Physics 142! Summer Final Exam. Solutions. Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points.

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1 Final Exam Solutions Part A. Multiple choice questions. Check the best answer. Each question carries a value of 4 points. 1.! The left sides of the four Maxwell equations are!!!! (1) E da (2) E dr (3) B da (4) B dr and the right sides are The correct matches of the two sides are: (a) 0! (1) = (a), (2) = (b), (3) = (c), (4) = (d).! (1) = (b), (2) = (a), (3) = (c), (4) = (d).! (1) = (d), (2) = (a), (3) = (c), (4) = (b). (b) 4πkQ enc (c) µ 0 I linked + µ 0 ε 0 (d) B t da E t da (1) = (b), (2) =(d), (3) = (a), (4) = (c). 2.! In the situation shown a grounded hollow conducting sphere has a positive charge +q at its center, and there is another positive charge +Q outside. +q +Q! The net charge on the outer surface of the sphere is zero. [Negative.] The charge +Q experiences a force to the left.! The charge +q experiences a force to the left. [No force.]! None of the above is true. 1

2 3.! Charges are fixed at two points. There is no point on the line between them where the total potential is zero (taking V( ) = 0 ). Which of these is wrong? There is no point on the line between them where the total E-field is zero.! There is no point at finite distance where the total potential is zero.! The potential energy of the two charges is positive.! The charges have the same sign. 4.! A small cylindrical bar magnet is dropped from rest down a long hollow cylindrical tube. It falls slowly to the floor. Which of the following is wrong?!! The tube is made of a conducting material.! The magnet s fall is slowed by forces arising from currents induced in the tube.! If the magnet is dropped with the orientation of its poles reversed, the induced currents in the tube will reverse direction. The force slowing the magnet s fall is independent of its speed. [Depends on the rate of change of the B-field, thus to the speed of the fall.] 5.! An astronomical telescope uses a mirror of diameter D and focal length f as its objective. When a distant double star is viewed using this telescope, the images of the two stars are not sufficiently resolved to tell that there are two stars. To get better resolved images, which of the following approaches will work?! Use a mirror of the same diameter but larger focal length.! Use an eyepiece of smaller focal length. Use a mirror of larger diameter.! Use a filter to let through only light of long wavelengths. [Short wavelengths might work.] 2

3 6.! White light is normally incident on a thin vertical soap film in air, and the reflected light is viewed. Because of gravity the liquid of the film sinks to the bottom so the top part becomes very thin. Just before it breaks the light reflected from the top part: Becomes dim for all wavelengths. [This was the demo in class.]! Becomes unusually bright for all wavelengths.! Becomes bluish in color.! Becomes reddish in color.! 7.! The yellow light from sodium lamps comes from from two wavelengths of about 600 nm, differing by 0.6 nm. Consider a grating with rulings 1500 nm apart. The two lines cannot be seen in 3rd order. [Requires sinθ = 1.2.]! To resolve these lines in 1st order, the grating must have at least 2000 rulings. [1000 will do.]! The angle at which these lines will be seen in 1st order is given by sinθ 0.6. [It is 0.4.]! None of the above is true. 8.! Which of the following is NOT a consequence of the uncertainty principle?! The state of an atomic electron cannot be specified in terms of its position and velocity. No state of an atomic electron can have a definite energy. [Stationary states are states of definite energy.]! The program of Newtonian mechanics can only be carried out as an approximation for large systems.! A particle in a bound state cannot have zero average kinetic energy. 3

4 Part B. True-false questions. Check T or F depending on whether the statement is true or false. Each question carries a value of 3 points. 1.! The total charge of an isolated system cannot be changed by any process. T F 2.! As one follows a line of the electrostatic field through space, the potential increases. T F [Decreases.] 3.! Motion of a conductor through a non-uniform static B-field is opposed by the interaction forces between the field and the induced currents in the conductor. T F 4.! When light of frequency f travels from air into a medium with refractive index n, the frequency becomes nf. T F [Frequency does not change.] 5.! AM radio waves diffract to follow the curve of the earth better than FM or TV waves, which have much higher frequencies. T F 6.! The supply of naturally occurring helium on earth comes from α-particles emitted in radioactive decays. T F 4

