Version 095 – MT4 – chiu – (58655) 1

This print-out should have 22 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

001 10.0 pointsTwo laser beams of equal intensity propagatealong parallel paths through a gas that makesthe beams visible to the eye. One laser hasa wavelength of λ1 = 638.19 nm and the sec-ond has a wavelength of λ2 = 434.81 nm. Youview these beams from the side, i.e. perpen-dicularly to the propagation path, so that allyou see is light scattered off the gas. Computethe ratio of the intensity of the light of eachwavelength that reaches your eye, I2/I1. As-sume the distance from your eye to each beamis the same.1. 5.986142. 4.461223. 3.537224. 4.233415. 3.838836. 4.831827. 4.64098. 5.213929. 3.9547110. 5.6154

Correct answer: 4.6409.

Explanation:The laser with the higher frequency is scat-

tered more. The scattered intensity varies likeω4.

I2/I1 = (ω2/ω1)4 = (λ1/λ2)

4 = 4.6409 .

002 (part 1 of 2) 5.0 pointsN closely spaced turns of wire are wound inthe direction indicated on a hollow, plasticring of radius R, with circular cross section,as in the figure below. The coil carries acurrent I with direction as shown in the figure.Assume the ring lies entirely within a regionof uniform electric field E that points into thepage and is decreasing: dE/dt = −|dE/dt|.

R

I

I

Find the magnitude of the displacementcurrent.

1. |ID| =1

µ0ǫ0πR2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

2. |ID| = I + ǫ0πR2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

3. |ID| = µ0ǫ0πR2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

4. |ID| = ǫ0πR2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

correct

5. |ID| =1

2ǫ0R

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

Explanation:The Ampere-Maxwell law can be under-

stood in the following way:∮

~B • d~ℓ = µ0I + µ0ǫ0dΦE

dt

= µ0 (I + ID) ,

where ID = ǫ0 dΦE/dt. Plugging in the givenvalues, we obtain

|ID| = ǫ0πR2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

.

003 (part 2 of 2) 5.0 pointsUsing the convention that clockwise aroundthe ring is positive and counterclockwise isnegative, determine ~B at the location marked×.

1. ~B = −µ0

2πRNI − µ0ǫ0

R

2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

2. ~B =µ0

2πRNI − µ0ǫ0

R

2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

correct

Version 095 – MT4 – chiu – (58655) 2

3. ~B = −µ0

2πRNI + µ0ǫ0

R

2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

4. ~B =µ0

2πRNI + µ0ǫ0

R

2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

Explanation:First we choose the orientation of the area.

It is simpler to work with positive fluxes, solet the area vector point into the page; thisorientation means the path of integration isclockwise about the ring.

By inspection, there are N wires piercingthe defined area with current flow in the samedirection as the area vector, so the enclosedcurrent is NI. The flux of E through the areais positive, but it is decreasing, so the displace-ment current is −ǫ0πR

2|dE/dt|. Pluggingthese into the Ampere-Maxwell equation,

∮

~B • d~ℓ = µ0NI − µ0ǫ0πR2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

~B =µ0

2πRNI − µ0ǫ0

R

2

∣

∣

∣

∣

dE

dt

∣

∣

∣

∣

004 (part 1 of 2) 5.0 pointsIn Professor Feynman’s lecture on the EMwaves modeled in the figures below, a con-stant, uniform B field points in the −z-direction and propagates outward at a con-stant speed v in the ±x-directions. The Bfield originates from a current sheet locatedin the yz-plane.

Applying Faraday’s law to loop 12341, onefinds that the the magnitude of the RHS ofFaraday’s law is given by Bhv. What is themagnitude of the LHS?

1. Bh

2. Eh correct

3. E(h+ 2vt)

Explanation:The emf about the loop must be equal to

the line integral of the electric field aroundthe loop. In the neighborhood of segment23, B is not changing, so we may reason thatE is zero along it. For segments 12 and 34,note that E is independent of x, so E alongthese segments must either be zero or the summust cancel. The emf must be entirely alongsegment 41, so Eh is the correct choice.

