physics 102 final examination - Princeton University
PRINTED NAME: SOLUTIONS Circle your Class Section 1
Problem
9:00am Sohn
Score
1
/30
2
10:00am Klebanov
2
/20
3
11:00am Galbiati
3
/20
4
12:30pm Muthukumar
4
/20
5
12:30pm Nappi
5
/15
6
/15
Total
/120
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 the numbered pages from 2 through 21. The exam contains 6 problems with unequal point values as listed above. Do not panic or be discouraged if you cannot do every problem; there are both easy and 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 your work—the grade you get depends on how well the grader can understand your solution even when you write down the correct answer. Partial credit is given for relevant material only 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 Honor Code during this examination.
Signature c 2002, Princeton University Physics Department, Edward J. Groth Copyright
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.
0 = 8.85 × 10−12 C2 N−1 m−2 k = 1/4π0 = 9 × 109 Nm2 C−2 h = 6.6 × 10−34 Js 1 yr = 3.13 × 107 s hc = 1240 eV nm mp = 1.67 × 10−27 kg = 938.2 MeV/c2 F = kQ1 Q2 /r2 V = kQ/r U = 21 CV 2 R = ρL/A V = IR Ns Φs = M Ip B = µ0 IN/L XL = ωL
µ0 = 4π × 10−7 TmA−1 e = 1.6 × 10−19 C c = 3.0 × 108 m/s 1 hr = 3600 s 1 u = 931.5 MeV/c2 me = 9.11 × 10−31 kg = 511 keV/c2 F = qE U = qV E = σ/0 B = µ0 I/2πr P = IV ; P = IrmsVrms NΦ = LI XC =q1/ωC
R2 + (XL − XC )2 √ fr = 1/2π LC F = ILB sin θ P B|| ∆l = µ0 I λ = ln(2)/T1/2 m = −di /d0 R=1.2 × 10−15 A1/3 m E = hf E = cB p E = mc2 / 1 − (v/c)2 vAB = vAC + vCB /(1 + vAC vCB /c2 ) sin(θ) = 1.22λ/d sin(θ) = λ/d λ = h/p Z=
Prms = Irms Vrms cos φ ~ ×B ~ F~ = I L P E⊥ ∆A = Q/0 N = N0 exp(−λt) 1/f = 1/di + 1/d0 E=−13.6Z 2/n2 eV rn = 5.3 × 10−11 n2 /Z m S = c0 E 2 p ∆t = ∆t0 / 1 − (v/c)2 p p = mv/ 1 − (v/c)2 p L = L0 1 − (v/c)2 Q = CV C = 0 A/d
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 3
1. In the following there are 15 multiple choice questions. Circle the correct choice. There are 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 sphere contains three times the charge as the left sphere. Which diagram correctly represents the magnitudes and directions of the electric forces acting on the spheres due to the charges on the spheres?
(b) A uniform electric field points to the right as shown. A charge enters the region of uniform field with an initial velocity 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.
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 4
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 four light bulbs are identical and the two batteries are identical. Which circuit puts out more light? A. Circuit 1. B. They’re the same. C. Circuit 2.
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 5
Problem 1. Continued (e) The figures below show three regions of uniform magnetic field, each containing a different 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 the same direction. Which of the following is true? A. The magnetic field from the left wire cancels the field from 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 we put a current through them.
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 6
Problem 1. Continued (g) The diagrams show two regions of uniform, vertical magnetic field. In case 1, a coil of 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 the loop area are fixed.
(h) A long straight wire carries a steady current I in the vertical direction as shown. A rectangular conducting loop lies in the same plane as the wire, with two sides parallel to the wire and two sides perpendicular to the wire. Suppose the loop is pushed towards the wire as shown. The direction of the induced current in the loop is A. clockwise. B. counterclockwise. C. indeterminant with the given information.
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 7
Problem 1. Continued (i) A light bulb and capacitor are connected in series and powered by a 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 long time. 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 long enough.
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 8
Problem 1. Continued (k) A fish swims below the surface of the water. An observer 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 pattern seen on a screen behind the grating consists of three spots, one at zero degrees (straight through) and one each at ±45◦ . Now, green light of equal intensity and coming from the 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.
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 9
Problem 1. Continued (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 with v = 0.999c, they can only travel about 0.6 km in one lifetime before they decay. Yet muons 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 by d = 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. Diffraction spots are seen with a regular spacing of 0.1 radians. The wavelength of the electrons is about A. 100 nanometers. B. 1 nanometer. C. 0.01 nanometers. D. 0.0001 nanometers.
