Physics 9 Fall 2009
Homework 10 - Solutions 1. Chapter 35 - Exercise 14The magnetic field of an electromagnetic wave in a vacuum is
Bz = (3.00µT) sin 1.00×107x−ωt,
where x is in m and t is in s. What are the wave’s (a) wavelength, (b) frequency, and (c) electric field amplitude?
———————————————————————————————————— Solution
The general expression for the general wave is
B(x, t) =B0sin (kx−ωt),
whereB0 is the amplitude, k= 2π/λis the wave number, and ω = 2πf is the angular
frequency. So,
(a) The wavelength λ = 2πk = 102π7 = 2π×10
−7 = 6.28×10−7 m.
(b) The wave is moving at the speed of light, c, and in general the speed of a wave is
v =λf. So, λf =c, or f = λc. So, f = λc = 2π3××10108−7 = 4.77×1014 Hz.
(c) For any electromagnetic wave, E = cB, and so the amplitude E0 = cB0. Thus, E0 =cB0 = 3×108×3.00×10−6 = 900 V/m.
The electric field of an electromagnetic wave in a vacuum is
Ey = (20.0V /m) cos 6.28×108
x−ωt,
where x is in m and t is in s. What are the wave’s (a) wavelength, (b) frequency, and (c) magnetic field amplitude?
———————————————————————————————————— Solution
Again, the general expression for the general wave is
E(x, t) =E0sin (kx−ωt),
whereE0 is the amplitude, k= 2π/λis the wave number, and ω = 2πf is the angular frequency. So, everything is defined as before.
(a) The wavelength λ = 2πk = 6.282π×108 = 1.00×10
−8 m.
(b) The wave is moving at the speed of light, c, and in general the speed of a wave is
v =λf. So, λf =c, or f = λc. So, f = λc = 1.003××10108−8 = 3.00×1016 Hz.
(c) For any electromagnetic wave, E =cB, and so the amplitude B0 = E0/c. Thus, B0 = Ec0 = 3×20108 = 6.67×10
−8 T.
3. Chapter 35 - Problem 30.
What electric field strength and direction will al-low the electron in the figure to pass through this region of space without being deflected?
———————————————————————————————————— Solution
The magnetic force pointsdown by the right-hand-rule (noting that the electron carries anegative charge). So, the applied electric field has to cancel this force, and so E~ has to point downward, pulling the electron up. The force balances when FE = FM, or
qE =qvB, which gives E =vB.
The magnetic field inside a 4.0 cm-diameter superconducting solenoid varies sinu-soidally between 8.0 T and 12.0 T at a frequency of 10 Hz.
(a) What is the maximum electric field strength at a point 1.5 cm from the solenoid axis?
(b) What is the value of B the instantE reaches its maximum value?
———————————————————————————————————— Solution
Faraday’s law says that
I
~
E·d~s=−d
dt
Z
~ B·d ~A.
For the solenoid, the electric field circles around inside the solenoid, so at a distance
r < R, H ~
E·d~s=E(2πr) =−d
dt(B(πr
2)), and so E =−r
2B˙. So, we just needB(t).
The field varies sinusoidally from 8 to 12 T, so the correct expression is
B(t) = BC +B0sin (2πf t) = 2 + 10 sin (20πt),
where BC is a constant magnetic field. (a) So, E =−r
2B˙, and taking the derivative gives ˙B = 40πcos (20πt), and so E =−20πrcos (20πt) V/m,
which has it’s maximum value when 20πt = π,3π,· · ·. In this case the cosine term is 1, and soEmax = 20πr V/m. So, at r= 1.5 cm, then
Emax = 20π(0.015) = 0.94 V/m.
(b) The electric field reaches a maximum when 20πt = π,3π,5π,· · ·, for which the sine function is zero. So, when the electric field reaches it’s maximum, the B =
BC+B0sin (π) =BC, or B = 10 T.
5. Chapter 35 - Problem 37.
A wire with conductivity σ carries current I. The current is increasing at the rate
dI/dt.
(a) Show that there is a displacement current in the wire equal to (0/σ) (dI/dt).
(b) Evaluate the displacement current for a copper wire in which the current is in-creasing at 1.0×106 A/s.
———————————————————————————————————— Solution
(a) The displacement current is Idisp = 0dΦdtE. Now, in general, I = J A, where J
is the current density. But, we also know from Ohm’s law, that J = σE. So,
I =J A= (σE)A = σ(EA). But, for this simple case, EA = ΦE is the electric
flux. So, we have that
I =σΦE.
Now, taking the time derivative of this expression gives dIdt = σdΦE
dt . Comparing
this to the definition for the displacement current we see that
Idisp = 0
σ dI dt.
(b) For copper, σ = 6.0×107Ω−1m−1, and so
Idisp= 0
σ dI dt =
8.85×10−12
6.0×107 1.0×10
The intensity of sunlight reaching the earth is 1360 W/m2. Assuming that all the
sunlight is absorbed, what is the radiation pressure force on the earth? Give your answer in newtons and as a fraction of the sun’s gravitational force on the earth. ————————————————————————————————————
Solution
The total force on the earth is Frad =pradA, where prad is the radiation pressure, and Ais the cross-sectional area of the earth, A=πR2
E. Here, RE = 6400 km is the radius
of the earth. Now, the radiation pressure is prad = Ic, where I is the intensity of the
radiation. So, the total radiation force is
Ffrad = πR
2 EI
c =
π(6.4×106)21360
2.99×108 = 5.78×10 8
N.