5 Part C. Problems. Work problem in space provided, using extra sheets if needed. Explain your method clearly. Problems carry the point values shown. 1.! A parallel plate capacitor is charged and kept connected to its charging source. The separation between the plates is then doubled. You are to express various quantities after the change in terms of their original values, and to explain how you know these are correct. Capacitance, voltage and charge, originally C 0, V 0 and Q 0. E-field, energy density and stored energy, originally E 0, u 0 and U 0. c.! [15 points] Explain carefully what agencies do work while the plates are being separated, and whether the work done is positive or negative.! C = 1 2 C 0. Formula for parallel plates (on formula sheet): C 1/d.! V = V 0. Remains connected to battery.! Q = 1 2 Q 0. Follows from Q = CV. E = 1 2 E 0. Follows from E = V /d for uniform field.! u = 1 4 u 0. Follows from u = 1 2 ε 0 E2.! U = 1 2 U 0. Follows from U = 1 2 CV 2, or from the fact that the volume between the plates is twice as large while u is four times smaller, or from U = Q 2 /2C. c.! The person who pulls the plates apart does positive work (against the attractive force between the plates). The battery, accepting charge from the plates, does negative work. [The work by the battery is 1 2 Q 0 V 0 = U 0. The work by the person is thus U 0.] 5

6 2 A positive charge q of mass m enters a region of uniform B-field out of the page as shown. The field region extends a distance d, after which the field drops to zero. What is the maximum speed v the charge can have at entry and not quite leave the field at the top side? d q B If it has that speed, how long does it stay in the field? [Sketch its path.] [10 points] It should move in a circle of radius d, so we have mv2 d = qvb, or v = qbd/m. Path shown. It is a semicircle of radius d, so the path length is πd and the time it spends in the field is πd/v = πm/qb. 2 A car battery is being recharged by the car s alternator. The emf of the battery is 12 V. When the current through the battery is 10 A the voltage across it is V. How much power from the alternator is being dissipated by the internal resistance of the battery?! [5 points] Since the terminal voltage is E + IR we see that I = 0.3 V and I 2 R = 10 (0.3) = 3 W. Or, the total power to the battery is (12.3) 10 = 123 W, of which EI = 120 W goes to charging the battery, so 3 W is dissipated. 6

7 3.! The jumping ring demonstration involves a solenoid with an iron core which creates a strong magnetic field in the z-direction given by B z (z,t) = B 0 (z) cosωt. Here B 0 decreases with z above the core so z db 0 /dz < 0. You will investigate the force on the ring. Assume the B-field is uniform across the area A of the ring. Find the induced emf E(t) in the ring. If the ring has resistance R, find the current I(t) and the magnitude µ(t) of the induced magnetic moment. What is its direction shortly after t = 0, when B z is positive but decreasing? c.! Write the potential energy U = µ B and the force F z = du/dz. d.! Show that the force, averaged over a cycle, is zero.! [Use for averages over a cycle: (sinθ) av = (cosθ) av = (sinθ cosθ) av = 0 ; (sin 2 θ) av = (cos 2 θ) av = 1 2.] [20 points] The flux is Φ = BA = A B 0 cosωt, so the emf is E = dφ/dt = ωa B 0 sinωt. c.! The current is I = E /R = (ωab 0 /R) sinωt and µ = IA = (ωa 2 B 0 /R) sinωt. When B is upward but decreasing, µ is also upward (to oppose the decrease in B.) We have U = µ B = (ωa 2 B 0 2 /R) sinωt cosωt. Since only B 0 depends on z we have F z = du/dz = (ωa 2 /R) 2B 0 (db 0 /dz) sinωt cosωt. d.! Because (sinωt cosωt) av = 0 the average force over a cycle is zero, as claimed. 7

