PHYSICS 102 FINAL EXAMINATION - Princeton Universitygrothserver.princeton.edu/~groth/phys102s02/quiz/finalsol.pdf · 0 = 4ˇ 10 7TmA 1 k= 1=4ˇ 0 = 9 109Nm ... In the following there

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3 11:00am Galbiati

4 12:30pm Muthukumar

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PHYSICS 102

FINAL EXAMINATION

May 20, 2002 7:30–10:30 pm McDonnell A02

When you are told to begin, check that this examination booklet contains all thenumbered pages from 2 through 21. The exam contains 6 problems with unequal pointvalues as listed above.

Do not panic or be discouraged if you cannot do every problem; there are both easyand hard parts in this exam. Keep moving and finish as much as you can!

Read each problem carefully. With the exception of problem 1, you must show yourwork—the grade you get depends on how well the grader can understand your solutioneven when you write down the correct answer. Partial credit is given for relevant materialonly if it is in a comprehensive context. Please BOX your answers.

DO ALL THE WORK YOU WANT GRADED IN THIS EXAMINATION BOOKLET!

Rewrite and sign the Honor Pledge: I pledge my honor that I have not violated the HonorCode during this examination.

Signature

Copyright c© 2002, Princeton University Physics Department, Edward J. Groth

Physics 102 Final Exam Solutions 20-May-2002 Page 2

*** Possibly useful constants and equations ***You may remove this page and use it as a reference, if you wish.

ε0 = 8.85× 10−12C2N−1m−2 µ0 = 4π × 10−7TmA−1

k = 1/4πε0 = 9× 109Nm2C−2 e = 1.6× 10−19C

h = 6.6× 10−34Js c = 3.0× 108m/s

1 yr = 3.13× 107 s 1 hr = 3600 s

hc = 1240 eV nm 1 u = 931.5 MeV/c2

mp = 1.67× 10−27kg = 938.2 MeV/c2 me = 9.11× 10−31kg = 511 keV/c2

F = kQ1Q2/r2 F = qE

V = kQ/r U = qV

U = 12CV

2 E = σ/ε0

R = ρL/A B = µ0I/2πr

V = IR P = IV ; P = IrmsVrms

NsΦs = MIp NΦ = LI

B = µ0IN/L XC = 1/ωC

XL = ωL Z =

√R2 + (XL −XC)2

Prms = IrmsVrms cosφ fr = 1/2π√LC

~F = I~L× ~B F = ILB sin θ∑E⊥∆A = Q/ε0

∑B||∆l = µ0I

N = N0 exp(−λt) λ = ln(2)/T1/2

1/f = 1/di + 1/d0 m = −di/d0

E=−13.6Z2/n2 eV R=1.2× 10−15A1/3 m

rn = 5.3× 10−11n2/Z m E = hf

S = cε0E2 E = cB

∆t = ∆t0/√

1− (v/c)2 E = mc2/√

1− (v/c)2

p = mv/√

1− (v/c)2 vAB = vAC + vCB/(1 + vACvCB/c2)

L = L0

√1− (v/c)2 sin(θ) = 1.22λ/d

Q = CV sin(θ) = λ/d

C = ε0A/d λ = h/p



1. In the following there are 15 multiple choice questions. Circle the correct choice. Thereare 2 points for each correct answer. Circling more than one is considered a wrong answer.There is no partial credit.

(a) Two uniformly charged spheres are shown in the diagrams below. The right spherecontains three times the charge as the left sphere. Which diagram correctly representsthe magnitudes and directions of the electric forces acting on the spheres due to thecharges on the spheres?

(b) A uniform electric field points to the right as shown. Acharge enters the region of uniform field with an initialvelocity upwards as shown. If the charge is negative,which of the trajectories are possible?

A. 1 and 2 B. 3 C. 4 D. 2 E. 6.



Problem 1. Continued

(c) If a parallel plate capacitor is charged up with a dielectric nearby, what will happen?

A. Nothing, dielectrics are electrically neutral.

B. The dielectric is “blown away.”

C. The dielectric is “sucked in.”

(d) In the circuits shown, all fourlight bulbs are identical andthe two batteries are identi-cal. Which circuit puts outmore light?