005 (part 2 of 2) 5.0 points

Now apply the Ampere-Maxwell law to theloop 12561. The magnitude of the LHS isfound to be Bb. What is the magnitude of theRHS?

1. ǫ0µ0Bvb

2. ǫ0µ0Evb correct

3. ǫ0µ0Bvh

Explanation:We are looking for a term proportional to

dE/dt. Since there is only one option con-taining E, it is the only choice. Physically,ΦE = Ebvt, so dΦ/dt = Ebv.

006 (part 1 of 2) 5.0 pointsA point source at O emits light isotropically.Denote the power which the point source ra-diates by P . A small flat surface is placed atD, which is a distance r from O. This sur-face has an area A and is perpendicular to theradial vector OD.

Version 095 – MT4 – chiu – (58655) 3

ODr

Find the intensity I of the incident light atD.

1. I =2P

3A

2. I =P

2A

3. I =P

6 π r2

4. I =P

2 π r2

5. I =P

3A

6. I =P

A

7. I =P

3 π r2

8. I =P

π r2

9. I =P

4A

10. I =P

4 π r2correct

Explanation:Notice that the emission is isotropic. So

the total power P is associated with theentire spherical surface enclosing the pointsource. In other words the average intensityof a spherical wave is given by

I =∆P

∆A=

P

4 π r2.

007 (part 2 of 2) 5.0 pointsConsider the setup described in Part 1, where

the surface absorbs1

3of the light.

If P = 116 W, r = 3.5 m, and A = 4 mm2,find the average pressure on the surface.1. 4.18929e-092. 7.84057e-093. 2.32016e-08

4. 4.23717e-095. 2.70859e-086. 9.67757e-097. 7.12857e-098. 4.08372e-089. 3.04025e-0910. 3.31584e-09

Correct answer: 4.18929× 10−9 N/m2.

Explanation:Note: The time-averaged energy density is

u =I

c=

P

4 c π r2.

Then the average pressure on the surface is

pressure =1

3u+ 2

2

3u

=5

3u =

5P

12 π c r2

=5 (116 W)

12 π (2.99792× 108 m/s) (3.5 m)2

= 4.18929× 10−9 N/m2 .

008 10.0 points

3.65µm

θ

Determine the maximum angle θ for whichthe light rays incident on the end of the lightpipe shown in the figure above are subjectto total internal reflection along the walls ofthe pipe. The pipe of diameter 3.65 µm hasan index of refraction of 1.22 and the outsidemedium is air.1. 63.12112. 37.39973. 53.03474. 65.08285. 28.80096. 61.2677. 44.33538. 71.9894

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9. 74.895710. 45.7392

Correct answer: 44.3353.

Explanation:

Given : nair = 1 ,

npipe = 1.22 , and

r = 1.825 µm .

d

θ

φφc

The critical angle φc for total internal re-flection is given by

sinφc =nair

npipe

.

φc = arcsin

(

nair

npipe

)

= arcsin

(

1

1.22

)

= 55.052 .

From the right triangle the angle φ at theleft-hand interface is

φ = 90 − φc

= 34.948 .

Applying Snell’s Law, we have

nair sin θ = npipe sinφ

θ = arcsin

(

npipe sinφ

nair

)

= arcsin

(

1.22 sin 34.948

1

)

= 44.3353 .

009 10.0 points

An electron is oscillating along the x-axisabout the origin, and its position is given by:

x(t) = x0 sin(ωt) ,

where x0 is a constant and ω is the angularfrequency of oscillation. What is the intensityof emitted radiation at some point x on thex-axis, |x| > x0?

