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21111 21111 Instructor: Profs. Paul Avery, Selman Hershfield PHYSICS DEPARTMENT PHY 2049 Final Exam April 23, 2011 Name (print, last first): Signature: On my honor, I have neither given nor received unauthorized aid on this examination. YOUR TEST NUMBER IS THE 5-DIGIT NUMBER AT THE TOP OF EACH PAGE. DIRECTIONS (1) Code your test number on your answer sheet (use 76–80 for the 5-digit number). Code your name on your answer sheet. DARKEN CIRCLES COMPLETELY. Code your student number on your answer sheet. (2) Print your name on this sheet and sign it also. (3) Do all scratch work anywhere on this exam that you like. At the end of the test, this exam printout is to be turned in. No credit will be given without both answer sheet and printout with scratch work most questions demand. (4) Blacken the circle of your intended answer completely, using a #2 pencil or blue or black ink. Do not make any stray marks or the answer sheet may not read properly. (5) The answers are rounded off. Choose the closest to exact. There is no penalty for guessing. Constants: e = 1.6 × 10−19 >>>>>>>>WHEN YOU FINISH <<<<<<<< Hand in the answer sheet separately. C mp = 1.67 × 10−27 kg me = 9.1 × 10−31 kg c = 3 × 108 m/s ²o = 8.85 × 10−12 C 2 /N · m2 k = 1/(4π²o ) = 9 × 109 N · m2 /C 2 µo = 4π × 10−7 T · m/A micro = 10−6 nano = 10−9 |q1 ||q2 | Coulomb’s Law: |F~ | = (point charge) 4π²o r2 ~ ~ =F Electric field: E q ~ = E ~A= Gauss’ law: Φ = n̂ · E H q r̂ (point charge) 4π²o r2 ~ = E R dq r̂ (general) 4π²o r2 E= ~ dA = qenc n̂ · E ²o R 1 1 F~ · d~s = mvf2 − mvi2 = Kf − Ki 2 2 R For conservative forces Uf − Ui = − F~ · d~s → Ki + Ui = Kf + Uf Energy: W = Electric potential: V = Vb − Va = − Rb a Ex dx = − U q Rb Capacitors: q = CV U= Resistors: i = a V = ~ · d~s E C= q2 2C dq = jA dt V = ∂V , ∂x ∂V , ∂y Ex = − Ey = − K²o A (parallel-plate) d u= R= q = CV (1 − e−t/RC ) (charging) q (point charge) 4π²o r V i 1 ²o E 2 2 R= R dq (general) 4π²o r Ez = − ∂V ∂z C = C1 + C2 (parallel) 1 1 1 = + (series) C C1 C2 ρL (wire) A P = iV q = qo e−t/RC (discharging) R = R1 + R2 (series) 1 1 1 = + (parallel) R R1 R2 ~ ~ ×B ~ ~ ~ Magnetism: F~ = q~v × B F~ = iL µ = N iA ~τ = µ ~ ×B U = −~ µ·B H µo i µo i µo iN ~ · d~s = µo ienc ~ = µo id~s × r̂ B B= , (wire) (loop center), (solenoid) dB 2 4π r 2πR 2R L σ (plane) 2²o 21111 21111 L = N ΦB /i (definition) UB = 1 2 Li 2 uB = B2 2µo AC Circuits: ω = √ tan φ = H i= I= Em Z q = Qo e−Rt/(2L) cos(ω 0 t + φ) First 2 Maxwell’s Eqs.: Last 2 Maxwell’s Eqs.: EM Waves: c = I = Io cos2 θ E= E (1 − e−t/τL ) R H Z= 1 1 1 2 + = = p i f r i = io e−t/τL E = Em sin(ωt) p E1 = −M τL = L R di2 dt i = I sin(ωt − φ) (driven RLC) R2 + (XL − XC )2 H ~ · n̂dA = qenc E ²0 ~ · n̂dA = 0 B H XL = ωL, XC = θc = sin−1 ~ =B ~ m sin(~k · ~r − ωt) B pr = 1 2 E cµo rms n2 θB = tan−1 n1 Interference: ∆L = mλn (constructive) Pavg = 1 IEm cos φ 2 1 di , vL = L , ωC dt n2 n1 ~m ⊥ B ~ m ⊥ ~k E p= id = ²o vC = cos φ = R Z q C dΦE dt Em Erms = √ 2 I = Savg = 2I (total reflection) c n1 n2 n2 − n1 + = p i r 1 1 1 = + (parallel) L L1 L2 ~ · d~s = −N dΦB E dt ~ · d~s = µo ²o dΦE + µo ienc B dt ~= 1E ~ ×B ~ S µo L = L1 + L2 (series) Vs Ns = Vp Np ¡ ¢1/2 ω 0 = ω 2 − (R/(2L))2 n1 sin θ1 = n2 sin θ2 I (total absorption) c Images: H E 1 =√ B µo ²o ~ =E ~ m sin(~k · ~r − ωt) E pr = ~ dA n̂ · B 1 (LC circuit) LC XL − XC R H ~ · d~s = − dΦB E dt di L = µo n2 Al (solenoid) E = −L dt ~A= Induction: ΦB = n̂ · B I= Ps 4πr2 c = ω/k = f λ U (mom. carried by EM radiation of energy U ) c 1 1 1 1 1 + = = (n − 1)( − ) p i f r1 r2 ∆L = (m + 21 )λn (destructive) m=− λn = λ/n i p n = 1 in air ∆L = d sin(θ) (2-slit), 2L1 − 2L2 (interferometer), 2t (thin film) with extra λ/2 when reflecting off higher index Diffraction: d sin(θ) = mλ (grating) 2d sin(θ) = mλ (X-ray) sin θ = 1.22λ/d (aperture) 1. Four charges are placed on the corners of a square as shown in the figure. All of the charges have the same magnitude, but as indicated in the figure some of the charges are positive and some are negative. In which of the three cases is the electric field at the center of the square zero? (1) A,B (2) A only (3) A,C (4) C only (5) B only 2. One of the two slits in a Young’s experiment is painted over so that it transmits only one-half the intensity of the other slit. As a result: (1) (2) (3) (4) (5) the the the the the dark fringes get brighter and the bright ones get darker fringes just get dimmer fringe system disappears bright fringes get brighter and the dark ones get darker dark fringes just get brighter 21111 21111 3. As shown in the figure, light is incident at an angle θ1 from a material with n1 = 1.1 to a material with n2 = 1.3. What is the smallest value of θ1 so that there is total internal reflection at the interface between n2 and n3 = 1? (1) 65.4◦ (2) 57.8◦ (3) 42.1◦ (4) 72.3◦ (5) 50.3◦ 4. A current moves from left to right in a horizontal copper wire. Which statement is true? (1) (2) (3) (4) (5) Power dissipation in the wire is directly proportional to both current and resistance The drift velocity of the electrons is slightly greater than the random motion speed of the electrons The electric field points from right to left The resistivity of the copper wire does not depend on its length The electrons in the wire move from left to right 5. In a television tube electrons are accelerated through a voltage of 5000 volts. What is the speed of the electrons assuming that they were initially at rest? (Since this is Physics 2, do not use special relativity to calculate the answer even though these speeds are a fraction of the speed of light.) (1) 2 × 107 m/s (2) 3 × 107 m/s (3) 1 × 107 m/s (4) 6 × 107 m/s (5) 4 × 107 m/s 6. An object is placed 30 cm to the left of a diverging lens of focal length −15 cm. A converging lens of focal length 30 cm is placed 30 cm to the right of the first lens. Where is the final image located relative to lens 2, and what is the overall magnification m? (1) (2) (3) (4) (5) 20 cm to the right; m = −0.5 30 cm to the right; m = −2.67 40 cm to the right; m = +1.5 120 cm to the right; m = −1.0 80 cm to the right; m = +2.5 7. An electron is launched with velocity v in a uniform magnetic field. The angle θ between the velocity vector ~v and the magnetic field direction is between 0 and 90◦ . As a result, the electron follows a helix, its velocity vector returning to its initial value in a time interval of: (1) 2πm/(eB sin θ) (2) 2πm/(eB) (3) 2πmv/(eB) (4) 2πmv sin θ/(eB) (5) 2πmv cos θ/(eB) 8. An RLC circuit is driven by a voltage of amplitude 12 V and has R = 4Ω, L = 2 H, and C = 100µF. At resonance what is the amplitude of the voltage oscillations across the inductor? (1) 3 V (2) 420 V (3) 6 V (4) 12 V (5) 24 V 9. Two concentric shells of negligible thickness have radius r1 = 10 cm and r2 = 20 cm. If the electric field at r = 15 cm is directed radially outwards with magnitude 1000 V /m and the electric field at r = 25 cm is directed radially inwards