Transcript notes 1 6340 Maxwell`s Equations
ECE 6340 Intermediate EM Waves
Fall 2013
Prof. David R. Jackson Dept. of ECE
Notes 1
1
Maxwell’s Equations E D C/m 2 , B H Wb/m 2 T A/m H J B
t
D
t
v
(Faraday's Law) (Ampere's Law) (Gauss's Law) 0 (Magnetic Gauss's Law) 2
Current Density J J
v v
2 [A/m ] The current density vector points in the direction that positive charges are moving. J
v v
v
Positive charges
v
Note: The charge density
v
free to move in the material).
is the "free charge density" (the charge density that is 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
number).
is the charge density of the electrons in the beam (a negative 4
Current Density A small surface
dS
current flow.
is introduced perpendicular to the direction of
dI
= current flowing through
dS
J J
v
ˆ
dI
dV v
ˆ ++++++ ++++++ ++++++ J
dS dl
Charge calculation:
dQ
v
v
v
v
J
dt dS
Hence
dI
J
dS
5
Current Density (cont.) A more general case of an arbitrary orientation of the surface is now considered.
J J
n
J
t
dS
J
t t
ˆ J
n
J
n
ˆ
dI
Integrating over the surface, we have
I
S
Surface
S
component of current density crossing
dS
6
Continuity Equation From Gauss’s Law,
v
t
D D
t
t v
From Ampere’s Law, so H H J D
t
J D
t
7
Continuity Equation (cont.) H J D
t
D
t
J From the last slide:
t v
D
t
Hence J
t v
“Continuity Equation” 8
Integral Form of Continuity Equation
V
J
dV
V
v
t dV V S
(closed)
n
ˆ Apply divergence theorem:
V
J Flux per unit volume
dV
S
J Net flux out of
S
9
Integral Form of Continuity Equation (cont.) Hence we have
S
J
V
v
t dV
Assume
V
is stationary (not changing with time):
V
v
t dV
d dt V
v dV
dQ encl dt
Hence
S
J
dQ encl dt
charge conservation 10
Integral Form of Continuity Equation (cont.)
dQ encl dt
S
J + + + +
Q encl
+ +
n
ˆ
S
( stationary ) J 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 v s
(
r
)
Q encl
J Note:
v s
denotes the velocity of the surface.
dQ encl dt
S
J
S
v s
Note :
dQ encl dt
S
v
v
v s
Difference in charge and surface velocities!
12
Generalized Continuity Equation (cont.) Define the current density in the moving coordinate system : J
v
s
S Q encl
J
v s
(
r
)
S
J
dQ encl 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
t
J Assume a homogeneous conducting medium that obeys Ohm’s law: J E We then have
v
t
v
Hence
t v
v
14
Relaxation Equation (cont.)
v
t
v
For an initial charge density
v
(
r
,0) at
t
= 0 we have:
v
v 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.
H J D
t
Ampere's law Stokes’s Theorem:
S
(open)
n
ˆ Note: right-hand rule for
C
.
C S
H Circulation per unit area of
S
C
H
dr
Circulation on boundary of
S
17
Integral Forms of Maxwell’s Eqs. (cont.) Integrate Ampere’s law over the open surface
S
:
S
H
S
J
S
D
t
Apply Stokes’s theorem: or
C
H
dr
S
J
S
D
t
C
H
dr
s S
D
t
Note: The current
i s
is measured in the frame of the observer.
(The current
i s
is the current though the surface
S.
) 18
Integral Forms of Maxwell’s Eqs. (cont.)
C
H
dr
s S
D
t
Ampere’s law
Note: In statics ,
i s
is independent of the shape of
S
, since the last term is zero and the LHS is independent of the shape of
S
.