A. Circuit 1.

B. They’re the same.

C. Circuit 2.




(e) The figures below show three regions of uniform magnetic field, each containing adifferent object. All vectors are in the plane of the paper. In which case(s) is there a

net force on the object?

A. 1 and 2.

B. 3 only.

C. 2 and 3.

D. No cases.

(f) The long parallel wires shown to the right carry equal currents in thesame direction. Which of the following is true?

A. The magnetic field from the left wire cancels the fieldfrom the right wire at the position of the right wire.

B. The wires are attracted to each other.

C. The wires repel each other.

D. The magnetic forces must cancel, otherwise wires would move whenever weput a current through them.




(g) The diagrams show two regions of uniform, vertical magnetic field. In case 1, a coilof wire is spun around a vertical axis. In case 2, a coil of wire is spun around a

horizontal axis. Which of the following is true?

A. An EMF is generated in the coil in case 1.

B. An EMF is generated in the coil in case 2.

C. An EMF is generated in the coil in both cases.

D. An EMF is not generated in either case, because the magnetic field and theloop area are fixed.

(h) A long straight wire carries a steady current I in the vertical directionas shown. A rectangular conducting loop lies in the same plane as thewire, with two sides parallel to the wire and two sides perpendicular tothe wire. Suppose the loop is pushed towards the wire as shown. Thedirection of the induced current in the loop is

A. clockwise.

B. counterclockwise.

C. indeterminant with the given information.




(i) A light bulb and capacitor are connected in series and powered bya variable frequency AC generator. Which of the following is true?

A. The light is brightest at low frequencies.

B. The light never lights up.

C. The light is brightest at high frequencies.

D. The intensity is the same at both high and low frequencies.

(j) In the circuit shown, switch S has been open for a longtime. It is closed, after which:

A. First light B comes on followed by light A.

B. Lights A and B come on at the same time.

C. Light A never comes on because the inductor impedes the current.

D. Light B goes out because the inductor acts just like a wire if one waits longenough.




(k) A fish swims below the surface of the water. An ob-server above the water sees the fish at

A. a greater depth than it really is.

B. its true depth.

C. a smaller depth than it really is.

(l) A diffraction grating is illuminated with red light at normal incidence. The patternseen on a screen behind the grating consists of three spots, one at zero degrees (straightthrough) and one each at ±45◦. Now, green light of equal intensity and coming fromthe same direction is added. The new pattern consists of:

A. Red spots at 0◦ and ±45◦.

B. Green spots at 0◦ and ±45◦.

C. Yellow spots at 0◦ and ±45◦.

D. A yellow spot at 0◦, red spots at ±45◦, and green spots slightly farther out.

E. A yellow spot at 0◦, red spots at ±45◦, and green spots slightly closer in.




(m) Muons are created at the top of the atmosphere (altitude 10 km) by cosmic rays.The lifetime of a muon is τ = 2.2 × 10−6 s. Even though they are travelling withv = 0.999c, they can only travel about 0.6 km in one lifetime before they decay. Yetmuons still manage to reach sea level. Why?

A. Muons decay into different particles than can make it to the Earth’s surface.

B. When moving at the speed of light the distance covered is no longer given byd = vt.

C. In the muon’s frame our clock runs faster so the distance is shorter.

D. In our frame, the muon’s clock runs slower and it has longer to traverse the

atmosphere before it decays.

(n) A beam of electrons hits a crystal with an interatomic spacing of 0.1 nm. Diffractionspots are seen with a regular spacing of 0.1 radians. The wavelength of the electronsis about

A. 100 nanometers.

B. 1 nanometer.

C. 0.01 nanometers.

D. 0.0001 nanometers.




(o) A radioactive source next to a Geiger counter producesclicks. When a lead shield is placed between the sourceand the counter, as shown in the figure, the clicks stop.If a uniform magnetic field out of the page is turnedon below the shield (as shown in the figure), clicks areheard again. If the field points into the page, no clicksare heard. What is the sign of the particles emitted inthe decay of the radioactive source?

A. Positive.

B. Negative.

C. Not enough information to tell.



2. Imagine that you are living in the distant future when technology has advanced to thepoint that it’s feasible to contemplate a trip to a star cluster which is 250 light years away.Even though you realize you’ll never again be able to see the friends and family you leavebehind, you volunteer to be one of the astronauts on the first mission to a distant starcluster.