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Physics 102 Final Exam Solutions
20-May-2002 Page 10
Problem 1. Continued (o) A radioactive source next to a Geiger counter produces clicks. When a lead shield is placed between the source and the counter, as shown in the figure, the clicks stop. If a uniform magnetic field out of the page is turned on below the shield (as shown in the figure), clicks are heard again. If the field points into the page, no clicks are heard. What is the sign of the particles emitted in the decay of the radioactive source? A. Positive. B. Negative. C. Not enough information to tell.
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Physics 102 Final Exam Solutions
20-May-2002 Page 11
2. Imagine that you are living in the distant future when technology has advanced to the point 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 leave behind, you volunteer to be one of the astronauts on the first mission to a distant star cluster. (a) Naturally, you want to get there in a reasonable amount of time, so you still have some of your life left for exploration. How fast must you be travelling (relative to the Earth) in order to get there in 10 years by your clock? Give your answer as a number times the speed of light. Be sure that you get enough digits from your calculator to see 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 cTp /v. However, your clocks run slower, so p 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 travelling to the star cluster at the speed calculated in part (a)? (4 points) The distance is length contracted by the factor the distance is 9.992 ≈ 10 light years .
p
1 − (v/c)2 = 1/
p
1 + (T /t)2 = 0.03997, so
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 12
Problem 2. Continued (c) The mass of you and your space ship is 30, 000 kg. What is the kinetic energy of you and your ship travelling at the speed calculated in part (a) as reckoned by an observer on Earth? (5 points) p
p
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) M c2 = 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. How long after you left the Earth do residents of the Earth receive this message (according to 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?
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Physics 102 Final Exam Solutions
20-May-2002 Page 13
3. In the lab you observe the light from a discharge lamp with a “crossbow” spectrometer as indicated in the diagram. Some data: distance from grating to meter stick, D = 0.60 m, grating contains n = 400 slits per millimeter. When you look through the grating, you see the light source directly in front; to one side, 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, bluegreen, 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; (1/2 − 1/42 )/(1/22 − 1/32 ) = 1.35. 2
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Problem 3. Continued (c) You notice the same set of lines appears on the other side of the lamp (at negative x). You decide to improve the accuracy of your measurements by averaging the two distances you can measure for each line. (Of course, you ignore the sign of x while averaging.) Give at least two reasons why this improves the accuracy of your measurements. (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 perpendicularly 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 the blue-green and blue lines). Another transmits only the blue-green line and a third transmits only the blue line. You have an unknown metal. Using these filters, you find that there is no photo-electric effect when the metal is illuminated by the red line, a strong photo-electric effect when the metal is illuminated by the blue line, and a 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 than K = 0.05 eV. What can you say about the work function (in electron volts) of the unknown 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 with the 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.
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Physics 102 Final Exam Solutions
20-May-2002 Page 15
4. An apparatus consists of two long concentric, conducting cylinders, sketched in cross section to the right. Each cylinder is very thin. The inner cylinder has radius a and the outer cylinder has radius b. Spread uniformly over the inner cylinder is a charge λ > 0 per unit length. (That is, each meter of the inner cylinder going into or out of the page contains a charge λ.) Similarly, a charge −λ per unit length is spread uniformly over the outer cylinder. In addition, the inner cylinder carries a current I > 0 coming out of the 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 inside the inner cylinder, r < a, between the cylinders, a < r < b, and outside the outer cylinder, b < r? Give your answers as formulae which may include physical or mathematical constants (such as µ0 , 0 , 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 to the right. Be sure to show in which direction the field points on each 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 depend only on r. This suggests we use Gauss’ law and take a Gaussian surface to be 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 the flux through the side. Φ = 2πrLE = Qinside /0 , where E is the magnitude of the radial component of the electric field. If r < a or r > b, there is no net charge inside the Gaussian surface and E = 0 . If a < r < b, the net charge inside is λL, so E = λ/(2π0 r) and points radially outward as shown in the diagram.
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Physics 102 Final Exam Solutions
20-May-2002 Page 16
Problem 4. Continued (b) What is the magnitude of the magnetic field in the region inside the inner cylinder, r < a, between the cylinders, a < r < b, and outside the outer cylinder, b < r? Again, you should give your answers in terms of formulae. Sketch some representative magnetic field lines on the diagram to the right. Be sure to show in 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 perpendicular to and concentric with the centerline. It’s different than the electric field case, because flipping the cylinders end for end reverses the direction of the current but not the sign of the charge! We apply Ampere’s law to a circular loop of radius r concentric with the centerline: 2πrB = µ0 Ithrough . If r < a or r > b, there is no net current through the Amperian loop and B = 0. For a < r < b, the current through the loop is I, out of the page, and B = µ0 I/(2πr) and the right hand rule tells us the direction of the magnetic field is counterclockwise as indicated on the diagram.