Similarly:
C
E
dr
S
B
t
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.)
v
Electric Gauss law
S
(closed)
V n
ˆ
Divergence theorem:
V
D Flux per unit volume
dV
S
D Net flux out of
S
Integrate the electric Gauss law throughout
V
and apply the divergence theorem:
S
D
V
v dV
Hence we have
S
D
Q encl
20
Integral Forms of Maxwell’s Eqs. (cont.) 0
Magnetic Gauss law
S
B Apply divergence theorem 0 21
Boundary Form of Maxwell’s Eqs.
1 , , 1 1 ˆ
C
C
n
ˆ J
s
2 , , 2 2 We allow for a surface current density . The narrow rectangular path
C
is chosen arbitrarily oriented in the tangential direction as shown.
0 The unit normal points towards region 1.
C
H
dr
s S
D
t
C
Evaluate the LHS of Ampere’s law on the path shown: lim 0
C
H
dr
lim 0
C
H
dr
C
H
dr
C
H
dr
lim 0
C
H
dr
H 1 H 2 22
Boundary Form of Maxwell’s Eqs. (cont.)
n
ˆ
C
1 , , 1 1 ˆ
C
C
n
ˆ J
s
2 , , 2 2 Evaluate the RHS of Ampere’s law for path:
i s
lim 0 0 J
s
J
s
n
d
J
s
n
ˆ where we have used the following vector identity: Next, for the displacement current term in ampere's law, we have lim 0
S
D
t
0 since
S
0 23
Boundary Form of Maxwell’s Eqs. (cont.) 1 , , 1 1 ˆ
C
C
n
ˆ Hence, we have ˆ H 1 H 2 J
s
2 , , 2 2 J
s
This may be written as: Since ˆ
n
ˆ H is an
arbitrary
H 1 H 2
t
unit tangent vector , J
s
n
ˆ
t
1 H 2 J
s
J
s
Hence
n
ˆ H 1 H 2 J
s
points into region 1! 24
Boundary Form of Maxwell’s Eqs. (cont.) 1 , , 1 1 ˆ
C
C
n
ˆ J
s
2 , , 2 2 Similarly, from Faraday's law, we have ˆ E 1 E 2 0 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.
There could also be a surface current, but it does not affect the result.
+
S
n
ˆ 1 , , 1 1 +
s
+ + +
S
S
2 , , 2 2 A “pillbox” surface is chosen.
0 The unit normal points towards region 1.
S
D
Q encl
Evaluate LHS of Gauss’s law lim 0
S
D
S
lim 0
S
D over the surface shown:
S
D lim 0
S
D
S
n
ˆ D 1 D 2
S
S
D
S
26
Boundary Form of Maxwell’s Eqs. (cont.) +
S
n
ˆ 1 , , 1 1 +
s
+ + +
S
2
S
2 2 RHS of Gauss’s law: lim 0
Q encl
lim 0
Q encl
lim 0
S
S
s dS
Hence Gauss’s law gives us D 1 D 2
S
s S
Hence we have D 1 D 2
s
Similarly , from the magnetic Gauss law: ˆ B 1 B 2 0 27
Boundary Form of Maxwell’s Eqs. (cont.) We also have a boundary form of the continuity equation.
v
From before: D 1 D 2
s
J Noting the similarity, we can write:
t v
J 1 J 2
t s
Note: If a surface current exists, this may give rise to another surface charge density:
s
J
s
t s
In general:
n
ˆ J 1 J 2
s
J
s
t s
28
Summary of Maxwell Equations Point Form Integral Form Boundary Form H E J D
t
B
t
C C
H
dr
E
dr
i s
S
S
B D
t
t
n
ˆ H 1 H 2
n
ˆ E 1 E 2 0 J
s
D
v S
D
Q encl n
ˆ D 1 D 2
s
B 0
S
B 0
n
ˆ B 1 B 2 0 Continuity equation : J
t v S
J
dQ encl dt n
ˆ J 1 J 2
s
J
s
s
t
29
C
Faraday’s Law E
dr
S
B
t
If
S
is stationary :
S
B
t
d dt S
B
d
dt
= magnetic flux through
S S
(open) Hence:
C
E
dr
d
dt n
ˆ
C
30
Faraday’s Law (cont.) If
S
is moving (e.g., expanding):
v s
Previous form is still valid:
C C
E
dr
S
B
t
However,
S
B
t
!
