ECE 6340
Intermediate EM Waves
Fall 2013
Prof. David R. Jackson
Dept. of ECE
Notes 1
1
Maxwell’s Equations
E  r, t   V/m ,
B  r, t   Wb/m2  T 
D  r , t  C/m2  , H  r , t   A/m
B
 E  
(Faraday's Law)
t
D
 H  J 
(Ampere's Law)
t
D  v
(Gauss's Law)
 B  0
(Magnetic Gauss's Law)
2
Current Density
J  v v [A/m ]
2
The current density vector
points in the direction that
positive charges are moving.
J
J
v
v
Positive charges
v
v  r 
Note: The charge density v is the "free charge density" (the charge density that is
free to move in the material).
3
Current Density
J
The charge density v is the "free charge
density" (the charge density that is free to
move in the material).
v
v
 A copper wire: v is the charge density of the electrons in the conduction band that
are free to move (the wire is neutral).
 Saltwater: v is the charge density of the Na (+) or Cl (-) ions that are free to move.
(There will be two currents densities, one from the Na ions and one from the Cl ions.)
 A electron beam: v is the charge density of the electrons in the beam (a negative
number).
4
Current Density
A small surface dS is introduced perpendicular to the direction of
current flow.
dI = current flowing through dS
dQ   v  dV 
J  J vˆ
vˆ
dV
  v  dl  dS
dI  dQ / dt
++++++
++++++
++++++
dl
Charge calculation:
  v  v dt  dS
   v v  dt dS
J
 J dt dS
dS
Hence
dI  J dS
5
Current Density (cont.)
A more general case of an arbitrary orientation of the surface is now considered.

 

n
t
J  J  J  J  nˆ nˆ  J  tˆ tˆ

J
J
Surface S
n
Integrating over the surface, we have
t
dS
J
nˆ
 J  nˆ  

dI  J  nˆ dS
tˆ


I   J  nˆ dS
S
component of current density crossing dS
6
Continuity Equation
From Gauss’s Law,
 D  v

 D  v

 D       
t
 t  t
From Ampere’s Law,
D
 H  J 
t
so
 D 
    H     J    


t


7
Continuity Equation (cont.)
 D 
    H     J    


t


 D 


    J
 t 
From the last slide:
Hence
v
 D 
 

t
 t 
 v
 J  
t
“Continuity Equation”
8
Integral Form of Continuity Equation
v
V   J dV  V t dV
S (closed)
Apply divergence theorem:
nˆ
V

V
  J dV 
Flux per
unit volume
 J  nˆ dS
S
Net flux
out of S
9
Integral Form of Continuity Equation (cont.)
Hence we have

S
v
J  nˆ dS  
dV
t
V
Assume V is stationary (not changing with time):
v
dQencl
d
V t dV  dt V v dV  dt
Hence

S
dQencl
J  nˆ dS  
dt
 charge conservation
10
Integral Form of Continuity Equation (cont.)
dQencl
   J  nˆ dS
dt
S
+
+
S (stationary)
+
J
Qencl +
+
+
 nˆ
Charge that enters the region must stay there, and this results in an
increase of total charge inside the region.
11
Generalized Continuity Equation
Assume S is moving (e.g., expanding)
S
vs (r)
Qencl
J
Note: vs denotes the
velocity of the surface.
dQencl
   J  nˆ dS   v v s  nˆ dS
dt
S
S
Note :
dQencl
dt
    v  v  v s   nˆ dS
S
Difference in charge and surface velocities!
12
Generalized Continuity Equation (cont.)
Define the current density in the moving coordinate system:
S
J   v  v  v s 
vs (r)
Qencl
 J   nˆ dS  
S
J
dQencl
dt
Note: The current J  is the current density seen by an observer on the moving surface.
13
Relaxation Equation
Start from the continuity equation:
v
   J
t
Assume a homogeneous conducting medium that obeys Ohm’s law:
J  E
We then have
v