8 4 The previous problem s conclusion contradicts the facts of the demonstration, in which the ring always jumps upward. The flaw is in part (b) where we assumed the ring had only resistance R; it also has self-inductance L. This makes the current in the ring lag behind the induced emf by the phase angle φ. Redo parts (b) and (c) with the proper current to find the correct expression for the force. Show that φ > 0. Use the identity sin(α β) = sinα cosβ cosα sin β to find the average force over a cycle. Show that it is upward. [Remember db 0 /dz < 0 because the field is weaker above the core.] [10 points] The current should be I = (ωab 0 /Z) sin(ωt φ), where Z = R 2 + X 2 L and tanφ = X L /R > 0. This leads to µ = IA = (ωa 2 B 0 /Z) sin(ωt φ), to U = µ B = (ωa 2 B 2 0 /Z) sin(ωt φ) cosωt, and finally to F z = du/dz = (ωa 2 /Z) 2B 0 (db 0 /dz) sin(ωt φ) cosωt. Use the identity: F z = (ωa 2 /Z) 2B 0 (db 0 /dz) [sinωt cosφ cosωt sinφ] cosωt. The average over a cycle is zero for the term in cosφ, but the other term survives. The average force is (F z ) av = (ωa 2 /Z) B 0 (db 0 /dz) sinφ. Since db 0 /dz < 0, this is positive (upward) and the ring always rises. 4 To supply power P 0 to a distant city a utility company uses wires of resistance R. The potential difference between these wires as they enter the city is V. Show that the power lost to resistance in the wires is (P 0 /V) 2 R. Use this to explain why utility companies make use of AC rather than DC.! [10 points] The power is P 0 = IV, so I = P 0 /V and the power lost is I 2 R = (P 0 /V) 2 R as claimed. To minimize the loss, V should be made large. This is possible with transformers and AC but not with DC. 8

9 5 The battery shown is part of a circuit with current as indicated. The battery has emf E, and is a circular cylinder of length l and cross-section radius r. The positive terminal is indicated. c.! [15 points] At the two points shown, draw with arrows the directions of E, B and S. Calculate the magnitude of S, assuming the points are on the battery s surface. Calculate the flux of S through the battery s surface and verify that it is the power supplied by the battery to the circuit. + S B B E E I S Drawing shown. E is toward lower potential, B is like that of a long wire. We have E = E /l, B = µ 0 I/2πr, so S = EB/µ 0 = EI/2πrl. c.! The magnitude of S is the same over the outer surface of the battery so we have P = S A = S 2πrl = EI, as expected. 5 A person on the ground is viewing the sun just above the horizon at sunset, while directly above him in an airplane at a high altitude the pilot also views the sun. The color of the sun as seen by the person on the ground is considerably redder than as seen by the pilot. Explain with a drawing. [5 points] See drawing. The sun s ray to the person on the ground, a, passes through much more air than that to the pilot, b. More of the blue light is scattered away from the light seen by a than by b, so the sun looks redder to a. b a 9

10 6 An elderly person wears bifocal eyeglasses. The upper part, to correct distant vision, has power 0.25 D, while the lower part, to produce an effective near point of 25 cm, has power +1.5 D. Without the eyeglasses: [10 points] What is the most distant point at which the person can focus clearly? What is the nearest point at which the person can focus clearly? Without the glasses the relaxed eye of this person has power P D, where P 0 is the power of the normal relaxed eye. At the far point p we have P = (1/p) + P 0, so p = 4 m. Alternative: The upper part of the glasses takes an object at infinity and makes a virtual image at the far point, so 0.25 = (1/ ) (1/p) ; this leads to the same answer. The normal eye focusing at 25 cm has power P 0 + 4, so this person has power P Thus we have, without the glasses, P = (1/p) + P 0, or p = 0.4 m. Alternative: The lower part of the glasses makes a virtual image at 25 cm of an object at the near point, so 1.5 = (1/0.25) (1/p) ; this leads to the same answer. 6 A microscope and a refracting telescope both use two positive lenses, both form a real image of the object between the lenses, and both have a virtual final image. Why is the magnification of the microscope so much larger?! [5 points] Since the object for a microscope can be placed where the viewer wishes, it can be placed close to the objective so as to make a greatly enlarged real image between the two lenses. The object for a telescope is beyond the viewer s control and far away, so the real image formed by the objective is always reduced. 10