d dt S
B
d
dt
31
Faraday’s Law (cont.) Vector identity for a moving path:
d dt S
B
S
B
t
C
v s
B
dr
“Helmholtz identity”
v s C
(See appendix for derivation) 32
Faraday’s Law (cont.) Start with
C
E
dr
S
B
t
Then use Hence
d dt S
B
C
E
dr
d
dt
S
B
t
C
v s
B
C
v s
B
dr
dr
or
C
E
v s
B
dr
d
dt
33
Define Faraday’s Law (cont.)
C
E
v s
B
dr
d
dt EMF
C
E
v s
B
dr
Then
EMF
d
dt
34
Two Forms of Faraday’s Law
C
E
dr S
B
t
Faraday’s law
C
E
v s
B
dr
d
dt
“Generalized” Faraday’s law Note: The path
C
is,
C
= means
C
(
t
), a fixed path that is evaluated at a given time
t
. That
C
(
t
) is a “snapshot” of the moving path. The same comment for
S
=
S
(
t
).
35
Two Forms of Ampere’s Law
C
H
dr
s S
D
t
Ampere’s law
C
H
dr
s d dt S
D
C
v s
D
d r
“Generalized” Ampere’s law 36
Example
y
B
z
ˆcos
t
t
Note:
t
is in [ s ], is in [ m ].
x 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) V
C
E
dr
S
S
B z
t
B
t dS
B z
t
S dS
B
t z
sin B
z
is independent of (
x
,
y
).
V
C
E
dr
t
2 sin
t
[V]
y
Choose unit normal:
n
ˆ B
z
ˆcos
t
t
Would this result change if the path were not moving? No!
x
38
Example (cont.) (2) Electric field (from voltage drop) E 2
t
2 sin
t
E
t
2 sin 2
t
E
t
2 2 sin
t
so E
t
sin
t
2 E ˆ
t
2 sin
t
[V/m]
y
dr
d
Note: There is no or
z
component (seen from the curl of the magnetic field): H
x
39
Example (cont.) (3) EMF (from Generalized Faraday’s law)
C
E
v s
B
S
B
dr
d
dt
S
B B
z dS z
cos B
y
z
ˆ cos
t x
40
cos
t
Example (cont.)
y d
2
t
cos
t dt
path moving
t
2 sin
t
field changing
EMF
d
dt EMF
2
t
cos
t
t
2 sin
t
[V]
x
41
Example (cont.) Electric field and voltage drop (Alternative method: from the Generalized Faraday’s law )
y
2 E
v s
B
d
dt
2 E
v s
B
z
d
dt
B Since
v s
2 E = 1 [ m/s ]: B
z
d
dt
E B
z
1 2
d
dt
so E cos 1
d
2
t dt
z
ˆ cos
t x
42
Example (cont.) E cos 1 2
d
t dt
cos cos
t
1 2
t
cos
t
2
t
cos
t
2 sin
t
t
t
sin
t
E
t
sin
t
2 [V/m] V 2
t
2 sin
t
[V] E ˆ
t
2 sin
t
[V/m] V
t
2 sin
t
[V] 43
Example (cont.) EMF (Alternative method: Faraday’s law ) V
t
2 sin
t
(Faraday’s law)
EMF
V
C
E
v s
B
C
v s
dr
B
dr C
v s
B
dr
2 0
z
ˆ cos 2 0 cos cos
d
2
d
B
z
ˆ cos
t y
v s
= 1 [ m/s ]
x
44
Hence
EMF
t
2 sin
t
cos Example (cont.)