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Physics 102 Final Exam Solutions
20-May-2002 Page 17
Problem 4. Continued (c) A charged particle is moving in the region between the inner and outer cylinders parallel to the axis of the cylinders. In other words, it’s travelling into or out of the page. Suppose its charge is q > 0 and its speed is v. Show that if the particle is moving 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 fields between the cylinders. Find the direction and find an expression for the speed. Show that the speed doesn’t depend on where the particle is (as long as it’s between the cylinders), nor on the particle’s charge (as long as it’s positive). If you were unable to do parts (a) and (b), then for partial credit, you may take the electric field to point to the right and have magnitude E and the magnetic field to point up and have magnitude 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/(2π0 r)]/[µ0 I/(2πr)] = (λ/I) × (1/(0 µ0 )) , 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 dielectric constant κ = 2.1. Nothing else is changed. What is the magnitude of the electric field between the cylinders in terms of the magnitude E 0 without the dielectric? (2 points) E = E0 /κ = E0 /2.1
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 18
5. Radium decays by emitting an alpha particle according to 226 88 Ra
→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 .
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Physics 102 Final Exam Solutions
20-May-2002 Page 19
Problem 5. Continued (c) If all this energy is carried away by the alpha particle, what is its de Broglie wavelength? (Hint: the alpha particle may be treated non-relativistically.) (4 points) p √ λ = h/p = hc/ 2mc2 E = 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 .
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
20-May-2002 Page 20
6. The diagram below shows an object (the arrow) which is located d o = 12 units (each tick mark on the horizontal line is a unit) in front of a lens. The lens is a converging lens with 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 determining its distance from the lens in “units” and stating whether it is to the left or right of the 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 geometrically determine 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. Rays parallel to the axis go through the focal points.
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20-May-2002 Page 21
Problem 6. Continued (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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THIS PAGE INTENTIONALLY LEFT ALMOST BLANK.
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Problem
9:00am Sohn
Score
1
/30
2
10:00am Klebanov
2
/20
3
11:00am Galbiati
3
/20
4
12:30pm Muthukumar
4
/20
5
12:30pm Nappi
5
/15
6
/15
Total
/120
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 the numbered pages from 2 through 21. The exam contains 6 problems with unequal point values as listed above. Do not panic or be discouraged if you cannot do every problem; there are both easy and 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 your work—the grade you get depends on how well the grader can understand your solution even when you write down the correct answer. Partial credit is given for relevant material only 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 Honor Code during this examination.
Signature c 2002, Princeton University Physics Department, Edward J. Groth Copyright
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.
0 = 8.85 × 10−12 C2 N−1 m−2 k = 1/4π0 = 9 × 109 Nm2 C−2 h = 6.6 × 10−34 Js 1 yr = 3.13 × 107 s hc = 1240 eV nm mp = 1.67 × 10−27 kg = 938.2 MeV/c2 F = kQ1 Q2 /r2 V = kQ/r U = 21 CV 2 R = ρL/A V = IR Ns Φs = M Ip B = µ0 IN/L XL = ωL
µ0 = 4π × 10−7 TmA−1 e = 1.6 × 10−19 C c = 3.0 × 108 m/s 1 hr = 3600 s 1 u = 931.5 MeV/c2 me = 9.11 × 10−31 kg = 511 keV/c2 F = qE U = qV E = σ/0 B = µ0 I/2πr P = IV ; P = IrmsVrms NΦ = LI XC =q1/ωC
R2 + (XL − XC )2 √ fr = 1/2π LC F = ILB sin θ P B|| ∆l = µ0 I λ = ln(2)/T1/2 m = −di /d0 R=1.2 × 10−15 A1/3 m E = hf E = cB p E = mc2 / 1 − (v/c)2 vAB = vAC + vCB /(1 + vAC vCB /c2 ) sin(θ) = 1.22λ/d sin(θ) = λ/d λ = h/p Z=
Prms = Irms Vrms cos φ ~ ×B ~ F~ = I L P E⊥ ∆A = Q/0 N = N0 exp(−λt) 1/f = 1/di + 1/d0 E=−13.6Z 2/n2 eV rn = 5.3 × 10−11 n2 /Z m S = c0 E 2 p ∆t = ∆t0 / 1 − (v/c)2 p p = mv/ 1 − (v/c)2 p L = L0 1 − (v/c)2 Q = CV C = 0 A/d
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Physics 102 Final Exam Solutions
20-May-2002 Page 3
1. In the following there are 15 multiple choice questions. Circle the correct choice. There are 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 sphere contains three times the charge as the left sphere. Which diagram correctly represents the magnitudes and directions of the electric forces acting on the spheres due to the charges on the spheres?