    E     D   v
t


Hence
v

  v
t

14
Relaxation Equation (cont.)
v

  v
t

For an initial charge density v (r,0) at t = 0 we have:
v  r , t   v  r ,0 e
 /  t
In a conducting medium the charge density dissipates very quickly and
is zero in steady state.
15
Time-Harmonic Steady State
Charge Density
An interesting fact: In the time-harmonic (sinusoidal) steady state, there is
never any charge density inside of a homogenous, source-free* region.
(You will prove this in HW 1.)
*Source-free means that there are no current
sources that have been placed inside the body.
Hence, the only current that exists inside the body
is conduction current, which is given by Ohm’s law.
16
Integral Forms of Maxwell’s Eqs.
D
 H  J 
t
Ampere's law
Stokes’s Theorem:
S (open)
nˆ
Note: right-hand rule for C.
C
    H   nˆ dS   H  dr
S
Circulation per
unit area of S
C
Circulation on
boundary of S
17
Integral Forms of Maxwell’s Eqs. (cont.)
Integrate Ampere’s law over the open surface S:
D
S  H   nˆ dS  S J  nˆ dS  S t  nˆ dS
Apply Stokes’s theorem:
D
C H  dr  S J  nˆ dS  S t  nˆ dS
or
D
C H  dr  is  S t  nˆ dS
Note: The current is is
measured in the frame
of the observer.
(The current is is the current though the surface S.)
18
Integral Forms of Maxwell’s Eqs. (cont.)
D
C H  dr  is  S t  nˆ dS
Ampere’s law
Note: In statics, is is independent of the shape of S, since the
last term is zero and the LHS is independent of the shape of S.
Similarly:
B
C E  dr  S t  nˆ dS
Faraday’s law
Note: In statics, the voltage drop around a closed path is always zero
(the voltage drop is unique).
19
Integral Forms of Maxwell’s Eqs. (cont.)
 D  v
S (closed)
Electric Gauss law
Divergence theorem:
nˆ

V Flux per
unit volume
V
Integrate the electric Gauss law throughout V
and apply the divergence theorem:
 D  nˆ dS   
S
  D dV 
 D  nˆ dS
S
Net flux
out of S
Hence we have
 D  nˆ dS  Q
encl
v
dV
S
V
20
Integral Forms of Maxwell’s Eqs. (cont.)
B  0
Magnetic Gauss law
Apply divergence theorem
 B  nˆ dS  0
S
21
Boundary Form of Maxwell’s Eqs.
nˆ
1,1, 1 
ˆ

C
We allow for a surface current density. The
narrow rectangular path C is chosen arbitrarily
oriented in the tangential direction  as shown.
 0

C


Js
The unit normal points towards region 1.
 2,2, 2 
D
C H  dr  is  S t  nˆ C dS
Evaluate the LHS of Ampere’s law on the path shown:


lim  H  dr  lim   H  dr   H  dr   H  dr 
 0
 0  


C
C
C
C



lim  H  dr  ˆ   H 1  H
 0
2
 
C
22
Boundary Form of Maxwell’s Eqs. (cont.)
nˆ C  nˆ  ˆ
nˆ
1,1, 1 
ˆ

Evaluate the RHS of Ampere’s law for path:
is  lim 
 0 0
C
C



J s   nˆ  ˆ  d


 J s   nˆ  ˆ    ˆ  J s  nˆ 
Js
 2,2, 2 
where we have used the following vector identity:
A   B  C   C   A B  B  C  A
Next, for the displacement current term in ampere's law, we have
D
lim 
 nˆ dS  0
 0
t
S
 since S  0
23
Boundary Form of Maxwell’s Eqs. (cont.)
nˆ
1,1, 1 
ˆ

Hence, we have
ˆ   H 1  H
C
C


Js
2
  ˆ   J s  nˆ 
 2,2, 2 
Since ˆ is an arbitrary unit tangent vector,
 H 1  H 2 t   J s  nˆ t
This may be written as:
Hence
nˆ   H 1  H
ˆ H 1  H
n
2
2
  nˆ   J s  nˆ   J s
  Js
Note nˆ always
points into region 1!
24
Boundary Form of Maxwell’s Eqs. (cont.)
nˆ
1,1, 1 
ˆ

C
C


Js
 2,2, 2 
Similarly, from Faraday's law, we have
ˆ E1  E2   0
n
Note: The existence of an electric surface current does not affect this result.
25
Boundary Form of Maxwell’s Eqs. (cont.)
A surface charge density is now allowed.
nˆ 1,1, 1 

S + +
+
S
+ +
s
There could also be a surface current, but it does not affect the result.
A “pillbox” surface is chosen.
 0

S
 2,2, 2 
The unit normal points towards region 1.
 D  nˆ dS
 Qencl
S
Evaluate LHS of Gauss’s law over the surface shown:


lim  D  nˆ S dS  lim   D  nˆ dS   D    nˆ  dS   D  nˆ S dS 
 0
 0  


S
S
S
S



lim  D  nˆ S dS  nˆ  D 1  D 2  S
 0
S
26
Boundary Form of Maxwell’s Eqs. (cont.)
nˆ
S + +
+
S
RHS of Gauss’s law:
1,1, 1 
+ +
s
lim Qencl