(a) Naturally, you want to get there in a reasonable amount of time, so you still havesome of your life left for exploration. How fast must you be travelling (relative to theEarth) in order to get there in 10 years by your clock? Give your answer as a numbertimes the speed of light. Be sure that you get enough digits from your calculator tosee that the number is not 1! (5 points)

If T = 250 years is the time it takes light to get to the cluster, then its distance is cT . If

you travel at speed v, then the time it takes you to get to the cluster according to an observer on

Earth is cT/v. However, your clocks run slower, so the time according to your clocks (10 years) is

t = (cT/v)√

1− (v/c)2. Solving for v, v = c(T/t)/√

1 + (T/t)2 = c× 25/√

1 + 252 = 0.9992c .

(b) What is the Earth to star cluster distance as it appears to you as you are travellingto the star cluster at the speed calculated in part (a)? (4 points)

The distance is length contracted by the factor√

1 − (v/c)2 = 1/√

1 + (T/t)2 = 0.03997, so

the distance is 9.992 ≈ 10 light years .




(c) The mass of you and your space ship is 30, 000 kg. What is the kinetic energy of youand your ship travelling at the speed calculated in part (a) as reckoned by an observeron Earth? (5 points)

K = mc2/√

1 − (v/c)2−mc2 = mc2(√

1 + (T/t)2−1) = 24×mc2 = 24×30000×9×1016 =

6.5× 1022 J , something like 650 times the annual United States energy consumption! (It’s the

distant future after all!)

(d) How much mass must be converted into energy to provide the energy of part (c)?(3 points)

Mc2 = 24mc2, M = 24m = 72× 104 kg

(e) Of course you are expected to send back reports once you arrive at the star cluster.Immediately on arrival, you send back an “I’ve arrived safely” radio message. Howlong after you left the Earth do residents of the Earth receive this message (accordingto their clocks)? (3 points)

According to residents of Earth, you travelled 250 light years at essentially the speed of light

and the radio wave travelled back 250 light years, again at the speed of light. So, altogether, it

took 500 years from the time you left until your first message was received. (Since you were

only travelling at 0.9992 times the speed of light, your trip actually took 250.2 years, so a more

correct answer is 500.2 years.) If this were a sociology exam, we would probably ask for an essay

about what human civilization would be like after 500 years. Aren’t you glad it’s a physics exam?



3. In the lab you observe the light from a dischargelamp with a “crossbow” spectrometer as indicated inthe diagram. Some data: distance from grating tometer stick, D = 0.60 m, grating contains n = 400slits per millimeter. When you look through the grat-ing, you see the light source directly in front; to oneside, you see a blue, a blue-green, and a red line.You measure their positions along the meter stick(from the 0 order image of the lamp) as x1 = 0.106 m,x2 = 0.119 m, and x3 = 0.163 m, for the blue, blue-green, and red lines, respectively.

(a) What are the wavelengths (in nanometers) of the light in the three lines? (5 points)

d sin θ = mλ. For our case, m = 1, d = 1/400 = 2.5 × 10−6 m = 2500 nm and sin θ =

x/√x2 +D2. Then λ = d · x/

√x2 +D2. So, λ1 = 435 nm , λ2 = 486 nm , and λ3 = 655 nm

for the blue, blue-green, and red lines, respectively.

(b) What gas is in the discharge lamp? (3 points)

You should recognize the Balmer series of hydrogen . Just to check, λ3/λ2 = 655/486 = 1.35;

(1/22 − 1/42)/(1/22 − 1/32) = 1.35.




(c) You notice the same set of lines appears on the other side of the lamp (at negativex). You decide to improve the accuracy of your measurements by averaging thetwo distances you can measure for each line. (Of course, you ignore the sign of xwhile averaging.) Give at least two reasons why this improves the accuracy of yourmeasurements. (4 points)

First, you make a slight random error when you determine the positions of the lines and

averaging reduces the random error . Second, if the grating isn’t aligned exactly perpendicularly

to the line from the grating to the lamp, or the meter stick isn’t aligned exactly perpendicu-

larly to this line, the pattern of lines will appear shifted relative to the lamp. In other words,

the lines on one side of the lamp will appear slightly farther away from the lamp than they

should, and the lines on the other side will appear slightly closer than they should. Averaging

reduces this systematic error .