(b) A uniform electric field points to the right as shown. A charge enters the region of uniform field with an initial velocity 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.
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Physics 102 Final Exam Solutions
20-May-2002 Page 4
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 four light bulbs are identical and the two batteries are identical. Which circuit puts out more light? A. Circuit 1. B. They’re the same. C. Circuit 2.
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Physics 102 Final Exam Solutions
20-May-2002 Page 5
Problem 1. Continued (e) The figures below show three regions of uniform magnetic field, each containing a different 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 the same direction. Which of the following is true? A. The magnetic field from the left wire cancels the field from 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 we put a current through them.
c 2002, Princeton University Physics Department, Edward J. Groth Copyright
Physics 102 Final Exam Solutions
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Problem 1. Continued (g) The diagrams show two regions of uniform, vertical magnetic field. In case 1, a coil of 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 the loop area are fixed.
(h) A long straight wire carries a steady current I in the vertical direction as shown. A rectangular conducting loop lies in the same plane as the wire, with two sides parallel to the wire and two sides perpendicular to the wire. Suppose the loop is pushed towards the wire as shown. The direction of the induced current in the loop is A. clockwise. B. counterclockwise. C. indeterminant with the given information.
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Problem 1. Continued (i) A light bulb and capacitor are connected in series and powered by a 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 long time. 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 long enough.
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Physics 102 Final Exam Solutions
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Problem 1. Continued (k) A fish swims below the surface of the water. An observer 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 pattern seen on a screen behind the grating consists of three spots, one at zero degrees (straight through) and one each at ±45◦ . Now, green light of equal intensity and coming from the 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.
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Physics 102 Final Exam Solutions
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Problem 1. Continued (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 with v = 0.999c, they can only travel about 0.6 km in one lifetime before they decay. Yet muons 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 by d = 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. Diffraction spots are seen with a regular spacing of 0.1 radians. The wavelength of the electrons is about A. 100 nanometers. B. 1 nanometer. C. 0.01 nanometers. D. 0.0001 nanometers.
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Physics 102 Final Exam Solutions
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Problem 1. Continued (o) A radioactive source next to a Geiger counter produces clicks. When a lead shield is placed between the source and the counter, as shown in the figure, the clicks stop. If a uniform magnetic field out of the page is turned on below the shield (as shown in the figure), clicks are heard again. If the field points into the page, no clicks are heard. What is the sign of the particles emitted in the decay of the radioactive source? A. Positive. B. Negative. C. Not enough information to tell.
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Physics 102 Final Exam Solutions
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2. Imagine that you are living in the distant future when technology has advanced to the point 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 leave behind, you volunteer to be one of the astronauts on the first mission to a distant star cluster. (a) Naturally, you want to get there in a reasonable amount of time, so you still have some of your life left for exploration. How fast must you be travelling (relative to the Earth) in order to get there in 10 years by your clock? Give your answer as a number times the speed of light. Be sure that you get enough digits from your calculator to see 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 cTp /v. However, your clocks run slower, so p 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 travelling to the star cluster at the speed calculated in part (a)? (4 points) The distance is length contracted by the factor the distance is 9.992 ≈ 10 light years .
p
1 − (v/c)2 = 1/
p
1 + (T /t)2 = 0.03997, so
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Problem 2. Continued (c) The mass of you and your space ship is 30, 000 kg. What is the kinetic energy of you and your ship travelling at the speed calculated in part (a) as reckoned by an observer on Earth? (5 points) p
p
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) M c2 = 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. How long after you left the Earth do residents of the Earth receive this message (according to 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?