 0
S
 2,2, 2 
lim Qencl
 0


 lim    s dS 
 0
 S

  s S
Hence Gauss’s law gives us
nˆ  D1  D 2  S  s S
Hence we have
nˆ  D1  D 2   s
Similarly, from the magnetic Gauss law:
nˆ  B 1  B 2   0
27
Boundary Form of Maxwell’s Eqs. (cont.)
We also have a boundary form of the continuity equation.
From before:
 D  v
nˆ  D1  D 2   s
Noting the similarity, we can write:
 J
 v
 
t

nˆ  J1  J 2

 s

t
Note: If a surface current exists, this may give rise to another surface charge
density:
s  Js
 s
 
t


In general: nˆ  J1  J 2   s  J s
 s
 
t
28
Summary of Maxwell Equations
Point Form
H  J 
 E  
Integral Form
D
H

dr

i

s
C
S t  nˆ dS
D
t
B
t
 E  dr   
C
 D  nˆ dS
  D  v
S
Boundary Form
nˆ   H1  H 2   J s
B
 nˆ dS
t
nˆ  E1  E 2   0
 Qencl
nˆ  D 1  D 2    s
 0
nˆ   B 1  B 2   0
S
 B  nˆ dS
 B  0
S
Continuity equation :
 J
 
v
t

S
J   nˆ dS  
dQencl
dt


nˆ  J1  J 2   s  J s  
s
t
29
Faraday’s Law
B
C E  dr   S t  nˆ dS
If S is stationary:
B
d
d
S t  nˆ dS  dt S B  nˆ dS  dt
Hence:
 = magnetic flux through S
S (open)
d
C E  dr   dt
nˆ
C
30
Faraday’s Law (cont.)
If S is moving (e.g., expanding):
vs
C
Previous form is still valid:
However,
B
C E  dr  S t  nˆ dS
! d
B
d
S t  nˆ dS  dt S B  nˆ dS  dt
31
Faraday’s Law (cont.)
Vector identity for a moving path:
d
B
B  nˆ dS  
 nˆ dS    v s  B   dr

dt S
t
S
C
vs
“Helmholtz identity”
(See appendix for derivation)
C
32
Faraday’s Law (cont.)
Start with
Then use
B
C E  dr  S t  nˆ dS
d
B
ˆ
B  n dS  
 nˆ dS    v s  B   dr

dt S
t
S
C
d
C E  dr   dt  C  vs  B   dr
Hence
or
d
C E  vs  B   dr   dt
33
Faraday’s Law (cont.)
d
C E  vs  B   dr   dt
EMF 
Define
 E  v
s
 B   dr
C
Then
d
EMF  
dt
34
Two Forms of Faraday’s Law
B
C E  dr  S  t  nˆ dS
d
C E  vs  B   dr   dt
Faraday’s law
“Generalized”
Faraday’s law
Note: The path C means C(t), a fixed path that is evaluated at a given time t. That
is, C = C(t) is a “snapshot” of the moving path. The same comment for S = S(t).
35
Two Forms of Ampere’s Law
D
C H  dr  is  S t  nˆ dS
Ampere’s law
d
C H  dr  is  dt S D  nˆ dS  C  v s D   d r
“Generalized” Ampere’s law
36
Example
y
B t   zˆ cost
 t   t

x
Note: t is in [s],  is in [m].
C
1) Find the voltage drop V(t) around the closed path.
2) Find the electric field on the path
3) Find the EMF drop around the closed path.
Practical note: At a
nonzero frequency the
magnetic field must have
some radial variation, but
this is ignored here.
37
Example (cont.)
(1) Voltage drop (from Faraday’s law)
B
V   E  dr   
 nˆ dS
t
C
S
B z
 
dS
t
S
B z

t
Choose unit normal:
nˆ

x
Bz is independent
of (x,y).
 dS
S
B z
 2 

t
  sin t  2 

V  t    E  dr   t sin t [V]
2
C
y
B t   zˆ cost
 t   t
Would this result change if the
path were not moving? No!
38
Example (cont.)
(2) Electric field (from voltage drop)
E  2    t 2 sin t
  t 2 sin t
E 
2 
dr  ˆ   d 
y