(d) You have several filters, one of which transmits only the red line (and absorbs theblue-green and blue lines). Another transmits only the blue-green line and a thirdtransmits only the blue line. You have an unknown metal. Using these filters, youfind that there is no photo-electric effect when the metal is illuminated by the redline, a strong photo-electric effect when the metal is illuminated by the blue line, anda feeble photo-electric effect when the metal is illuminated by the blue-green line.“Feeble” means the ejected electrons have almost no kinetic energy, not more thanK = 0.05 eV. What can you say about the work function (in electron volts) of theunknown metal? (Be as quantitative as you can!) (5 points)

Since the photons in the blue-green line have just enough energy to eject electrons, the work

function must be just a little less than the energy of these photons. The energy is E2 = hc/λ2 =

1240 eV nm/486 nm = 2.55 eV. Therefore the work function must be W = 2.55 − 0.05 = 2.5 eV

(e) You crank up the lamp intensity to see if you can observe a photoelectric effect withthe same metal and the red line. Do you succeed? Why or why not? (3 points)

You don’t succeed . The photons in the red line have an energy E3 = hc/λ3 = 1.89 eV

which is not enough to give an electron the 2.5 eV needed to overcome the work function and

escape from the metal.



4. An apparatus consists of two long concentric, conducting cylin-ders, sketched in cross section to the right. Each cylinder is verythin. The inner cylinder has radius a and the outer cylinder hasradius b. Spread uniformly over the inner cylinder is a charge λ > 0per unit length. (That is, each meter of the inner cylinder goinginto or out of the page contains a charge λ.) Similarly, a charge−λ per unit length is spread uniformly over the outer cylinder. Inaddition, the inner cylinder carries a current I > 0 coming out ofthe page and a current I flows into the page on the outer cylinder.The currents are uniformly spread over the cylinders.

(a) What is the magnitude of the electric field in the region insidethe inner cylinder, r < a, between the cylinders, a < r < b, andoutside the outer cylinder, b < r? Give your answers as formulaewhich may include physical or mathematical constants (such asµ0, ε0, π, etc., and the variables defined in the problem such as a,b, λ, I, and r, the distance from the centerline of the two cylinders.Sketch some representative electric field lines on the diagram tothe right. Be sure to show in which direction the field points oneach line! (6 points)

First of all, by the cylindrical symmetry, the electric field must point

radially outward or inward from the centerline of the cylinders and dependonly on r. This suggests we use Gauss’ law and take a Gaussian surface tobe a cylinder of radius r, length L and concentric with the other cylinders.There is no electric flux through the ends, so we only need to worry about theflux through the side. Φ = 2πrLE = Qinside/ε0, where E is the magnitudeof the radial component of the electric field. If r < a or r > b, there isno net charge inside the Gaussian surface and E = 0 . If a < r < b, the

net charge inside is λL, so E = λ/(2πε0r) and points radially outward asshown in the diagram.




(b) What is the magnitude of the magnetic field in the region insidethe inner cylinder, r < a, between the cylinders, a < r < b,and outside the outer cylinder, b < r? Again, you should giveyour answers in terms of formulae. Sketch some representativemagnetic field lines on the diagram to the right. Be sure to showin which direction the field points on each line! (6 points)

In this case, the cylindrical symmetry means that the magnetic field

must again depend only on r, but now it is directed in circles perpendicularto and concentric with the centerline. It’s different than the electric fieldcase, because flipping the cylinders end for end reverses the direction ofthe current but not the sign of the charge! We apply Ampere’s law to acircular loop of radius r concentric with the centerline: 2πrB = µ0Ithrough .If r < a or r > b, there is no net current through the Amperian loop andB = 0. For a < r < b, the current through the loop is I , out of the page,and B = µ0I/(2πr) and the right hand rule tells us the direction of themagnetic field is counterclockwise as indicated on the diagram.




