Download Week 13 - Electromagnetic Waves

Week 13 - Electromagnetic Waves
November 25, 2012
Exercise 13.1: Ultraviolet Radiation
There are two categories of ultra-violet light. Ultraviolet A (UVA) has a wavelength ranging from 320 nm
to 400 nm. It is not so harmful to the skin and is necessary for the production of vitamin D. UVB however,
with a wavelength between 280 nm and 400 nm, is much more dangerous because it causes skin cancer.
a) Find the frequency ranges of UVA and UVB.
Answer: For UVA: 750 THz to 940 THz.
Solution:
Frequency is related to wavelength trough the relation c = λf . So we get a frequency band of
UV A
fmax
=
c
VA
λU
min
=
3 × 108
s−1 = 940 × 1012 Hz = 940 THz
320 × 10−9
(1)
where the T stands for ’tera’ = 1012 . In other words almost a thousand terahertz. Similarly
UV A
fmin
=
c
VA
λU
max
=
3 × 108
s−1 = 750 × 1012 Hz = 750 THz.
400 × 10−9
1
(2)
b) What are the ranges of the wave numbers of UVA and UVB?
Answer: For UVA: 157 × 105 m−1 to 196 × 105 m−1 .
Solution:
The wavenumber is like a ’spatial frequency’, i.e. how many oscillations of the wave per unit space.
It is related to it’s wavelength trough k = 2π/λ and
UV A
kmax
=
2π
VA
λU
min
=
2π
m−1 = 196 × 105 m−1 .
320 × 10−9
(3)
Similarly
UV A
kmin
= 157 × 105 m−1 .
(4)
Exercise 13.2: Fields from a Light Bulb
We can reasonably model a 75 W incandescent light-bulb as a sphere 6.0 cm in diameter. Typically, only
about 5% of the energy goes to visible light; the rest goes largely to nonvisble infrared radiation.
a) What is the visible-light intensity (in W/m2 ) at the surface of the bulb?
Solution:
We have the power (energy per unit time) of the bulb. The intensity is power per unit area. The
area of the bulb with diameter d is A = 4πr2 = πd2 . Assuming the bulb is radiating uniformly in all
directions and that only 5% of the energy goes in to light
I=
P
0.05 ∗ 75
2
2
=
2 W/m = 332 W/m .
−2
A
π × (6.0 × 10 )
(5)
b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with
this intensity?
Solution:
The intensity at a point is the average value of the Poynting vector at that point and the average
value of the Poynting vector is related to E and B trough
I = Sav =
Emax Bmax
1
2
= cε0 Emax
.
2µ0
2
(6)
Therefore
r
Emax =
2I
=
ε0 c
r
2 × 332
= 500 V/m
8.85 × 10−2 × 3 × 108
2
(7)
and
Bmax = Emax cε0 µ0 =
Emax
= 1.67 × 10−6 T
c
(8)
Exercise 13.3: Discussion Questions
a) By measuring the electric and magnetic fields at a point in space where there is an electromagnetic
wave, can you determine the direction from which the wave came? Explain.
Solution:
Yes you can. Assuming you measured both the direction of the electric and the mangnetic field you
can find the propagation direction by taking the cross product E × B and the wave must have come
from −E × B.
b) Most automobiles have vertical antennas for recieving radio broadcasts. Explain what this tells you
you about the direction of polarization for E in the radio waves used for broadcast.
Solution:
The signal is recieved by the electromagnetic wave doing work on the charges in the antenna so that
they start to oscillate, i.e. they produce a current. For the charges to be able to move significantly they
must have some space to move on. Therefore the vertical antennas indicate that the electromagnetic
waves are vertically polarized so the electric field in the wave is able to do work on those charges.
c) Is polarization a property of all electromagnetic waves, or is it unique to visible light? Can sound
waves be polarized? What fundamental disctinction in wave properties is involved?
Solution:
Light is just another elecromagnetic wave within a certain frequency band and there is nothing,
except it’s frequency, that districts it from other electromagnetic wave phenomena. The polarization
property comes about because electromagnetic waves are transverese waves. The quantity that is
waving is doing so perpendicular to the direction of propagation. Therefore all electromagnetic waves
can be polarized and sound waves can not, because they are longitudinal waves.
d) The magnetic-field amplitude of the electromagnetic wave from carbon diode lasers is about 100 times
greater than the earth’s magnetic field. If you illuminate a compass with the light from this laser,
would you expect the compass to deflect? Why or why not?