This seems large, but let’s compare it to the gravitational force, Fg =
GmEM
r2 E
. Here
G= 6.672×10−11 is Newton’s gravitational constant, mE = 6×1024 kg is the mass of
the earth, M = 2×1030 kg is the solar mass, and r
E = 1.50×1011 m is the orbital
radius of the earth. So,
Frad
Fg =
πR2 Er2EI
GcmEM = 1.64×10
−14.
So, the radiation force is ≈10−14 times as large as the gravitational force. Of course,
it makes sense that this force should be much weaker since we are in a stable orbit around the sun and aren’t being blown away by the radiation pressure.
7. Chapter 35 - Problem 52.
A laser beam shines straight up onto a flat, black foil with a mass of 25µg. What laser power is needed to levitate the foil?
———————————————————————————————————— Solution
The foil will be balanced when the gravitational force is exactly canceled by the radi-ation pressure force, Fg = FR. Now, in terms of the radiation pressure, pR, we have FR =pRA = P AAc = Pc, where P is the power of the radiation source, and we recalled
that pR=P/Ac. So, FR = Pc =mg =Fg, or P =mcg.
Thus,
You’ve recently read about a chemical laser that generates a 20 cm-diameter, 25 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20 cm-diameter, 100 kg, perfectly absorbing block. What speed would such a block have if pushed horizontally 100 m along a frictionless track by such a laser?
———————————————————————————————————— Solution
The radiation pressure creates a force on the block, FR = pRA, where A is the area
of the beam. This produces an acceleration, a= FR/m= pRmA, and is constant. This
acceleration leads to a final velocityv2
f =vi2+ 2a∆x, which says that
vf = √
2a∆x= r
2pRA∆x
m =
r 2P∆x
mc ,
since vi = 0, and we have replaced pRA = P/c, where P is the power of the laser.
Plugging in the numbers gives
vf =
r 2P∆x
mc =
s
2 (25×106) (100)
100 (3×108) = 0.408 m/s,
which is pretty slow! This isn’t the best way to launch payloads!
9. Chapter 35 - Problem 55.
An 80 kg astronaut has gone outside his space capsule to do some repair work. Un-fortunately, he forgot to lock his safety tether in place, and he has drifted 5.0 m away from the capsule. Fortunately, he has a 1000 W portable laser with fresh batteries that will operate for 1.0 hr. His only chance is to accelerate himself toward the space capsule by firing the laser in the opposite direction. he has a 10 hr supply of oxygen. Can he make it?
———————————————————————————————————— Solution
The laser produces a radiation force, FR =pRA = P/c, where P is the power of the
laser. This produces a constant acceleration,a =FR/m = mcP , but it only lasts for an
hour (the battery life of the laser). After this time, the astronaut will have picked up a final velocity of vf =at= mcP t, after having gone a distancex= 12at2 = 2mcP t2. After
this he will just coast along at the same constant velocity. Let’s see if he can make it. After the first hour, the astronaut is traveling at a speed of
v = P t
mc =
(1000) (3600)
80×3×108 = 1.5×10
−4 m/s,
and he has travelled a distance
x= P
2mct 2
= 1000
2 (80) (3×108)(3600) 2
= 0.270 m.
So, he still has to travel a distance 5−0.270 = 4.73 m, and he only has 9 hours left to do it. Can he make it? Coasting along at the speedv, to travel the remaining distance would take
t= distance speed =
4.73
1.5×10−4 = 31,533 s = 8.76 hours.
Large quantities of dust should have been left behind after the creation of the solar system. Larger dust particles, comperable in size to soot and sand grains, are common. They creat shooting stars when the collide with the earth’s atmosphere. But very small dust particles are conspicuously absent. Astronmers believe that the very small dust particles have been blown out of the solar system by the sun. By comparing the forces on dust particles, determine the diameter of the smallest dust particles that can remain in the solar system over long periods of time. Assume that the dust particles are spherical, black, and have a density of 2000 kg/m3. The sun emits electromagnetic radiation with power 3.9×1026 W.
———————————————————————————————————— Solution
The sun gravitationally attracts the dust particles, while its radiation pressure pushes them away. In order for the dust to stay, Frad ≤ Fg. Now, the gravitational force is Fg = Gmdustr2M, where mdust is the mass of the dust particle, which we can write in
terms of the density and radius as mdust = 43πρR3, and M is the solar mass. Thus, Fg = 4π3
GMρR3
r2 .
Now, the radiation force is given in terms of the radiation pressure and cross-sectional area of the dust particle as Frad = pradAdust = IcπR2. Here we have re-expressed the
force in terms of the intensity of the radiation. The intensity is the power of the sun,
P, divided by the area over which the power is spread out. This area is just a sphere with radius equal to the earth’s orbital radius, so A= 4πr2. So, F
rad = P
4c Rdust
r2 .
In order for the dust to stay in the solar system,Frad ≤Fg, so
4π
3
GMρR3
r2 ≤
P
4c Rdust
r2
or, upon solving for Rdust, we have Rdust ≥
3P
16πGMcρ.
With numbers, we have
Rdust ≥
3P
16πGMcρ =
3 (3.9×1026)
16π(6.67×10−11) (2×1030) (3×108) (2000) = 2.93×10
−7m.
So, any particles bigger than 0.293 µm are more strongly attracted to the sun than they are repelled by the radiation pressure. Particles smaller than this will be pushed out of the solar system. So, if the biggest radius is 0.293µm, then the biggest diameter is 0.585µm.