(c) A charged particle is moving in the region between the inner and outer cylindersparallel to the axis of the cylinders. In other words, it’s travelling into or out of thepage. Suppose its charge is q > 0 and its speed is v. Show that if the particle ismoving in the correct direction (either into or out of the page) with the correct speed,it will travel in a straight line, undeflected by the combined electric and magnetic fieldsbetween the cylinders. Find the direction and find an expression for the speed. Showthat the speed doesn’t depend on where the particle is (as long as it’s between thecylinders), nor on the particle’s charge (as long as it’s positive). If you were unableto do parts (a) and (b), then for partial credit, you may take the electric field topoint to the right and have magnitude E and the magnetic field to point up and havemagnitude B. (6 points)

For the particle to be undeflected, the magnetic force and the electric force must be equal

and opposite. Note that the directions of E and B above are the same as in the true solution

for points to the right of the centerline. Due to cylindrical symmetry, whatever we find for

these points applies throughout the cylinder. The electric force is qE directed to the right. The

magnetic force is qvB directed to the left if the particle is coming out of the page and to the

right if the particle is headed into the page. It must be headed out of the page for the magnetic

and electric forces to cancel. Setting the magnitudes equal, qE = qvB or v = E/B and the

charge has dropped out . Now we plug in the magnitudes of E and B from parts (a) and (b):

v = [λ/(2πε0r)]/[µ0I/(2πr)] = (λ/I)× (1/(ε0µ0)) , and the radius has dropped out . This

device is the beginning of a velocity selector!

(d) Suppose the region between the cylinders is filled with teflon which has a dielectricconstant κ = 2.1. Nothing else is changed. What is the magnitude of the electric fieldbetween the cylinders in terms of the magnitude E0 without the dielectric? (2 points)

E = E0/κ = E0/2.1



5. Radium decays by emitting an alpha particle according to

22688Ra→ X + α ,

with a half life of 1600 years. Mass data: mRa = 226.02544 u, mX = 222.01761 u, mα =4.00260 u.

(a) Identify the nucleus X by giving its atomic mass number, A, and its atomic number,Z. (3 points)

An alpha particle contains 2 neutrons and 2 protons. Thus A = 226 − 4 = 222 and

Z = 88− 2 = 86 . X is radon!

(b) How much energy (in MeV) is released by this decay? (3 points)

The mass defect is 226.02544 − 222.01761 − 4.00260 = 0.00523 u. Converting to MeV,

0.00523 u× 931.5 MeV/u = 4.87 MeV .




(c) If all this energy is carried away by the alpha particle, what is its de Broglie wave-length? (Hint: the alpha particle may be treated non-relativistically.) (4 points)

λ = h/p = hc/√

2mc2E = 1240 eV nm/√

2× 4.00260 u× 931.5 MeV/u× 4.87 MeV This

gives λ = 6.51× 10−15 m , comparable to the size of a nucleus!

(d) How many decays per second occur in one gram of radium? (5 points)

The number of atoms remaining at time t starting with N0 at t = 0 is N(t) = N0 exp(−λt).The decay constant is λ = ln 2/thalf = 0.693/(1600 × 365.25 × 24 × 60 × 60) = 1.37 × 10−11 s−1

The number of atoms in 1 gram is 1 g×6.02×1023 atoms/mole/(226 g/mole) = 2.66×1021 atoms.

Finally, the number of decays per second is λN0 = 3.7× 1010 s−1 .



6. The diagram below shows an object (the arrow) which is located do = 12 units (eachtick mark on the horizontal line is a unit) in front of a lens. The lens is a converging lenswith a focal length f = 3 units. The focal points are indicated on the diagram.

(a) Algebraically determine the location of the image. Specify the location by determiningits distance from the lens in “units” and stating whether it is to the left or right ofthe lens. (5 points)

1/do + 1/di = 1/f , 1/di = 1/f − 1/do = 1/3 − 1/12 = 1/4, di = 4 units and it’s to the

right of the lens (since it’s positive).

(b) On the diagram above, carefully trace at least two rays that allow you to geometricallydetermine the location of the image of the tip of the arrow. (5 points)

The diagram shows three easy-to-draw rays. The ray through the center is undeflected. Raysparallel to the axis go through the focal points.




(c) What is the magnification? (3 points)

m = −di/do = −1/3 .

(d) Is the image real or virtual? Is it erect or inverted? (2 points)

Real and inverted .


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PHYSICS 102 FINAL EXAMINATION - Princeton Universitygrothserver.princeton.edu/~groth/phys102s02/quiz/finalsol.pdf · 0 = 4ˇ 10 7TmA 1 k= 1=4ˇ 0 = 9 109Nm ... In the following there