x
t 2
E 
sin t
2
so
E 
t
2
sin t
t
ˆ
E   sin t [V/m]
2
Note: There is no  or z component (seen
from the curl of the magnetic field):
H  zˆ f    cos t 
39
Example (cont.)
(3) EMF (from Generalized Faraday’s law)
d
C E  vs  B   dr   dt
   B  nˆ dS
y

x
S
  B z dS
S
 B z  2 
 cos t  t
2

B t   zˆ cos t 
 t   t
40
Example (cont.)
y
   t 2  cos t
path moving
d
 2 t cos t
dt
  t 2 sin t

x
field changing
d
EMF  
dt
EMF  2 t cos t
  t 2 sin t [V]
41
Example (cont.)
Electric field and voltage drop
B t   zˆ cos t 
(Alternative method: from the Generalized Faraday’s law)
 2 E  v s  B 
 t   t
y
d

dt
d
 2  E  vs Bz   
dt

x
Since vs = 1 [m/s]:
d
 2  E  Bz   
dt
1 d
E  Bz 
2 dt
so
1 d
E  cos t  
2 t dt
42
Example (cont.)
1 d
2 t dt
1
 2 t cos t   t 2 sin t 
 cos t  
2 t
t


 cos t  cos t  sin t 
2


E  cos t  
E 
t
2
sin t [V/m]
t
E  ˆ sin t [V/m]
2
V   2 
t
2
sin t [V]
V   t 2 sin t [V]
43
Example (cont.)
B t   zˆ cos t 
EMF
 t   t
(Alternative method: Faraday’s law)
V t    t sin t
2
EMF 
 E  v
s
y
(Faraday’s law)
 B   dr
C
 V  t     v s  B   dr

x
C
 v
s
 B   dr 
C
2
 1ˆ    zˆ cos t    ˆ   d  
0
2

  cos t   d
0
  cos t  2 
vs = 1 [m/s]
44
Example (cont.)
B t   zˆ cos t 
 t   t
y
Hence
EMF   t 2 sin t  cos t  2 
so

x
EMF   t sin t  cos t  2 t 
2
vs = 1 [m/s]
45
Example Summary
Voltage drop V(t) around the closed path:
V t    t 2 sin t [V]
t
sin t [V/m]
Electric field on the path: E  ˆ
2
EMF drop around the closed path: EMF  2 t cos t   t 2 sin t [V]
46
Example
Find the voltage on the voltmeter
Perfectly conducting leads
Vm
+
y
a
V0
Voltmeter
Applied magnetic field:
+
x
-
B
B t   zˆ cos t 
Note: The voltmeter is assumed to have a very high internal resistance,
so that negligible current flows in the circuit. (We can neglect any
magnetic field coming from the current flowing in the loop.)
47
Example (cont.)
B
V   E  dr   
 nˆ dS
t
C
S
Practical note: In such a measurement, it
is good to keep the leads close together
(or even better, twist them.)
B z
 
dS
t
S

B z
t

dS
S
B z
2


a
 
t
  sin t  a 2 
Vm V0   a2 sin t 
C
Vm
+
y
a
V0
Voltmeter
+
x
-
B
B t   zˆ cos t 
B z
  sin t 
t
Vm  V0   a2 sin t 
48
Generalized Ohm's Law
A resistor is moving in a magnetic field.
v
R
A
B

B
i
+
V
EMF  Ri
B
 E  v
s
-
You will be proving this
in the homework.
 B   dr  Ri
A
B
V    v s  B   dr  Ri
A
49
Perfect Electric Conductor (PEC)
A PEC body is moving in the presence of a magnetic field.
B
v
Inside the PEC body:
E  v  B
You will be proving this
in the homework.
50
Example
Find the voltage Vm on the voltmeter
Sliding bar (perfect conductor)
Velocity v0
Vm
W
+
+
V0
Voltmeter
-
dW
 v0
dt
-
L
B  t   zˆ
We neglect the magnetic field coming from the current in the loop itself.
51
Example (cont.)
d
EMF   E  v s  B   dr  
dt
C
(Generalized Faraday’s law)
  Bz ( LW )
d
dW
 Bz L
dt
dt
 Bz Lv0
 Lv0
Velocity v0
Vm
+
+
V0
Voltmeter
-
-
C
Hence
EMF  Lv0
L
B  t   zˆ
52
Example (cont.)
EMF 
 E  v
s
 B   dr
C
 Vm  0  (V0 )  0
PEC bar
PEC wires
EMF  Ri ( R  0)
Hence
Velocity v0
Vm  V0  Lv0
so
Vm
Vm  V0  Lv0
+
+
V0
Voltmeter
-
-
C
L
53
Example (cont.)
Now let's solve the same problem using Faraday's law.
B
C E  dr  S  t  nˆ dS
Velocity v0
  0 dS
S
0
Vm
+
+
V0
Voltmeter
-
-
C
 E  dr  Vm  0  (V0 )   E  dr
C
top
0
L
B  t   zˆ
54
Example (cont.)
For the top wire we have
0
0
ˆ    E dx   LE
 E  dr   E   xdx
x
top
L
x
L
   
