiates
E  80cos(6.277 108 t  2.092 y)a z V/m.
b) a pair of 100 cm2 area plates separated by a 1.0 mm thick layer
of lossy dielectric characterized by r = 50 and  = 1.010-4
S/m given the voltage across plates V(t) = 1.0cos(2103t) V.