y
B
z
ˆ cos
t
2 so
EMF
t
2 sin
t
cos
x v s
= 1 [ m/s ] 45
Example Summary
Voltage drop V (
t
) around the closed path: V
t
2 sin
t
[V] Electric field on the path: E ˆ
t
2 sin
t
[V/m] EMF drop around the closed path:
EMF
2
t
cos
t
2 sin
t
[V] 46
Example Find the voltage on the voltmeter Voltmeter + V
m y B x a
Perfectly conducting leads
V
0 + Applied magnetic field: B
z
ˆ cos 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.) V
C
E
dr
S
S
B
t z
B
t dS
B
t z
S dS
B z
t
sin V
m
0
a
2 sin V
m
0
a
2 sin Practical note: In such a measurement, it is good to keep the leads close together (or even better, twist them.)
C
+ V
m
Voltmeter
y B
B
z
ˆ cos
B z
t
sin
x a V
0 + 48
Generalized Ohm's Law A resistor is moving in a magnetic field.
R A
i
+ V
EMF
Ri B A
E
v s
B
dr
Ri
V
B A
v s
B
dr
Ri v B
B You will be proving this in the homework.
49
Perfect Electric Conductor (PEC) A PEC body is moving in the presence of a magnetic field. B
v
Inside the PEC body: E B You will be proving this in the homework.
50
Example Find the voltage V
m
on the voltmeter Sliding bar (perfect conductor) Velocity
v
0
W
V
m
+ Voltmeter -
V
0 +
dW
v
0
dt L
B
z
ˆ We neglect the magnetic field coming from the current in the loop itself.
51
Example (cont.)
EMF
C
E
v
B
z
(
LW
) B
dr
d
dt d
dt
B
z L dW dt
B
z Lv
0
Lv
0 V
m
+ Voltmeter Hence
EMF
Lv
0 (Generalized Faraday’s law) Velocity
v
0 B
L
z
ˆ
C V
0 + 52
Example (cont.)
EMF
C
V
m
E
v s
B
V
0
dr
PEC bar PEC wires Hence V
m
V
0
Lv
0
EMF
Ri
(
R
0) Velocity
v
0 so V
m
V
0
Lv
0 V
m
Voltmeter + -
C V
0 +
L
53
Example (cont.) Now let's solve the same problem using Faraday's law .
C
E
dr
S
B
t
0
S
0
dS
V
m
Voltmeter + Velocity
v
0
C C
E
dr
V
m
0
V
0 )
top
E
dr
B
L
z
ˆ
V
0 + 54
Example (cont.) For the top wire we have
top
E
dr
0
L
E E
x x
ˆ B 0
L
E
x
dx
L
E
x
z
ˆ
v
0 B
z
ˆ Hence
top
E
dr
Lv
0 so
C
E
dr
V
m
0
Lv
0 V
m
Voltmeter + -
L
Velocity
v
0
C V
0 + 55
Example (cont.)
C
E
dr
V
m
0
Lv
0 0 Hence, we have V
m
V
0
Lv
0 V
m
Voltmeter + B
z
ˆ Velocity
v
0
C L V
0 + 56
Time-Harmonic Representation Assume
f
(
r,t
) is sinusoidal: Re cos
j
e
Denote
j
(Phasor form of
f
) arg Then Re From Euler’s identity:
e jz
cos
z
j
sin
z
cos
z
Re if
z
real , 57
Time-Harmonic Representation (cont.) cos
Re
j
Notation: for scalars F for vectors 58
Time-Harmonic Representation (cont.) Derivative Property:
f
t
t
Re Re Re
j
t
Hence: phasor for the derivative F
t
t
j
j
for scalars for vectors 59
Time-Harmonic Representation (cont.) Consider a differential equation such as Faraday’s Law: E B
t
This can be written as Re Re
j
or Re
j
0 60
Time-Harmonic Representation (cont.) Let choose
c
r jc i
Then Re
c e
0
j
B
Does this imply that
c
= 0 ?