ˆ 0  zˆ  v0
Ex  xˆ    v  B   xˆ   yv
B  t   zˆ
Hence
 E  dr  Lv
Velocity v0
0
top
Vm
so
+
+
V0
Voltmeter
 E  dr  V
m
 V0  Lv0
-
-
C
C
0
L
55
Example (cont.)
 E  dr  V
m
 V0  Lv0  0
C
B  t   zˆ
Velocity v0
Hence, we have
Vm  V0  Lv0
Vm
+
+
V0
Voltmeter
-
-
C
L
56
Time-Harmonic Representation
Assume f (r,t) is sinusoidal:
f  r , t   A  r  cos t    r  

 Re A  r  e
Denote
j  r 
F  r   A r  e
e jt

j  r 
From Euler’s identity:
e jz  cos z  j sin z
 cos z  Re  e jz  ,
if z  real
(Phasor form of f )
F r   Ar 
arg F  r     r 
Then
f  r , t   Re F  r  e jt 
57
Time-Harmonic Representation (cont.)
f  r, t   A r  cos t    r 
F  r   A r  e
j  r 
f  r , t   Re F  r  e
Notation:
jt

f  r , t   F  r  for scalars
F  r , t   F  r  for vectors
58
Time-Harmonic Representation (cont.)
Derivative Property:
f

 Re  F  r  e jt 
t t


 Re   F  r  e jt  
 t

 Re  j  F  r  e jt 
phasor for the derivative
Hence:
f  r , t 
 j  F  r  for scalars
t
F  r , t 
 j  F  r  for vectors
t
59
Time-Harmonic Representation (cont.)
Consider a differential equation such as Faraday’s Law:
B  r , t 
 E  r , t   
t
This can be written as
  Re  E  r  e jt    Re  j  B  r  e jt 
or
Re    E  j  B  e jt   0
60
Time-Harmonic Representation (cont.)
Let
c  cr  jci     E  j  B  x , y , z
Then
Re c e jt   0
Phasor
interpretation:
Does this imply that c = 0?
choose
choose
t  0
t   / 2
cr  0
ci  0
ce
jt
Im
t
Re
Re  c e jt 
Hence: c = 0
61
Time-Harmonic Representation (cont.)
Therefore
so
   E  j  B x, y , z  0
 E  j  B  0
 E   j  B
Hence
B  r , t 
 E  r , t   
t
 E   j  B
Time-harmonic (sinusoidal) steady state
62
Time-Harmonic Representation (cont.)
We can also reverse the process:
 E   j  B
Re    E  j  B  e jt   0
  Re  E  r  e jt    Re  j  B  r  e jt 
B  r , t 
 E  r , t   
t
63
Time-Harmonic Representation (cont.)
Hence, in the sinusoidal steady-state, these two equations are
equivalent:
B
 E  
t
 E   j B
64
Maxwell’s Equations in
Time-Harmonic Form
 E   j  B
 Η  J  j  D
  D  v
 B  0
65
Continuity Equation
Time-Harmonic Form
 v
 J  
t
 J   jv
Surface (2D)
s  J s   js
Wire (1D)
dI
  jl
d
I is the current in
the direction l.
66
Frequency-Domain Curl Equations (cont.)
At a non-zero frequency, the frequency domain curl equations imply
the divergence equations:
Start with Faraday's law:
 E   j B
     E    j  B
Hence
 B  0
67
Frequency-Domain Curl Equations
Start with Ampere's law:
 Η  J  j D
     Η     J  j    D 
  j   v    D 
 J   jv
(We also assume the
continuity equation here.)
Hence
 D  v
68
Frequency-Domain Curl Equations (cont.)
 Η  J  j  D
 E   j B
0
 B  0
With the
continuity
equation
assumed
 D  v
Hence, we often consider only the curl equations in the
frequency domain, and not the divergence equations.
69
Time Averaging of Periodic Quantities
Define:
f t 
T
1
  f  t  dt
T 0
T  period
s 
Assume a product of sinusoidal waveforms:
f  t   A cos  t     F  Ae j
g  t   B cos  t     G  Be j
f  t  g  t   A B cos  t    cos  t   
1

