Transcript Phys2102 Spring 2002 - Louisiana State University

Physics 2102
Jonathan Dowling
James Clerk Maxwell (1831-1879)
Lecture 34: MON 13 APR
Ch.33.1–3,5–7: E&M Waves
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Q1/P3 K. Schafer Office hours: MW 1:30-2:30 pm 222B Nicholson
P1/Q2 J. Dowling Office hours: MWF 10:30-11:30 am 453 Nicholson
P2/Q3 M. Gaarde Office hours: TTh 2:30-3:30 pm 215B Nicholson
P3/Q2 C. Buth Office hours: MF 2:30-3:30 pm 222A Nicholson
A: 90-100
B: 80-89
C: 60-79
D: 50-59
Problem 1 [18 points]
In the figure below, two semicircular arcs (I & II) have radii R1 = 4.10 cm and
R2 = 6.60 cm, carry current i = 0.281 A, and share the same center of curvature C.
The same current i also flows through the straight sections of wire labeled III & IV.
(a) (5 pts) What is the cont ribution to the magnitude
of the magnetic field at point C from the two straight
sections of wire III and IV? Explain your answer!
Question 2 [8 points]
The figure below shows four arrangements in which long parallel wires carry equal
currents i directly into or out of the page at the corners of identical squares.
BIII  BIV  0
r
r r
dB  d s  r  0
Biot-Savart
(b) (8 pts) Calculate the magnitude of the magnetic field at the point C due t o all four
sections of wire.
(i) (4 pts) For square B, what is the direction of the magnetic field, with respect t o the
page, at the center of the square? Circle one:
In 
Out ()
r
r
 i  1
1 
BTOT = BI  BII  0  
 8.14  10 7 T

4  RI RII 
    half a circle
Right 
Down
Field is Zero ( B  0)
Right Hand Rule & Vector Addition!
Right Hand Rule & BI>BII Since I is Closer!
(c) (5 pts) What is the direction of the total magnetic field at the point C due to all four
sections of wire? Circle one:
Out of the page () .
Int o the page  .
Up t owards the t op of the page  .
Down t owards the bot tom of the page  .
To the right of the page  .
Left 
Up 
To the left of the page  .
The t otal magnetic field at C is zero and has no direction.
(ii) (4 pts) Rank the sections according to the magnitude of the magnitude of the
magnetic field at the center of each square, greatest first. Circle one:
BA > B B > B C
BC > B B > B A
BB > B C > B A
BA = B B = B C
B=
0i
;
2 r
BA  0;
BB  2 2B;
BA > B C > B B
BC  2B;
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  10 8 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 propagatio n.
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: k

EM Wave Simulation
Frequency:

2

2
T


0
Vacuum Permeability:
0
Magnitude Ratio:
E t 
c
B t 
c
 2 f
Vacuum Permittivity:
Fig. 33-5
E
Amplitude Ratio: m  c
Bm
1
(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.
1
1
____________
2
1
The average of sin2 over
I S 
E sin (kx   t) 
Em
E 
one cycle is ½:
c
c
2c
___
2
2
m
Em2
1
I
Em 2 or,
2c
Both fields have the
same energy density.
I
1
c
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
2
Erms
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


0.8mW/m
4 r 2 4 (1km)2
Bm  Em / c  2.58nT
1
I
Em2  Em  2c I  0.775V/m
2c
Received S  P  U /t  U  SAt  2.4 mJ
energy:
A
A
Radiation Pressure
Waves not only carry energy but also momentum. The effect is
very small (we don’t ordinarily feel pressure from light). If light
is completely absorbed during an interval Dt, the momentum
transferred is given by
u
p 
and twice as much if reflected.
p
F
t
Newton’s law:
c
F
A
I
Now, supposing one has a wave that hits a surface
of area A (perpendicularly), the amount of energy
transferred to that surface in time Dt will be
U  IAt
Radiation
pressure:
pr 
IAt
therefore p 
c
IA
F
c
I
2I
(total absorption), pr 
(total reflection)
c
c
[N/m2]
Radiation Pressure & Comet Tails
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Solar Sails: Photons Propel Spacecraft!
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StarTrek DS9
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NASA Demo
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NASA Concept
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EM waves: polarization
Radio transmitter:
If the dipole antenna
is vertical, so will be
the electric fields. The
magnetic field will be
horizontal.
The radio wave generated is said to be “polarized”.
In general light sources produce “unpolarized
waves”emitted by atomic motions in random
directions.
EM Waves: Polarization
Completely unpolarized light will have
equal components in horizontal and vertical
directions. Therefore running the light through
a polarizer will cut the intensity in half: I=I0/2
When polarized light hits a polarizing sheet,
only the component of the field aligned with the
sheet will get through.
E y  E cos( 
And therefore:
I  I 0 cos2 
Example
Initially unpolarized light of
intensity I0 is sent into a system of
three polarizers as shown. What
fraction of the initial intensity
emerges from the system? What
is the polarization of the exiting
light?
• Through the first polarizer: unpolarized to polarized, so I1=½I0.
• Into the second polarizer, the light is now vertically polarized. Then, I2 =
I1cos260o = 1/4 I1 = 1/8 I0.
• Now the light is again polarized, but at 60o. The last polarizer is
horizontal, so I3 = I2cos230o = 3/4 I2 =3 /32 I0 = 0.094 I0.
• The exiting light is horizontally polarized, and has 9% of the original
amplitude.