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Physics 102 Final Exam Solutions
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3. In the lab you observe the light from a discharge lamp with a “crossbow” spectrometer as indicated in the diagram. Some data: distance from grating to meter stick, D = 0.60 m, grating contains n = 400 slits per millimeter. When you look through the grating, you see the light source directly in front; to one side, 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, bluegreen, 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; (1/2 − 1/42 )/(1/22 − 1/32 ) = 1.35. 2
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Problem 3. Continued (c) You notice the same set of lines appears on the other side of the lamp (at negative x). You decide to improve the accuracy of your measurements by averaging the two distances you can measure for each line. (Of course, you ignore the sign of x while averaging.) Give at least two reasons why this improves the accuracy of your measurements. (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 perpendicularly 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 the blue-green and blue lines). Another transmits only the blue-green line and a third transmits only the blue line. You have an unknown metal. Using these filters, you find that there is no photo-electric effect when the metal is illuminated by the red line, a strong photo-electric effect when the metal is illuminated by the blue line, and a 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 than K = 0.05 eV. What can you say about the work function (in electron volts) of the unknown 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 with the 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.
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Physics 102 Final Exam Solutions
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4. An apparatus consists of two long concentric, conducting cylinders, sketched in cross section to the right. Each cylinder is very thin. The inner cylinder has radius a and the outer cylinder has radius b. Spread uniformly over the inner cylinder is a charge λ > 0 per unit length. (That is, each meter of the inner cylinder going into or out of the page contains a charge λ.) Similarly, a charge −λ per unit length is spread uniformly over the outer cylinder. In addition, the inner cylinder carries a current I > 0 coming out of the 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 inside the inner cylinder, r < a, between the cylinders, a < r < b, and outside the outer cylinder, b < r? Give your answers as formulae which may include physical or mathematical constants (such as µ0 , 0 , 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 to the right. Be sure to show in which direction the field points on each 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 depend only on r. This suggests we use Gauss’ law and take a Gaussian surface to be 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 the flux through the side. Φ = 2πrLE = Qinside /0 , where E is the magnitude of the radial component of the electric field. If r < a or r > b, there is no net charge inside the Gaussian surface and E = 0 . If a < r < b, the net charge inside is λL, so E = λ/(2π0 r) and points radially outward as shown in the diagram.
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Problem 4. Continued (b) What is the magnitude of the magnetic field in the region inside the inner cylinder, r < a, between the cylinders, a < r < b, and outside the outer cylinder, b < r? Again, you should give your answers in terms of formulae. Sketch some representative magnetic field lines on the diagram to the right. Be sure to show in 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 perpendicular to and concentric with the centerline. It’s different than the electric field case, because flipping the cylinders end for end reverses the direction of the current but not the sign of the charge! We apply Ampere’s law to a circular loop of radius r concentric with the centerline: 2πrB = µ0 Ithrough . If r < a or r > b, there is no net current through the Amperian loop and B = 0. For a < r < b, the current through the loop is I, out of the page, and B = µ0 I/(2πr) and the right hand rule tells us the direction of the magnetic field is counterclockwise as indicated on the diagram.
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Problem 4. Continued (c) A charged particle is moving in the region between the inner and outer cylinders parallel to the axis of the cylinders. In other words, it’s travelling into or out of the page. Suppose its charge is q > 0 and its speed is v. Show that if the particle is moving 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 fields between the cylinders. Find the direction and find an expression for the speed. Show that the speed doesn’t depend on where the particle is (as long as it’s between the cylinders), nor on the particle’s charge (as long as it’s positive). If you were unable to do parts (a) and (b), then for partial credit, you may take the electric field to point to the right and have magnitude E and the magnetic field to point up and have magnitude 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/(2π0 r)]/[µ0 I/(2πr)] = (λ/I) × (1/(0 µ0 )) , 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 dielectric constant κ = 2.1. Nothing else is changed. What is the magnitude of the electric field between the cylinders in terms of the magnitude E 0 without the dielectric? (2 points) E = E0 /κ = E0 /2.1
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Physics 102 Final Exam Solutions
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5. Radium decays by emitting an alpha particle according to 226 88 Ra
→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 .
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Problem 5. Continued (c) If all this energy is carried away by the alpha particle, what is its de Broglie wavelength? (Hint: the alpha particle may be treated non-relativistically.) (4 points) p √ λ = h/p = hc/ 2mc2 E = 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 .
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6. The diagram below shows an object (the arrow) which is located d o = 12 units (each tick mark on the horizontal line is a unit) in front of a lens. The lens is a converging lens with 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 determining its distance from the lens in “units” and stating whether it is to the left or right of the 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 geometrically determine 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. Rays parallel to the axis go through the focal points.
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Problem 6. Continued (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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