class17A.ppt

Chapter 34
The Laws of Electromagnetism
Maxwell’s Equations
Displacement Current
Electromagnetic Radiation

The Equations of Electromagnetism
(at this point …)
E dA
q
  
0
B dA
   0
E dl d
dt
B
    
B dl I
   0
Gauss’ Law for Electrostatics
Gauss’ Law for Magnetism
Faraday’s Law of Induction
Ampere’s Law

1
2
E dA
q
  
0
B dA
   0
..monopole..
?
...there’s no
magnetic monopole....!!
Gauss’s Laws

4
E dl d
dt
B
    
B dl I
   0
3
.. if you change a
magnetic field you
induce an electric
field.........
.......is the reverse
true..?
Faraday’s Law
Ampere’s Law

...lets take a look at charge flowing into a capacitor...
...when we derived Ampere’s Law
we assumed constant current...
E
B
B dl I
   0

...lets take a look at charge flowing into a capacitor...
E
...when we derived Ampere’s Law
we assumed constant current...
.. if the loop encloses one
plate of the capacitor..there
is a problem … I = 0
B
Side view:(Surface
is now like a bag:)
E
B
B dl I
   0

Maxwell solved this problem
by realizing that....
B E
Inside the capacitor there must
be an induced magnetic field...
How?.

B E
x
x x x x
x x x x x
x x
A changing
electric field
induces a
magnetic field
How?. Inside the capacitor there is a changing E 
E
B

B E
x
x x x x
x x x x x
x x
A changing
electric field
induces a
magnetic field
where Id is called the
displacement current
B dl
d
dt
I
E
d
   
  
0 0 0

E
B

B E
B dl I d
dt
E
   
  
0 0 0

x
x x x x
x x x x x
x x
A changing
electric field
induces a
magnetic field
where Id is called the
displacement current
Therefore, Maxwell’s revision
of Ampere’s Law becomes....
B dl
d
dt
I
E
d
   
  
0 0 0

E
B

Derivation of Displacement Current
q EA I
dq
dt
d EA
dt
  
 
0 0
( )
For a capacitor, and .
I
d
dt
E
 0
( )

Now, the electric flux is given by EA, so: ,
where this current , not being associated with charges, is
called the “Displacement current”, Id.
Hence:
and: B ds I I
B ds I d
dt
d
E


  
   

  
0
0 0 0
( )

I
d
dt
d
E
  
0 0


Derivation of Displacement Current
q EA I
dq
dt
d EA
dt
  
 
0 0
( )
For a capacitor, and .
I
d
dt
E
 0
( )

Now, the electric flux is given by EA, so: ,
where this current, not being associated with charges, is
called the “Displacement Current”, Id.
Hence:
and: B dl I I
B dl I d
dt
d
E


  
   

  
0
0 0 0
( )

I
d
dt
d
E
  
0 0


Maxwell’s Equations of Electromagnetism
E dA
q
  
0
B dA
   0
E dl d
dt
B
    
Gauss’ Law for Electrostatics
Gauss’ Law for Magnetism
Faraday’s Law of Induction
Ampere’s Law B dl I
d
dt
E
   
  
0 0 0


in Vacuum (no charges, no masses)
Consider these equations in a vacuum.....
......no mass, no charges. no currents.....
B dl
d
dt
E
    
0 0

E dl
d
dt
B
   

E dA
q
  
0
B dA
   0
B dl I d
dt
E
   
  
0 0 0

E dA
   0
E dl
d
dt
B
   

B dA
   0

in Vacuum (no charges, no masses)
E dA
   0
B dA
   0
E dl
d
dt
B
   

B dl
d
dt
E
    
0 0


B dl
d
dt
E
    
0 0
 E dl
d
dt
B
   

Electromagnetic Waves
Faraday’s law: dB/dt electric field
Maxwell’s modification of Ampere’s law
dE/dt magnetic field
These two equations can be solved simultaneously.
The result is:
E(x, t) = EP sin (kx-t)
B(x, t) = BP sin (kx-t) ẑ
ĵ

B dl
d
dt
E
    
0 0

E dl
d
dt
B
   

B
E
dE
dt

dB
dt


B dl
d
dt
E
    
0 0

E dl
d
dt
B
   

B
E
Special case..PLANE WAVES...
satisfy the wave equation
  
 
A t
sin( )
Maxwell’s Solution
v
 
 
 

2
2 2
2
2
1
x t

 
E E x t j B B x t k
y z
 
( , ) ( , )
dE
dt

dB
dt


Plane Electromagnetic Waves
x
Ey
Bz
ĵ
c

Static wave
F(x) = FP sin (kx + )
k = 2  
k = wavenumber
 = wavelength
F(x)
x

Moving wave
F(x, t) = FP sin (kx - t )
 = 2  f
 = angular frequency
f = frequency
v =  / k
F(x)
x

v

x
v Moving wave
F(x, t) = FP sin (kx - t )
What happens at x = 0 as a function of time?
F(0, t) = FP sin (-t)
F
For x = 0 and t = 0  F(0, 0) = FP sin (0)
For x = 0 and t = t  F (0, t) = FP sin (0 – t) = FP sin (– t)
This is equivalent to: kx = - t  x = - (/k) t
F(x=0) at time t is the same as F[x=-(/k)t] at time 0
The wave moves to the right with speed /k

Plane Electromagnetic Waves
x
Ey
Bz
Notes: Waves are in Phase,
but fields oriented at 900.
k=2.
Speed of wave is c=/k (= f)
At all times E=cB.
c m s
  
1 3 10
0 0
8
/ /
 
ĵ
c

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