1. 0 correct

2.µ0

c

(

kqωx0 cos(ωt)

x

)2

3.1

µ0c6

(

kqω2x0 sin(ωt)

x

)2

4.1

2ǫ0

(

kqω2x0 sin(ωt)

c2x

)2

5.1

µ0c5

(

kqω2x0 sin(ωt)

x

)2

Explanation:Radiation will only be emitted in direc-

tions for which the perpendicular componentof the acceleration of the electron is nonzero.Since the electron is oscillating along the x-axis, its acceleration will always be in the±x-directions. Since a⊥ = 0 for all pointsalong the x-axis, 0 is the correct choice.

010 (part 1 of 2) 5.0 pointsAn ideal circuit contains an inductor of induc-tance 135 mH and a capacitor of capacitance38.3 µF. At what frequency will this circuitoscillate? Answer in units of Hz.1. 63.24192. 69.99293. 75.774. 60.52885. 57.06496. 103.477. 119.7878. 62.52589. 111.95710. 56.2473

Correct answer: 69.9929 Hz.

Explanation:

Version 095 – MT4 – chiu – (58655) 5

Let : L = 135 mH = 0.135 H , and

C = 38.3 µF = 3.83× 10−5 F .

The natural frequency is given by

f =1

2 π

√

1

LC

=1

2 π

√

1

(0.135 H) (3.83× 10−5 F)

= 69.9929 Hz .

011 (part 2 of 2) 5.0 pointsThe capacitor is first charged to 360 µC andthen connected to the inductor, forming anideal LC circuit. An ammeter measures cur-rent flow in the circuit — when the current is76.7853 mA, how much charge remains on thecapacitor? Answer in units of µC.1. 187.3732. 90.66883. 177.8634. 350.095. 328.6686. 314.8257. 191.6018. 144.0439. 282.04410. 308.049

Correct answer: 314.825.

Explanation:

Let : Q0 = 360 µC and

I = 76.7853 mA .

Since there are no dissipative elements in thecircuit (i.e. resistors), energy is conservedduring the oscillation:

E =1

2

Q20

C=

1

2

Q(t)2

C+

1

2LI2 .

By solving this equation for Q(t), we candetermine the charge on the capacitor for anymeasured current I:

Q(t) =

√

Q20−

(

I

ω

)2

.

In this case, we obtain:

Q = 314.825µC .

012 (part 1 of 3) 3.0 pointsYou spend the day at the beach. At noon

you look at sky a little bit above the horizonover the ocean. Describe the polarization oflight that comes from that portion of the skyto your eye. What is the best choice?

1. It is predominantly unpolarized.

2. It is predominantly polarized roughly ver-tical to the horizon.

3. It is predominantly polarized roughly par-allel to the horizon. correct

Explanation:For the case of blue noon sky, the incident

sunlight is unpolarized (or randomly polar-ized in a plane perpendicular to its propaga-tion). However, since you are looking awayfrom the Sun at the horizon the direct sunlightnever reaches your eye. It is the rescatteredlight from the air molecules in that portionof the sky that does reach your eye. Themolecules reradiate in the same plane as theplane of the polarization of the sunlight. Nearthe horizon, that plane is roughly parallel tothe horizon. From your view on the beachyou see light that is predominantly horizon-tally polarized.

013 (part 2 of 3) 3.0 pointsLater in your day at the beach, the Sun setsin the same portion of the sky a little bitabove the horizon over the ocean. Describethe polarization of light that comes from theSun to your eye. What is the best choice?

1. It is predominantly polarized roughly par-allel to the horizon.

2. It is predominantly unpolarized. correct

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3. It is predominantly polarized roughly ver-tical to the horizon.

Explanation:When sunlight is viewed head-on, one sees

mostly unpolarized light, since there is no pre-ferred direction of polarization. Sunset lightpartly consists of an uninterrupted light beamcoming directly from the sun, which is neces-sarily unpolarized. However, some of it mightalso be rescattered light produced throughscattering of the unpolarized sunlight by airmolecules in the atmosphere. Even for thiscomponent of light, there is no preferred az-imuthal direction since the incident light ray(from the sun to the scattering molecule) andthe rescattered light ray (from the scatteringmolecule to the eye) travel in the same direc-tion. Hence even this rescattered componentwill be unpolarized, to the extent that we as-sume scattering has no azimuthal preference.