11 7 Shown are waves resulting from a light wave of amplitude E 0 incident normally on an air film between plates of identical glass. The film s thickness is 1/2 the wavelength of the light. The reflectivity of the surfaces is R. Show that amplitudes E a and E b are of opposite E a b c d 0 sign, i.e., the waves are out of phase by an odd multiple of π. Also show that the amplitudes of waves b, c, d,... have the same sign as E b. Write the amplitudes for these four waves in terms of E 0 and R. Use this to show that if all the waves, including the ones not shown, are added the total reflected amplitude is zero. [Use 1+ x + x 2 + x 3 + = 1 for x < 1.] 1 x [10 points] The reflection of a at the bottom of the upper plate gives no phase change, while the reflection of b at the top of the bottom plate gives a phase change of π. The path difference gives a phase difference of 2π, so b is out of phase by 3π with a. This means the amplitudes have opposite signs. Waves c, d, etc. are reflected from both glass plates so the extra phase changes are always multiple of 2π, and their extra path differences always add multiples of 2π. Their phase difference with b is a multiple of 2π so their amplitudes have the same sign as b. Thus the amplitudes of b, c, d etc. all have the same sign, and the amplitude of a has the opposite sign. Waves b, c, d, etc. are transmitted twice, so they are proportional to T. We have E a = RE 0, E b = T RE 0, E c = T R RE 0, E d = T R 2 RE 0, etc. Adding the amplitudes we find E refl = RE 0 [1 T(1 + R + R )]. Using the given series this becomes E refl = RE 0 [1 T/(1 R)]. But T = 1 R, so this is zero, as claimed 7 A pair of objects are 5 cm apart vertically and are 10 km horizontally from a camera with a lens of diameter 12.2 cm. If resolution is limited only by diffraction, show that the image made by the camera resolves these objects with light of λ = 400 nm, but not with light of λ = 600 nm. [5 points] Here Δθ = 0.05/10 4 = For 400 nm, δθ = /0.122 = , which is less than Δθ. For 600 nm, δθ = /0.122 = , more than Δθ. 11

12 8 A radioactive parent nucleus with average lifetime τ 1 decays into a daughter nucleus, also radioactive, with average lifetime τ 2 < τ 1. We start at t = 0 with a sample of N 1 (0) atoms of the parent and none of the daughter. Write the equation giving dn 2 /dt in terms of N 1 (t) and N 2 (t). [Remember that the rate of decay is N(t)/τ.] The solution to this equation that satisfies the initial conditions is!! N 2 (t) = N τ 1 τ 1 (0) e t/τ 1 e t/τ 2. 2 [10 points] τ 2 Show that the ratio N 2 (t)/n 1 (t) approaches a constant value as t and evaluate that limit. The rate of increase of the daughter is the rate of decay of the parent minus the rate of decay of the daughter, so dn 2 /dt = N 1 /τ 1 N 2 /τ 2. Since N 1 (t) = N 1 (0)e t/τ 1, we find by dividing N 2 (t) N 1 (t) = τ 2 1 e t/τ 2 e t/τ 1 τ 1 τ 2. Since τ 2 < τ 1 the last term in [ ] goes to zero as t. The limit is N 2 N 1 τ 2 τ 1 τ 2. 8 Hoping to discredit Darwin, Lord Kelvin made an estimate of how long ago the earth was too hot to support life. Because the rate of flow of heat to the earth s surface is so great he arrived at a figure of only 20 million years, much too short for natural selection to have done its work. He did this before the discovery of radioactivity. Explain why that discovery invalidates Kelvin s conclusion. [5 points] Much of the heat flowing to the surface comes from energy released in radioactive decays, so one cannot attribute it to a simple cooling of the earth. In fact well over half of the heat flow is from nuclear decays, from the long-lived isotopes of thorium, uranium and potassium. 12

13 9.! You have completed two courses on the description of nature given by physics. Pick three important things you learned that you didn t already know, and explain briefly why you chose these. [15 points] I enjoy reading these and am sometimes surprised by the choices. 13