choose
t
0 / 2
c r
0
c i
0
c e
t
Phasor interpretation: Im Re Re
c e
Hence:
c
= 0 61
Time-Harmonic Representation (cont.) Therefore so
j
B
j
B
0
j
B
0 Hence E B
t
Time-harmonic (sinusoidal) steady state
j
B
62
Time-Harmonic Representation (cont.) We can also reverse the process:
j
B
Re
j
0 Re Re
j
E B
t
63
Time-Harmonic Representation (cont.) Hence,
in the sinusoidal steady-state
, these two equations are equivalent: B
j
t
B
64
Maxwell’s Equations in Time-Harmonic Form
Η
v
0
j
B
65
Continuity Equation Time-Harmonic Form J
t v 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
Hence 0 67
Frequency-Domain Curl Equations Start with Ampere's law:
Η Η
j
Hence
v j
D
D
j
v
(We also assume the continuity equation here.)
v
68
Frequency-Domain Curl Equations (cont.) 0 0
J
v
With the continuity equation assumed Hence, we often consider only the curl equations in the frequency domain, and not the divergence equations.
69
Time Averaging of Periodic Quantities Define:
T
1
T T
0
period Assume a
product
of sinusoidal waveforms:
A
cos
B
cos
t t
F
Ae j
G
Be j
A B
cos
t
A B
1 2 cos
t
t
70
Time Average (cont.)
A B
1 2 cos The time-average of a constant is simply the constant.
t
The time-average of a sinusoidal wave is zero.
Hence: 1 2
A B
cos 71
Time Average (cont.) The phasors are denoted as:
F
Ae j
G
B e j
Consider the following: so Re
FG
*
A B e j
A B
cos Hence 1 Re 2 The same formula extends to vectors as well.
72
Example: Stored Energy Density U
E
1 2 D 1 2 D E
x x
D E
y y
D E
z z
Re
D e
etc . D E
x x
D E
y y
D E
z z
1 1 2 Re
D E x x
* 1 2 Re
D E y
*
y
2 Re
D E x x
*
D E y
*
y
D E z z
* 1 2 Re
D E z z
* or 1 2 Re * 73
Example: Stored Energy Density (cont.) From electrostatics: U
E
1 2 Hence, U
E
1 2 1 4 Re * Similarly, U
H
1 2 1 4 Re * 74
Example: Stored Energy Density (cont.) U
E
U
H
1 4 Re 1 4 Re * * 75
Example: Power Flow S (instantaneous Poynting vector) S Define
S
2 1 Then S Re * 1 2 Re * (complex Poynting vector [ VA/m 2 ]) 2 [W / m ] This formula gives the time-average power flow.
76
Appendix: Proof of Moving Surface (Helmholtz) Identity
S n S
(
t
)
S
(
t+
t
)
d dt
Note:
S
(
t
) is fist assumed to be planar, for simplicity.
S
B lim
t
0 1
t
B
t
B
t
B
t
B
t
S
B
t
77
Proof of Moving Surface Identity (cont.) Hence:
d dt S
B lim
t
0 1 B
t
so
d dt S
B B
t
B
lim
t
0 1
t
S
B
S
B
t
t
For the last term:
S
B
t
S
B 78
Proof of Moving Surface Identity (cont.)
dS n
ˆ
v s
t
Examine term:
S
B
dr S t t+dt
Hence B
C dS
dr
v s
B
s
t
79
Proof of Moving Surface Identity (cont.) B
n
ˆ
s
B
s
Hence B B
s
t
Therefore, summing all the
dS
contributions: lim
t
0 1
t
s
B
C
B
s
80
Proof of Moving Surface Identity (cont.) Therefore
d dt S
B Vector identity: B B
t
C
B
v s
B
dr s s
Hence:
d dt S
B B
t
C
v s
B
dr
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. 82