014 (part 3 of 3) 4.0 pointsNow consider the sky colors in the proximityof the sun at sunset. One day it is red andanother day it is orange. Which case corre-sponds to having a stronger scattering effecton the original white beam?

1. Red correct

2. Neither — sunset color is not dependenton scattering effects

3. Orange

Explanation:Let us assume the frequency composition

of original sunlight to be such that it is whitelight. The more the scattering that has takenplace, the more the higher frequency compo-nents of light will be removed from the orig-inal white beam and hence the redder it willbe.

015 (part 1 of 3) 3.0 pointsAn electron is initially at rest. At time t = 0,it is accelerated at an angle θ = 18 with re-spect to the vertical for a very short time. The

magnitude of the acceleration is 2× 1018 m/s2

(this large acceleration is possible because theelectron has a very small mass). We makeobservations at location A at a horizontal dis-tance d = 10 m from the electron as in thefigure below.

θA

dElectron

Take the +x direction to be toward theright, the +y direction to be upward, andthe +z direction to be out of the page (to-ward you). Throughout this problem, use thefollowing constants:

c = 3× 108 m/s

e = 1.6× 10−19 C

1

4πǫ0= 9× 109 N ·m2/C2 .

When the radiative electric field reacheslocation A, what is the direction of that ra-diative electric field at location A? Choose theunit vector corresponding to this direction.

1. 〈0, 0,−1〉

2. 〈1, 0, 0〉

3. 〈−1, 0, 0〉

4. 〈0,−1, 0〉

5. 〈0, 1, 0〉 correct

6. 〈0, 0, 1〉

Explanation:The radiative electric field is proportional

to −q~a⊥ = +e~a⊥, which is upward. Thereis an accompanying magnetic field out of thepage because the direction of propagation ~E×~B is to the right.

016 (part 2 of 3) 3.0 points

Version 095 – MT4 – chiu – (58655) 7

What is the magnitude of this radiative elec-tric field?1. 3.49306e-092. 1.52266e-093. 1.94547e-094. 3.04338e-095. 2.07846e-096. 1.26611e-097. 2.22524e-098. 1.39032e-099. 2.3236e-0910. 3.15138e-09

Correct answer: 3.04338× 10−9 N/C.

Explanation:The magnitude of the upward radiative

electric field is

∣

∣

∣

~E∣

∣

∣

rad

=1

4πǫ0

e~a⊥c2 r

= 3.04338× 10−9 N/C .

017 (part 3 of 3) 4.0 pointsJust after the radiative field has reachedlocation A, what is the direction of themagnetic force on a positive charge that hadbeen at rest before the radiative field arrivedat location A? Note that the static electricfield is much smaller than the radiative elec-tric field at this location so that only theradiative electric field will play a role in deter-mining your answer. Choose the unit vectorcorresponding to this direction.

1. 〈−1, 0, 0〉

2. 〈1, 0, 0〉 correct

3. 〈0, 0, 1〉

4. 〈0,−1, 0〉

5. 〈0, 1, 0〉

6. 〈0, 0,−1〉

Explanation:A positive charge is accelerated upward

(the +y direction) by the upward radiative

electric field. Since Erad is in the +y-directionwhile the Poynting vector ~E× ~B points in the+x-direction, it must be that Brad points inthe +z-direction, out of the page; F = q~v× ~Btells us the force must be in the +x-direction(to the right).

018 10.0 pointsTwo light pulses are emitted simultaneouslyfrom a source. The pulses take parallel pathsto a detector 7.05 m away, but one movesthrough air and the other through plastic.Determine the difference in the pulses’

times of arrival at the detector. (Assumethe index of refraction of plastic is 1.271 andthat of air is 1.)1. 9.161072. 8.5123. 6.301334. 6.22725. 6.36856. 8.91257. 12.04978. 7.80889. 5.352110. 7.3656

Correct answer: 6.3685 ns.

