Download Phys2102 Spring 2002

Physics 2102
Jonathan Dowling
James Clerk Maxwell (1831-1879)
Lecture 33: FRI 03 APR
Ch.33.1–3,5: E&M Waves
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Maxwell, Waves and Light
A solution to the Maxwell equations in empty space is
a “traveling wave”…
d
C B  ds 0 0 dt S E  dA
d
C E  ds   dt S B  dA
electric and magnetic “forces” can travel!
d2E
d2E
  0 0 2  E  E0 sin k(x  ct)
2
dx
dt
c
1
0  0
 3  108 m/s
The “electric” waves travel
at the speed of light!
Light itself is a wave of
electricity and magnetism!
Electromagnetic Waves
A solution to Maxwell’s equations in free space:
E  Em sin( k x   t )

B  Bm sin( k x   t )
k
 c, speed of propagation.
Em
c

Bm
1
0 0
m
=299,462,954
= 187,163 miles/sec
s
Visible light, infrared, ultraviolet,
radio waves, X rays, Gamma
rays are all electromagnetic waves.
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Radio waves are reflected by the layer of the Earth’s
atmosphere called the ionosphere.
This allows for transmission between two points which are
far from each other on the globe, despite the curvature of the
earth.
Marconi’s experiment discovered the ionosphere! Experts
thought he was crazy and this would never work.
Maxwell’s Rainbow
The wavelength/frequency range in which electromagnetic (EM) waves (light)
are visible is only a tiny fraction of the entire electromagnetic spectrum.
Fig. 33-2
Fig. 33-1
(33-2)
The Traveling Electromagnetic (EM) Wave, Qualitatively
An LC oscillator causes currents to flow sinusoidally, which in turn produces
oscillating electric and magnetic fields, which then propagate through space as
EM waves.
Next slide
Fig. 33-3
Oscillation Frequency:

1
LC
(33-3)
Mathematical Description of Traveling EM Waves
Electric Field:
Magnetic Field:
E  Em sin  kx  t 
B  Bm sin  kx  t 
Wave Speed:
c
0 0
All EM waves travel a c in vacuum
Wavenumber:
EM Wave Simulation
1
k
2

Angular frequency: 

2

Vacuum Permittivity:  0
Vacuum Permeability:
Fig. 33-5
Em
Amplitude Ratio:
c
Bm
0
E t 
c
Magnitude Ratio:
B t 
(33-5)
The Poynting Vector:
Points in Direction of Power Flow
Electromagnetic waves are able to transport energy from transmitter
to receiver (example: from the Sun to our skin).
The power transported by the wave and its
direction is quantified by the Poynting vector.
 1  
S
EB

John Henry Poynting (1852-1914)
For a wave, since
E is perpendicular to B:
Units: Watt/m2

EB 
1
c
E2
In a wave, the fields change
with time. Therefore the
Poynting vector changes
too!!
E
S
B
| S |
1
The direction is constant, but
the magnitude changes from
0 to a maximum value.
EM Wave Intensity, Energy Density
A better measure of the amount of energy in an EM wave is obtained by
averaging the Poynting vector over one wave cycle.
The resulting quantity is called intensity. Units are also Watts/m2.
I S 
1
c 
___
2
E 
1
c 
____________
2
2
Em sin (kx  t )
1
I
Em 2 or,
2c
Both fields have the
same energy density.
I
1
c
The average of sin2 over
one cycle is ½:
Erms 2
1
1
1
B2
1 B2
2
2
uE    E    (cB)   0

 uB
2
2
2    2 
The total EM energy density is then
u   0 E  B / 0
2
2
Solar Energy
The light from the sun has an intensity of about 1kW/m2. What
would be the total power incident on a roof of dimensions
8m x 20m ?
I = 1kW/m2 is power per unit area.
P=IA=(103 W/m2) x 8m x 20m=0.16 MegaWatt!!
The solar panel shown (BP275) has dimensions 47in x
29in. The incident power is
then 880 W. The actual solar
panel delivers 75W (4.45A at
17V): less than 10%
efficiency….
The electric meter on a solar home
runs backwards — Entergy Pays YOU!
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EM Spherical Waves
The intensity of a wave is power per unit area. If one has a
source that emits isotropically (equally in all directions) the
power emitted by the source pierces a larger and larger
sphere as the wave travels outwards: 1/r2 Law!
I
Ps
4r
2
So the power per
unit area decreases
as the inverse of
distance squared.
Example
A radio station transmits a 10 kW signal at a frequency of
100 MHz. At a distance of 1km from the antenna, find the
amplitude of the electric and magnetic field strengths, and
the energy incident normally on a square plate of side 10cm
in 5 minutes.
Ps
10kW
2
I


0
.
8
mW
/
m
4r 2 4 (1km) 2
1
2
I
Em  Em  2c  I  0.775V / m
2c 
Bm  Em / c  2.58 nT
Received S  P  U / t  U  SAt  2.4 mJ
energy:
A
A