with magnitude 500 V /m, what is the sum of the charges on the two shells? (1) +2.5 nC (2) −3.5 nC (3) −1.0 nC (4) +3.5 nC (5) +6.0 nC 21111 21111 10. A current of 3A flows into circular parallel plate capacitor as shown in the figure. The plates have radius 1.0 cm. What is the magnitude of the magnetic field at point P in between the plates at a radial distance of 0.5 cm from the center of the plates? (1) 12 × 10−5 T (2) 3.0 × 10−5 T (3) 6.0 × 10−5 T (4) 1.5 × 10−5 T (5) 0.75 × 10−5 T 11. Sunlight is incident on a diffraction grating with 400 rulings/mm . What is the highest order, m, that a complete visible spectrum 400 nm < λ < 700 nm can be seen? (1) 1 (2) 5 (3) 4 (4) 2 (5) 3 12. A concave shaving mirror has a radius of curvature of 30 cm. It is positioned so that the (upright) image of a man’s face is 3 times the size of the face. How far is the mirror from the face (in cm)? (1) 40 (2) 20 (3) 10 (4) 3.8 (5) 80 13. A circular loop of wire of radius 2 cm is in the plane of the page. A magnetic field perpendicular to the page is changing with time according to B(t) = 1 + 3t, where t is measured in seconds, B is measured in Tesla, and positive field is going out of the page. If the loop has resistance 5Ω, what is the magnitude and direction of the induced current at t = 1 second? (1) (2) (3) (4) (5) 0.5 mA clockwise 0.75 mA counterclockwise 1.0 mA clockwise 0.75 mA clockwise 1.0 mA counterclockwise 14. In the figure shown, two long straight wires with separation d carry currents i1 and i2 = 2i1 out of the page, with i1 located at x = 0. At what point on the x-axis shown is the net magnetic field due to the currents equal to zero? y i2 i1 x d (1) 2d/3 (2) 2d (3) d/2 (4) d/3 (5) 3d/2 15. Resistances of 2.0 Ω, 4.0 Ω, and 6.0 Ω and a voltage source with 24 V are all in series. The potential difference across the 2.0 Ω resistor is: (1) 8 V (2) 24 V (3) 12 V (4) 2 V (5) 4 V 16. A satellite is 20,000 km above the surface of the Earth. It has a lens of diameter 50 cm. For light of wavelength 500 nm, what is the smallest resolvable distance based on Rayleigh’s criterion? (1) 25 m (2) 2.5 cm (3) 2.5 m (4) 25 cm (5) 250 m 17. A small insect starts walking away from a concave mirror along its central axis. Which statement is true? (1) (2) (3) (4) (5) The The The The The image is initially erect, then flips upside down image grows monotonically larger as the insect moves away distance from the insect to its image remains constant image is always upside down image is always erect 21111 21111 18. Three charges are placed on the corners of a rectangle as shown in the figure. If Q1 = −2µC, Q2 = 1µC, and Q3 = 3µC, what is the x component of the force on Q3 due to Q1 and Q2 ? (1) +9.8 N (2) +17 N (3) +34 N (4) −4.7 N (5) −0.4 N 19. A battery is used to charge a series combination of two identical capacitors. If the potential difference across the battery terminals is V and total charge Q flows through the battery during the charging process then the charge on the positive plate of each capacitor and the potential difference across each capacitor are, respectively: (1) Q and V (2) Q and V /2 (3) Q/2 and V (4) Q/2 and V /2 (5) 2Q and 2V 20. A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. The intensities of wavelengths 480 nm and 800 nm and no wavelengths in between are to be maximized in the reflected beam. (Take the index of refraction of air to be 1.) The thickness of the film is: (1) 240 nm (2) 600 nm (3) 400 nm (4) 360 nm (5) 150 nm