Physics 9 Fall 2009Homework 10 - Solutions

1. Chapter 35 - Exercise 14The magnetic field of an electromagnetic wave in a vacuum is

Bz = (3.00µT ) sin((

1.00× 107)x− 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?

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Solution

The general expression for the general wave is

B (x, t) = B0 sin (kx− ωt) ,

where B0 is the amplitude, k = 2π/λ is the wave number, and ω = 2πf is the angularfrequency. So,

(a) The wavelength λ = 2πk

= 2π107 = 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 isv = λf . So, λf = c, or f = c

λ. So, f = c

λ= 3×108

2π×10−7 = 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.

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2. Chapter 35 - Exercise 15The 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?

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Solution

Again, the general expression for the general wave is

E (x, t) = E0 sin (kx− ωt) ,

where E0 is the amplitude, k = 2π/λ is the wave number, and ω = 2πf is the angularfrequency. So, everything is defined as before.

(a) The wavelength λ = 2πk

= 2π6.28×108 = 1.00× 10−8 m.

(b) The wave is moving at the speed of light, c, and in general the speed of a wave isv = λf . So, λf = c, or f = c

λ. So, f = c

λ= 3×108

1.00×10−8 = 3.00× 1016 Hz.

(c) For any electromagnetic wave, E = cB, and so the amplitude B0 = E0/c. Thus,B0 = E0

c= 20

3×108 = 6.67× 10−8 T.

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3. Chapter 35 - Problem 30.

What electric field strength and direction will al-low the electron in the figure to pass through thisregion of space without being deflected?

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Solution

The magnetic force points down by the right-hand-rule (noting that the electron carries

a negative charge). So, the applied electric field has to cancel this force, and so ~E hasto point downward, pulling the electron up. The force balances when FE = FM , orqE = qvB, which gives E = vB.

Plugging in the numbers gives E = (2.0× 107) (0.010) = 2.0×105 V/m, pointing down.

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4. Chapter 35 - Problem 35.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 solenoidaxis?

(b) What is the value of B the instant E reaches its maximum value?

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Solution

Faraday’s law says that ∮~E · d~s = − d

dt

∫~B · d ~A.

For the solenoid, the electric field circles around inside the solenoid, so at a distancer < R,

∮~E · d~s = E (2πr) = − d

dt(B (πr2)), and so E = − r

2B. So, we just need B (t).

The field varies sinusoidally from 8 to 12 T, so the correct expression is

B (t) = BC +B0 sin (2πft)= 2 + 10 sin (20πt) ,

where BC is a constant magnetic field.

(a) So, E = − r2B, and taking the derivative gives B = 40π cos (20πt), and so

E = −20πr cos (20πt) V/m,

which has it’s maximum value when 20πt = π, 3π, · · · . In this case the cosineterm is 1, and so Emax = 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 thesine function is zero. So, when the electric field reaches it’s maximum, the B =BC +B0 sin (π) = BC , or B = 10 T.

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5. Chapter 35 - Problem 37.A wire with conductivity σ carries current I. The current is increasing at the ratedI/dt.

(a) Show that there is a displacement current in the wire equal to (ε0/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.

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Solution

(a) The displacement current is Idisp = ε0dΦE

dt. Now, in general, I = JA, where J

is the current density. But, we also know from Ohm’s law, that J = σE. So,I = JA = (σE)A = σ (EA). But, for this simple case, EA = ΦE is the electricflux. So, we have that

I = σΦE.

Now, taking the time derivative of this expression gives dIdt

= σ dΦE

dt. Comparing

this to the definition for the displacement current we see that

Idisp =ε0σ

dI

dt.

(b) For copper, σ = 6.0× 107 Ω−1m−1, and so

Idisp =ε0σ

dI

dt=

8.85× 10−12

6.0× 1071.0× 106 = 1.48× 10−13 A.

