Transcript lecture16

Two questions:
(1) How to find the force, F on the electric charge, Q excreted by the


 
field E and/or B?
F  qE  qv  B
(2) How fields E and/or B can be created?
Maxwell’s equations
Gauss’s law for electric field
Electric charges create electric field:
 E   EA cos  Q /  0
For one not moving
(v<<c) charge:
Ek
Gauss’s law for magnetic field
Magnetic charges do not exist:
 B   BA cos  0
Q
r2
Amperes law
Faraday’s law
Electric current creates magnetic field: A changing magnetic field induces an EMF
 Bl cos  
0
(I c  I d )
 
 B
t
 E
Id  0
t
   El cos 
A changing electric flux induces
magnetic field !
A changing magnetic flux induces an
electric field !
13. Displacement current
1) Ideas (Maxwell)
Faraday law: Changing magnetic field (more exactly, flux) produces electric field
Maxwell’s idea: Changing electric field (more exactly, flux) produces magnetic field
2) Problem
The current I ≠ 0 throughout the flat surface below, but I = 0 throughout the
curved surface. This is in contradiction with the Ampere’ law:
Bl cos   0 I

 0 AV
Q
C   Q  CV 
  0 AE
V
d
Q
 AE 
 E
Ic 
 0
 0
t
t
t
Q(t)
Ic(t)
3) Solution
I c1  0
Ic2  0
I d1  0
Id 2  0
I c1  I d 1  I c 2  I d 2
Alternating current can flow
in a circuit with a capacitor
  EA
 E
Id  0
 0
t
t
 Bl cos  
0
(I c  I d )

  EA 
 Bl cos    0  I c   0 t 


Example: A circular parallel-plate capacitor with plates 2.0cm in diameter is
accumulating charge at the rate of 3.50 mC/s at some instant of time. What is
the magnitude of the induced magnetic field at the distance r measured
radially outward from the center of the plates? a) r=10.0 cm; b) r=1.0 cm
R  2.0cm
I c  3.50mA
ra  10.0cm
Q(t)
R
B 2r   0 I d ( encl )
Ic(t)
rb  1.0cm
B ?
a)
b)
r
rb
r  R  I d ( encl )
0 I c
 Ic  B 
2r
r  R  I d ( encl )
0 I c r
r 2
 Ic 2  B 
R
2R 2
14. Electromagnetic waves
1) Maxwell’s equations and electromagnetic waves
2) Properties of electromagnetic waves
c
Speed of light:
1
 0 0
 3 108 m / s
Wave length: 
c

f
Energy density:
U
B2 1
u 
 0E2
V 20 2
for any e.m. field
B  E/c
for e.m. wave
U B2
u 
 0E2
V 0
(V - volume)
Intensity:
U
I
At
Polarization:
uV uAx
I

At
At
I  cu  c 0 E 2  cB 2  0  EB  0
15. Polarization
(Polarizeation of transverse waves)
1) Waves on string (polarisation and polrizing filters)
2) Electromagnetic waves (polarized light)

E

E
y
This wave is polarized in y direction

v

B
x
z

B
e.m. waves are transverse waves
direction
of motion
of wave
Light is polarized when its electric fields oscillate in a single plane, rather than
in any direction perpendicular to the direction of propagation.
3) Unpolarized light
(a) Unpolarized light consist of waves with randomly directed
electric fields. Here the waves are all traveling along the same
axis, directly out of the page, and all have the same amplitude E.
(b) A second way of representing unpolarized light – the light is the
superposition of two polarized waves whose planes of oscillation
are perpendicular to each other.
I0
- intensity of unpolarized light
I  12 I 0
4) Polarisation of light (Malus’s law)
When light passes through a polarizer, only the component parallel to the
polarization axis is transmitted.
If the incoming light is plane-polarized, the outgoing intensity is:
I  c 0 E 2
Example (two sheets):
The light transmitted by polarizing sheet P1
is vertically polarized, as represented by the
vertical double arrow. The amount of that
light that is transmitted by polarizing sheet
P2 depends an the angle between the
polarization direction of that light and the
polarizing direction of P2
Polarized light will not be transmitted through a polarized film whose axis is
perpendicular to the polarization direction.
This means that if initially unpolarized light passes through crossed polarizers,
no light will get through the second one.
Example (three sheets):
I1  12 I 0
I 2  I1 cos 2 60  14 I1  18 I 0
I 3  I 2 cos 2 30  34 I 2 
I 3  0.094 I 0
3
32
I0
Classification of Electromagnetic Waves
Wavelength decreases 
Frequency increases 
Note: 1 nanometer = 10-9 meter
Example1: An electromagnetic wave in vacuum has a frequency of 1500 KHz.
What is the wavelength of the wave?
c
 
f
Example2: An electromagnetic wave in vacuum is moving in +y direction. At
time t=0 and at position (x,y,z)=(0,0,0), the electric field is pointing in the +z
direction. In what direction is pointing the magnetic field at that time and
position?

v
z

E
y

B
x