Transcript lecture17

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 flux induces an EMF
 Bl cos  
0
 B
 
t
(I c  I d )
  EA
 E
Id  0
 0
t
t
A changing electric field
induces magnetic field !
 El cos   
 BA cos  
t
A changing magnetic field
induces electric field !
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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

Q(t)
Q  CV 
Ic 
Ic(t)
3) Solution
I c1  0
Ic2  0
I d1  0
Id 2  0
I c1  I d1  I c 2  I d 2
Alternating current can flow
in a circuit with a capacitor
 0 AV
d
  0 AE   0  E
Q
 E
 0
t
t
  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 ?
rb
a) r  R  I d ( encl )  I c  B 
b)
r
r  R  I d ( encl )
0 I c
2r
0 I c r
r 2
 Ic 2  B 
R
2R 2
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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
(U – energy, V - volume)
Intensity:
U
I
A t
uV uAx
I

At
At
I  cu  c 0 E 2  cB 2  0  EB  0
I  12 c 0 E02  12 cB02  0  12 E 0 B0  0
2
2
I  c 0 E rms
 cBrms
 0  E rms Brms  0
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15. Polarization
(Polarizeation of transverse waves)
1) Waves on string (polarization and polrizing filters)
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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 - intensity of polarized component
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
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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
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Classification of Electromagnetic Waves
Wavelength decreases 
Frequency increases 
Note: 1 nanometer = 10-9 meter
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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
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