Download Ch. 34 - Maxwell`s equations

Chapter 34
The Laws of Electromagnetism
Maxwell’s Equations
Displacement Current
Electromagnetic Radiation
The Electromagnetic Spectrum
The Equations of Electromagnetism
(at this point …)
q
Gauss’ Law for Electrostatics
 E  dA  0
Gauss’ Law for Magnetism
 B  dA  0
Faraday’s Law of Induction
d

B
E

dl



dt
Ampere’s Law
 B  dl  0 I
The Equations of Electromagnetism
Gauss’s Laws
..monopole..
q
1
 E  dA  0
2
 B  dA  0
...there’s no
magnetic monopole....!!
?
The Equations of Electromagnetism
Faraday’s Law
3
d

B
E

dl



dt
Ampere’s Law
4
 B  dl  0 I
.. if you change a
magnetic field you
induce an electric
field.........
.......is the reverse
true..?
...lets take a look at charge flowing into a capacitor...
...when we derived Ampere’s Law
we assumed constant current...
 B  dl  0 I
B
E
...lets take a look at charge flowing into a capacitor...
...when we derived Ampere’s Law
we assumed constant current...
B
E
 B  dl  0 I
E
.. if the loop encloses one
plate of the capacitor..there
is a problem … I = 0
Side view: (Surface
is now like a bag:)
B
Maxwell solved this problem
by realizing that....
Inside the capacitor there must
be an induced magnetic field...
How?.
B
E
Maxwell solved this problem
by realizing that....
Inside the capacitor there must
be an induced magnetic field...
B
E
How?. Inside the capacitor there is a changing E 
B
x
x x x x
E
x x x x x
x x
A changing
electric field
induces a
magnetic field
Maxwell solved this problem
by realizing that....
Inside the capacitor there must
be an induced magnetic field...
B
E
How?. Inside the capacitor there is a changing E 
B
x
x x x x
E
x x x x x
x x
A changing
electric field
induces a
magnetic field
d E
 B  dl  00 dt  0 Id
where Id is called the
displacement current
Maxwell solved this problem
by realizing that....
Inside the capacitor there must
be an induced magnetic field...
B
E
How?. Inside the capacitor there is a changing E 
B
x
x x x x
E
x x x x x
x x
A changing
electric field
induces a
magnetic field
Therefore, Maxwell’s revision
of Ampere’s Law becomes....
d E
 B  dl  00 dt  0 Id
where Id is called the
displacement current
d

E
B

dl


I



0
0 0

dt
Derivation of Displacement Current
dq
d( EA )
For a capacitor, q  0 EA and I  dt  0 dt
.
d ( E )
Now, the electric flux is given by EA, so: I   0 dt ,
where this current , not being associated with charges, is
called the “Displacement current”, Id.
Hence:
and:
d E
I d  0  0
dt
 B  ds  0( I  Id )
d

  B  ds  0 I  00 E
dt
Derivation of Displacement Current
dq
d( EA )
For a capacitor, q  0 EA and I  dt  0 dt .
d ( E )
Now, the electric flux is given by EA, so: I   0 dt ,
where this current, not being associated with charges, is
called the “Displacement Current”, Id.
Hence:
d E
I d  0  0
dt
and:
 B  dl  0( I  Id )
d

  B  dl  0 I  00 E
dt
Maxwell’s Equations of Electromagnetism
q
Gauss’ Law for Electrostatics
 E  dA   0
Gauss’ Law for Magnetism
 B  dA  0
Faraday’s Law of Induction
d

B
E

dl



dt
Ampere’s Law
d E
 B  dl  0 I  00 dt
Maxwell’s Equations of Electromagnetism
in Vacuum (no charges, no masses)
Consider these equations in a vacuum.....
......no mass, no charges. no currents.....
q
 E  dA   0
 E  dA  0
 B  dA  0
 B  dA  0
d B
 E  dl   dt
d E
B

dl


I



0
0 0

dt
d B
 E  dl   dt
d E
 B  dl  0 0
dt
Maxwell’s Equations of Electromagnetism
in Vacuum (no charges, no masses)
 E  dA  0
 B  dA  0
d B
 E  dl   dt
 B  dl  0 0
d E
dt
Electromagnetic Waves
Faraday’s law:
dB/dt
electric field
Maxwell’s modification of Ampere’s law
dE/dt
magnetic field
d E
 B  dl  00 dt
d B
 E  dl   dt
These two equations can be solved simultaneously.
The result is:
E(x, t) = EP sin (kx-t) ĵ
B(x, t) = BP sin (kx-t) ẑ
Electromagnetic Waves
d E
 B  dl  0 0 dt
B

dE
dt
d B
 E  dl   dt
E

dB
dt
Electromagnetic Waves
d E
 B  dl  0 0 dt
B
v

dE
dt
d B
 E  dl   dt

Special case..PLANE WAVES... E  E y ( x ,t ) j
satisfy the wave equation
Maxwell’s Solution

dB
dt
E

B  Bz ( x ,t )k
 2
1  2
 2 2
2
x
 t
  A sin(t   )
Plane Electromagnetic Waves
Ey
Bz
c
E(x, t) = EP sin (kx-t) ĵ
B(x, t) = BP sin (kx-t) ẑ
x
F(x)

Static wave
F(x) = FP sin (kx + )
k = 2  
k = wavenumber
 = wavelength
x
F(x)

Moving wave
v
x
F(x, t) = FP sin (kx - t )
 = 2  f
 = angular frequency
f = frequency
v=/k
F
v
x
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)
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
Ey
Bz
E(x, t) = EP sin (kx-t) ĵ
B(x, t) = BP sin (kx-t) ẑ
Notes: Waves are in Phase,
but fields oriented at 900.
k=2.
Speed of wave is c=/k (= f)
c  1 / 00  3 108 m / s
At all times E=cB.
c
x