Explanation:

Let : nplastic = 1.271 ,

nair = 1 ,

ℓ = 7.05 m , and

c = 3× 108 m/s .

The time difference is

∆t = tplastic − tair

=ℓ

vplastic−

ℓ

vair.

Since the speed of light in a medium of

refractive index n is v =c

n,

∆t = ℓ(nplastic

c−

nair

c

)

Version 095 – MT4 – chiu – (58655) 8

=7.05 m

3× 108 m/s(1.271− 1)

×

(

1 ns

1× 10−9 s

)

= 6.3685 ns .

019 (part 1 of 2) 5.0 points

An EM wave polarized along the x-axis andpropagating in the +z direction is incidentupon a pair of polarizers. The long moleculesof the first polarizer make an angle θ withthe +x-axis, while the long molecules of thesecond polarizer are parallel to the y-axis, asshown in the figure.

After passing through the first polarizer,what is the polarization of the EM wave?

1. along the x-axis

2. along the z-axis

3. parallel to the direction of the longmolecules

4. none — no light gets through

5. along the y-axis

6. perpendicular to the direction of the longmolecules correct

Explanation:

The first polarizer transmits light that ispolarized perpendicular to the long molecules.

020 (part 2 of 2) 5.0 pointsAfter passing through the second polarizer,what is the polarization of the EM wave?

1. along the x-axis correct

2. along the y-axis

3. along sin θ x+ cos θ y

4. none — no light gets through

5. along the z-axis

6. along cos θ x+ sin θ y

Explanation:The first polarizer transmits light that is

polarized perpendicular to the long molecules;in this case, the polarization of the waveafter passing through the first polarizer isalong sin θ x + cos θ y. The second polarizertransmits light polarized along the x-axis, soonly the x-component of the EM wave passesthrough, leaving a wave that is once again po-larized along the x-axis (though reduced inintensity).

021 (part 1 of 2) 5.0 points

Consider the circuit shown in the figurewhere the switch has been initially in positiond for a long period of time so that a steadycurrent has been achieved. Then, at timet = 0 the switch is set to position c.

LR

E

S c

d

ab

Version 095 – MT4 – chiu – (58655) 9

What is the current through the inductorat time t = 0? Let positive current flow bedefined as from a to b.

1. −E

R

2.E

Rcorrect

3. 0

Explanation:The loop equation while S is at d is given

byE − (Va − Vb)− IR = 0 .

Immediately after the switch is thrown toposition c, the current remains at the max-imum amplitude it had immediately beforethe switch was thrown, due to the magneticinertia of the inductor. That amplitude is de-termined by the value of I when dI/dt = 0,or

E − IR = 0 → I =E

R.

The current direction must remain unchangedwhen the switch is thrown, so flow must bepositive, from a to b.

022 (part 2 of 2) 5.0 pointsConsider an inductance of 3 H, resistance of97.6 Ω, and initial current of 3.3 A at t = 0.Find the total energy dissipated through

the resistor during the time interval t = 0 to∞.1. 46.082. 144.063. 0.94. 1.085. 74.066. 1.2257. 1.628. 50.6259. 16.33510. 217.8

Correct answer: 16.335.

Explanation:

Let : L = 3 H ,

R = 97.6 Ω , and

I0 = 3.3 A .

After switching to c, the total stored energyis the magnetic energy at t = 0:

UL

∣

∣

∣

t=0

=1

2L I20 ,

which will be dissipated in the resistor be-tween 0 < t < ∞ . Applying conservation ofenergy,

UR

∣

∣

∣

t=0→∞

= UL

∣

∣

∣

t=0

=1

2L I20

=1

2(3 H) (3.3 A)2 = 16.335 J .