Transcript Slide 1

ECE 3317
Prof. Ji Chen
Spring 2014
Notes 5
Poynting Theorem
Adapted from notes by Prof. Stuart A. Long
1
Poynting Theorem
The Poynting theorem is one on the most important in EM theory. It
tells us the power flowing in an electromagnetic field.
John Henry Poynting (1852-1914)
John Henry Poynting was an English physicist. He was a
professor of physics at Mason Science College (now the University
of Birmingham) from 1880 until his death.
He was the developer and eponym of the Poynting vector, which
describes the direction and magnitude of electromagnetic energy
flow and is used in the Poynting theorem, a statement about energy
conservation for electric and magnetic fields. This work was first
published in 1884. He performed a measurement of Newton's
gravitational constant by innovative means during 1893. In 1903 he
was the first to realize that the Sun's radiation can draw in small
particles towards it. This was later coined the Poynting-Robertson
effect.
In the year 1884 he analyzed the futures exchange prices of
commodities using statistical mathematics.
(Wikipedia)
2
Poynting Theorem
E  
B
t
H  J 
D
t
From these we obtain:
H   E    H 
E   H

B
t
J E  E 
D
t
3
Poynting Theorem (cont.)
H   E    H 
E   H

B
t
J E  E 
D
t
Subtract, and use the following vector identity:
H   E   E    H
   E  H 
We then have:
 E  H
  J E  H

B
t
E
D
t
4
Poynting Theorem (cont.)
 E  H
  J E  H

B
t
E
D
t
Next, assume that Ohm's law applies for the electric current:
J  E
 E  H
    E  E   H

B
t
E
D
t
or
 E  H
  
E
2
H 
B
t
E
D
t
5
Poynting Theorem (cont.)
 E  H
  
E
2
H 
B
t
E
D
t
From calculus (chain rule), we have that
D
E 
1 

E
 E
E E

t
t 
2 t

B
H 
1 

H 
 H 
H H

t
t 
2 t


Hence we have
 E  H
  
E
2

1 
2 t
H
H 
1 
2 t
E E 
6
Poynting Theorem (cont.)
 E  H
  
E
2

1 
2 t
H
H 
1 
2 t
E E 
This may be written as
 E  H
  
2
E

1 
2 t
H
2

1 
2 t
E
2
or
 E  H

  E
2
 1

  H
t  2
2
  1

  E
 t  2
2



7
Poynting Theorem (cont.)
Final differential (point) form of Poynting theorem:
 E  H
  
E
2
 1

  H
t  2
2
  1

  E
 t  2
2



8
Poynting Theorem (cont.)
Volume (integral) form
Integrate both sides over a volume and then apply the divergence theorem:
 E  H
  
    E  H  dV     E
V
E
2
2
dV 
V
  E  H   nˆ dS
S
 1

  H
t  2

V
  E
V
2
dV 
2
 1
  H
t  2

V
  1

  E
 t  2
2
 1
  H
t  2
2



 1

dV


  E

t  2

V
2
2
 1

 dV  
  E
t  2

V

 dV

2

 dV

9
Poynting Theorem (cont.)
Final volume form of Poynting theorem:

S
 E  H   nˆ dS     E
2
dV 
V

V
 1
  H
t  2
2
 1

 dV  
  E
t  2

V
2

 dV

For a stationary surface:

S
 E  H   nˆ dS     E
V
2
dV 

t

V
1
  H
2
2


 dV 
t

1
  2  E
V
2

 dV

10
Poynting Theorem (cont.)
Physical interpretation:

  E  H   nˆ dS
S
  E
V
(Assume that S is stationary.)
2
dV 

t

V
1
  H
2
2


 dV 
t

1
  2  E
V
2

 dV

Power dissipation as heat (Joule's law)
Rate of change of stored magnetic energy
Rate of change of stored electric energy
Right-hand side = power flowing into the region V.
11
Poynting Theorem (cont.)
Hence

  E  H   nˆ dS
 pow er flow ing into the region
S
Or, we can say that
  E  H   nˆ dS
 pow er flow ing out of the region
S
Define the Poynting vector:
 S  nˆ dS
S  EH
 pow er flow ing out of the region
S
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Poynting Theorem (cont.)
Analogy:
 S  nˆ dS
 pow er flow ing out of the region
S
 J  nˆ dS
 current flow ing out of the region
S
J = current density vector
S = power flow vector
13
Poynting Theorem (cont.)
S  EH
direction of power flow
E
S
H
The units of S are [W/m2].
14
Power Flow
S  EH
surface S
nˆ
The power P flowing through the surface S (from left to right) is:
P
 S  nˆ dS
S
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Time-Average Poynting Vector
Assume sinusoidal (time-harmonic) fields)
S  x, y, z, t   E  x, y, z, t   H  x, y, z, t 
E  x, y , z , t   R e E  x, y , z  e
H  x, y , z , t   R e H  x, y , z  e
j t

j t

From our previous discussion (notes 2) about time averages, we know that
S t   E t   H t  
1
2
Re E  H
*

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Complex Poynting Vector
Define the complex Poynting vector:
S
1
2
E  H 
*
We then have that
S  x, y, z, t  = R e  S  x, y, z  
Note: The imaginary part of the complex Poynting vector corresponds to the
VARS flowing in space.
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Complex Power Flow
S 
1
EH
*
2
surface S
nˆ
The complex power P flowing through the surface S (from left to right) is:
P
 S  nˆ dS
S
R e  P   W atts
Im  P   V ars
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Complex Poynting Vector (cont.)
What does VARS mean?
  R e S  nˆ dS 
Equation for VARS (derivation omitted):
1
VARS  2     H
4
V 
W atts flow ing in
S
  Im S  nˆ dS 
 2
V A R S flow ing in

2

Wm  We
1
4
2

 E  dV


S
The VARS flowing into the region V is equal to the difference in the time-average
magnetic and electric stored energies inside the region (times a factor of 2).
Watts + VARS
S
E
VARS consumed
H
V
Power (watts)
consumed

19
Note on Circuit Theory
Although the Poynting vector can always be used to calculate power flow,
at low frequency circuit theory can be used, and this is usually easier.
Example (DC circuit):
P
I
R
V0
z
P 
  E  H   zˆ dS
S
S
P  V0 I
The second form is much easier to calculate!
20
Example: Parallel-Plate Transmission Line
y
V  z   V0 e
I
I  z   I0 e
h
, 
+
V -
x
w
z
 jkz
 jkz
The voltage and current have
the form of waves that travel
along the line in the z direction.
y
At z = 0:
E
V  0   V0
H
h
x
I  0   I0
w
21
Example (cont.)
At z = 0:
y
I
h
, 
 V0 
E ( x , y , 0)   yˆ 

h


+
V -
I 
H  x , y , 0   xˆ  0 
w
x
w
(from ECE 2317)
z
y
S 
E
2
H
h
1
x
S
E  H 
*
V  I 
zˆ  0   0 
2  h  w 
1
*
w
22
Example (cont.)
 V0   I 0 
S  zˆ 
 
2  h  w 
y
1
I
h
, 
+
V -
*
Hence
x
h w
Pf 
w
z
h w
  S  zˆ dx dy    S
0 0
E
dx dy
0 0
1  V0   I 0 
Pf  
 
2  h  w 
y
z
*
 wh 
H
h
x
Pf 
1
2
V0 I 0
*
w
23