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Chapter 29
Maxwell’s Equations and
Electromagnetic Waves
Coulomb Oersted
France
Denmark
Ampere
France
Faraday
England
Lenz
Russia
Maxwell
Scotland
All electric and magnetic phenomena could be
described using only 4 equations involving electric
and magnetic fields —— Maxwell’s equations
→ Culmination: Electromagnetic waves
2
Changing E produces B
Changing magnetic field
electric field
Changing electric field
magnetic field ?
A beautiful symmetry in nature:
Ampere’s law:
 B  dl 
0 I  S1
0  S2
E  B
S2
S1
?
I
3
Discontinuity of current
S2
Ampere’s law:
 B  dl 
S1
0 I  S1
I
0  S2
+
+
+
+
-
E
Contradiction? → discontinuity of the current

j  dS   I  0
Which result is right?
S1  S2
An extra term in Ampere’s law → changing E
4
Displacement current
Conduction current:
S1
dQ d
I

S
dt
dt
I
I d
dE
 j 
 0
S
dt
dt
Electric field between plates:
Displacement current:
Q
S2
+
+
+
+
-
Q
E

E
0
dE
dE
jD   0
, ID  0
dt
dt
5
Generalized Ampere’s law
ID
I
dE
dE
jD   0
, ID  0
dt
dt
Ampere’s law in a general form:
dE
 B  dl  0 (I  I D )in  0 Iin  0 0 dt
1) ID is produced by changing electric field
2) Continuity of total current:
(j j
D
)  dS  0
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Charging capacitor
Example1: A parallel-plate capacitor is charging
with dE / dt  2 109V / m  s. Determine (a) the
conduction current I ; (b) magnetic field.
Solution: (a)
dE
jD   0
dt
Continuity of total current:
I
0.1m
dE
I  I D  jD S   R  0
dt
2
 5.56 104 A
7
dE / dt  2 10 V / m  s
9
0.1m
(b) magnetic field ?
Generalized Ampere’s law:
 B  dl   (I  I
0
I
r
)
D in
0 0 dE
dE
2
r  0.1m : B  2 r  0   0
 r  B 
r
dt
2 dt
0 I
r  0.1m : B  2 r  0 I  B 
2 r
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Capacitor in LC circuit
Question: The voltage on a capacitor is changing
as V  V0 cos t . What is the EMF on the square
coil inside the capacitor?
dE  0 dV
jD   0

dt
d dt
μ0 jD
B
r  B
2
d
a
a
dB
 
dt
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Summary of electromagnetism
Electrostatic / induced electric / total electric field:
E
s
E
s
 dS 
Qin
0
 dl  0;
;
 E  dS  0
i

dB
 Ei  dl   dt 
Magnetic field created by IC or ID :
Q
 E  dS  
0
dB
 E  dl   dt
 B  dS  0
dE
 B  dl  0 (I  I D )in  0 I  0 0 dt
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Maxwell’s equations (1)
Finally we can state all 4 of Maxwell’s equations:
 E  dS 
Q
0
① source of E
 B  dS  0
② no magnetic
charges/monopoles
d B
 E  dl   dt
③ changing B → E
dE
 B  dl  0 I  0 0 dt
④ changing E → B
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Maxwell’s equations (2)
Differential form of Maxwell’s equations:

E 
0
 B  0
B
 E  
t
E
 B  0 j  0 0
t
1) Basic equations for
all electromagnetism
2) As fundamental as
Newton’s laws
3) Important outcome:
electromagnetic waves
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Production of EM waves
Maxwell’s prediction
Hertz’s experiment
“Antenna”
B
Near field & radiation field
E
→ electromagnetic wave
 (t )
Also by accelerating charges
E
Antenna
Ground
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Wave equations for EM wave
Free space: no charges or conduction currents
 E  0

 B  0

   E   B

t

   B  0 0 E

t
2

E
2
  E  0  0 2
t

   (  E )   (  B)
t
2 E
  0  0 2
t
 ( E)  （ E） 2 E
 2 E
Wave equation!
14
Speed of EM wave
2

E
2
 E  0  0 2
t
2

B
2
 B  0  0 2
t
3-D wave equation → 1-D (plane) wave equation
2 E
2 E
 0  0 2 ,
2
x
t
2 B
2 B
 0  0 2
2
x
t
Compare with standard wave equation:
2
2 y

y
1
2
8

v

v


3.0

10
m/s  c !
2
2
t
x
0 0
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Properties of EM wave
2
2

E

E
2
c
 2 ,
2
x
t
Particular solutions:
1) Transverse wave
2) In phase:
2
2

B

B
2
c
 2
2
x
t
x
E  E0 cos  (t  )
c
x
B  B0 cos  (t  )
c
E
c
B
3) E  B
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Energy in EM wave
Total energy stored per unit volume in EM wave:
2
2
1
1
B
B
u  0 E2 
 0 E2 
2
2 0
0
E
( c
B
1
0 0
)
Energy transports per unit time per unit area:
dU  udV  u  A  cdt
EB
dU
S
 uc 
0
A  dt
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Poynting vector
Consider the direction of energy transporting:
S
1
0
( E  B)
→ Poynting vector
1 E0 B0
Time averaged S is intensity: S 
2 0
Example 2: Show the direction
of energy transporting inside the
battery and resistor.
I
B
E
S
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Sunshine
Example3: Radiation from the Sun reaches the
Earth at a rate about 1350W/m2. Assume it is a
single EM wave, calculate E0 and B0.
Solution: Rate → time averaged S / intensity
2S
1 E0 B0 1
2
3

E


1.01

10
V /m
S
  0 cE0
0
 0c
2 0
2
1
 u  c  um c
2
E0
B0 
 3.37  106 T
c
19
*Radiation pressure
EM waves carry energy → also carry momentum
Be absorbed / reflected → radiation pressure
Absorbed: P  S
c
Reflected: P  2S
c
20