Solution:
It would not be deflected. At any moment, the needle feels a torque from the magnetic field which tries
to align it with the field lines. However, the frequency of these lasers are very high, so the direction
of the magnetic field changes rapidly causing the torque to also change direction very rapidly. The
average torque over a very tiny period is zero and therefore the needle won’t rotate.
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Exercise 13.4: Solar Sailing
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, lowmass sail and the energy and momentum of sunlight for propulsion. In this exercise you might need the
mass of the sun M = 3.9 × 1026 kg and the universal gravitational constant G = 6.67 × 10−11 Nm2 /kg2 .
a) Should the sail be absorbing or reflective? Why?
Answer: Reflective.
Solution:
The sail should be as reflective as possible. This is because a totally reflective surface experiences
twise the radiation pressure compared to a totally absorbing surface.
b) The total power output of the sun is P = 3.9 × 1026 W. How large a sail is necessary to propel a
10000 kg spacecraft against the gravitational force of the sun?
Answer:
A>
2cGmM
= 6.42 × 106 m2 = 6.42 km2 .
P
(9)
Solution:
Let m be the mass of the spacecraft and M be the mass of the sun. The magnitude of the gravitational
force exerted by the spacecraft by the sun is
Fg = G
mM
r2
(10)
where G is the universal gravitational constant and r is the distance to the spaceship. Now the
radiation pressure is related to the sun’s intensity at this distance by
I
.
c
(11)
P
4πr2
(12)
prad =
The intensity at distance r is
I=
giving a radiation pressure of
Prad = 2
P
.
4πcr2
(13)
where the factor 2 comes from the fact that our sail is reflective. Assuming the sail is of area A and
all the radiation falls perpendicular on to the sail the force from radiation pressure is
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Frad = 2A
P
.
4πcr2
(14)
The radiation pressure will be able to propel the ship at once Frad > Fg , i.e. when
2A
P
mM
>G 2
2
4πcr
r
(15)
or equivalently
A>
2cGmM
= 6.42 × 106 m2 = 6.42 km2 .
P
(16)
c) Explain why your answer to part (b) is independent of the distance from the sun.
Solution:
The answer is indepentent of distance because of the fact that both the gravitational force and the
radiation pressure are inversely proportional to the square of the distance.
Figure 1
Exercise 13.5: Flowing Energy
A cylindrical conductor with a circular cross section has a radius a and a resistivity ρ and carries a
constant current I as shown in figure 1.
a) What are the magnitude and direction of the electric-field vector E at a point just inside the wire at
a distance a from the axis? Express E in terms of the current I.
Answer: E = ρJ in the same direction as the current.
Solution:
We know that E = ρJ and the direction of E is the same as the direction of the current. Since J is
uniform throughout the conductor E = ρJ just inside at a radius a (and anywhere else).
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b) What are the magnitude and direction of the magnetic-field vector B at the same point?
Answer:
B=
µ0 I
2πa
(17)
circling the conductor.
Solution:
Since we have no changing electric fields we can use Ampere’s law in it’s simplest form without
displacement current
I
B · dl = B2πa = µ0 I
(18)
such that
B=
µ0 I
2πa
(19)
and by the right hand rule, since the current is going to the right, the magnetic field is circling around
the conductor such that it’s pointing out of the page at the top and into the page at the bottom.
c) What are the magnitude and direction of the Poynting vector S at the same point? (The direction
of S is the direction in which electromagnetic energy flows into or out of the conductor.)
Answer:
S=
1
ρI 2
|E × B| =
µ0
2π 2 a3
(20)
1
ρI 2
|E × B| =
µ0
2π 2 a3
(21)
towards the center of the conductor.
Solution:
The Poynting vector is given by
S=
and by the right hand rule it’s always pointing in towards the center of the conductor.
d) Use the reuslts in part (c) to find the rate of flow of energy into the volume occupied by a length l
of the conductor. (Hint: Integrate S over the surface of this volume.) Compare your result to the
rate of generation of thermal energy in the same volume. Discuss why the energy dissapated in a
current-carrying conductor, due to its resistance, can be thought of as enerting trough the sides of
the cylindrical conductor.
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Answer:
P = RI 2 .
(22)
Solution:
S is power per unit area. We must integrate it over whole area of the conductor of length l to get the
energy flow per unit time into the volume
Z
P=
S · dA = SA =
ρl 2
I ,
πa2
(23)
but ρl/πa2 is the resistance R! Therefore we get the familiar expression
P = RI 2 .
(24)
This indicates that the energy in dissapated in a conductor, which normaly we attribute to the kinetic
energy of the electrons, as energy being drained from the stored energy in the electric and magnetic
fields.
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