 A B  cos      cos  2  t     
2

70
Time Average (cont.)
1

f  t  g  t   A B  cos      cos  2  t     
2

The time-average of a constant
is simply the constant.
Hence:
f t  g t 
The time-average of a
sinusoidal wave is zero.
1
 A B cos    
2
71
Time Average (cont.)
F  Ae j
The phasors are denoted as:
G  Be
Consider the following:
so
Hence
FG  A B e
*
j
j    
Re  FG   A B cos    
*
f t  g t 
1
 Re  FG * 
2
The same formula extends to vectors as well.
72
Example: Stored Energy Density
1
1
U E  D  E  DxEx  DyEy  DzEz 
2
2
Dx , y , z  Re  Dx , y , z e jt  etc.
D  E  DxEx  DyEy  DE
z z
1
1
1
*
*
 Re  Dx Ex   Re  Dy E y   Re  Dz Ez* 
2
2
2
1
 Re  Dx Ex*  Dy E *y  Dz Ez* 
2
or
D E 

1
*
Re D  E
2

73
Example: Stored Energy Density (cont.)
From electrostatics:
Hence,
UE
1
UE  D E
2

1
1
*
 D  E  Re D  E
2
4

Similarly,
UH

1
1
*
 B  H  Re B  H
2
4

74
Example: Stored Energy Density (cont.)
UE
1
*
 Re D  E
4


UH
1
*
 Re B  H
4


75
Example: Power Flow
S  E H
(instantaneous Poynting vector)

1
*
S  E  H  Re E  H
2
Define
Then

1
*
S  EH
2


(complex Poynting vector [VA/m2])
S  Re  S  [W / m2 ]
This formula gives the time-average power flow.
76
Appendix:
Proof of Moving Surface (Helmholtz) Identity
S
n
S(t)
S(t+t)
Note: S(t) is fist assumed to be planar, for simplicity.

d
1 


ˆ
ˆ
ˆ
B

n
dS

lim
B
t


t

n
dS

B
t

n
dS



 




t

0
dt S
t 
S t 

S t t 


S  t t 
B  t  t   nˆ dS 
 B  t  t   nˆ dS   B t  t   nˆ dS
S t 
S
77
Proof of Moving Surface Identity (cont.)
Hence:

d
1 


ˆ
ˆ
ˆ
B

n
dS

lim
B
t


t

B
t

n
dS

B
t


t

n
dS

 


 




t

0
dt S
t 
S

S t 

so
d
B
1
ˆ
ˆ
B  n dS  
 n dS  lim
B  t  t   nˆ dS


t 0 t
dt S
t
S t 
S
For the last term:
 B  t  t   nˆ dS   B  t   nˆ dS
S
S
78
Proof of Moving Surface Identity (cont.)
dS
Examine term:
 B  t   nˆ dS
S
nˆ
v s t
dr
t
t+dt
S
C
dS    dr   vs t   nˆ
Hence
B t   nˆ dS   B t   nˆ   dr  vs   nˆ t
79
Proof of Moving Surface Identity (cont.)
Since
 dr  vs  only has an
nˆ
component, we can write
B t   nˆ   dr  v   nˆ   B t    dr  v 
s
s
Hence
B t   nˆ dS  B t    dr  vs  t
Therefore, summing all the dS contributions:
1
lim
B  t   nˆ dS    B   dr  v s 

t 0 t
s
C
80
Proof of Moving Surface Identity (cont.)
Therefore
d
B
B  nˆ dS  
 nˆ dS   B   dr  v s 

dt S
t
S t 
C
Vector identity:
B   dr  vs    vs  B   dr
Hence:
d
B
B  nˆ dS  
 nˆ dS    v s  B   dr

dt S
t
S t 
C
81
Proof of Moving Surface Identity (cont.)
Notes:
 If the surface S is non-planar, the result still holds, since the magnetic
flux through any two surfaces that end on C must be the same, since
the divergence of the magnetic field is zero (from the magnetic Gauss
law). Hence, the surface can always be chosen as planar for the
purposes of calculating the magnetic flux through the surface.
 The identity can also be extended to the case where the contour C is
nonplanar, but the proof is omitted here.
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