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6. Chapter 35 - Problem 51.The intensity of sunlight reaching the earth is 1360 W/m2. Assuming that all thesunlight is absorbed, what is the radiation pressure force on the earth? Give youranswer 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, andA is the cross-sectional area of the earth, A = πR2

E. Here, RE = 6400 km is the radiusof the earth. Now, the radiation pressure is prad = I

c, where I is the intensity of the

radiation. So, the total radiation force is

Ffrad =πR2

EI

c=π (6.4× 106)

21360

2.99× 108= 5.78× 108N.

This seems large, but let’s compare it to the gravitational force, Fg =GmEM

r2E. Here

G = 6.672× 10−11 is Newton’s gravitational constant, mE = 6× 1024 kg is the mass ofthe earth, M = 2 × 1030 kg is the solar mass, and rE = 1.50 × 1011 m is the orbitalradius of the earth. So,

Frad

Fg=

πR2Er

2EI

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 orbitaround the sun and aren’t being blown away by the radiation pressure.

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7. Chapter 35 - Problem 52.A laser beam shines straight up onto a flat, black foil with a mass of 25 µg. What laserpower is needed to levitate the foil?

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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 haveFR = pRA = PA

Ac= P

c, 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,P = mcg =

(25× 10−6

) (3× 108

)(9.8) = 73.5 W.

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8. Chapter 35 - Problem 54.You’ve recently read about a chemical laser that generates a 20 cm-diameter, 25 MWlaser beam. One day, after physics class, you start to wonder if you could use theradiation pressure from this laser beam to launch small payloads into orbit. To seeif 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 ifpushed horizontally 100 m along a frictionless track by such a laser?

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Solution

The radiation pressure creates a force on the block, FR = pRA, where A is the areaof the beam. This produces an acceleration, a = FR/m = pRA

m, and is constant. This

acceleration leads to a final velocity v2f = v2

i + 2a∆x, which says that

vf =√

2a∆x =

√2pRA∆x

m=

√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 =

√2P∆x

mc=

√2 (25× 106) (100)

100 (3× 108)= 0.408 m/s,

which is pretty slow! This isn’t the best way to launch payloads!

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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 awayfrom the capsule. Fortunately, he has a 1000 W portable laser with fresh batteriesthat will operate for 1.0 hr. His only chance is to accelerate himself toward the spacecapsule by firing the laser in the opposite direction. he has a 10 hr supply of oxygen.Can he make it?

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Solution

The laser produces a radiation force, FR = pRA = P/c, where P is the power of thelaser. This produces a constant acceleration, a = FR/m = P

mc, but it only lasts for an

hour (the battery life of the laser). After this time, the astronaut will have picked upa final velocity of vf = at = P

mct, after having gone a distance x = 1

2at2 = P

2mct2. 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 =Pt

mc=

(1000) (3600)

80× 3× 108= 1.5× 10−4 m/s,

and he has travelled a distance

x =P

2mct2 =

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 todo it. Can he make it? Coasting along at the speed v, to travel the remaining distancewould take

t =distance

speed=

4.73

1.5× 10−4= 31, 533 s = 8.76 hours.

This time is just barely under the remaining 9 hours, so he can just make it!

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10. Chapter 35 - Problem 60.Large quantities of dust should have been left behind after the creation of the solarsystem. 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 smalldust particles are conspicuously absent. Astronmers believe that the very small dustparticles have been blown out of the solar system by the sun. By comparing theforces on dust particles, determine the diameter of the smallest dust particles that canremain in the solar system over long periods of time. Assume that the dust particlesare spherical, black, and have a density of 2000 kg/m3. The sun emits electromagneticradiation with power 3.9× 1026 W.

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Solution

The sun gravitationally attracts the dust particles, while its radiation pressure pushesthem away. In order for the dust to stay, Frad ≤ Fg. Now, the gravitational force is

Fg =GmdustM

r2, 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-sectionalarea of the dust particle as Frad = pradAdust = I

cπR2. 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, Frad =P4c

Rdust

r2.

In order for the dust to stay in the solar system, Frad ≤ Fg, so

4π

3

GMρR3

r2≤P4c

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 thanthey are repelled by the radiation pressure. Particles smaller than this will be pushedout of the solar system. So, if the biggest radius is 0.293 µm, then the biggest diameteris 0.585 µm.

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