PHY 2054 Exam 2 Spring 2011 Key

PHY 2054 Exam 2 Spring 2011 Key - 77777 Instructor(s):...

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77777 77777 Instructor(s): Reitze/Kumar PHYSICS DEPARTMENT PHY 2054 Exam 2 March 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. (1) Code your test number on your answer sheet (use lines 76–80 on the answer sheet for the 5-digit number) . Code your name on your answer sheet. DARKEN CIRCLES COMPLETELY . Code your UFID 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. Circle your answers on the test form. 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. (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 some answers may be counted as incorrect. (5) The answers are rounded off. Choose the closest to exact. There is no penalty for guessing. If you believe that no listed answer is correct, leave the form blank. (6) Hand in the answer sheet separately. Useful Constants: k e = 8 . 99 × 10 9 Nm 2 /C 2 ² 0 = 8 . 85 × 10 - 12 C 2 /(Nm 2 ) V=volt N=newton μ 0 = 4 π × 10 - 7 Tm/A k 0 = μ 0 / (4 π ) = 10 - 7 Tm/A c = 3 × 10 8 m/s electron charge = - 1 . 6 × 10 - 19 C electron mass = 9 . 11 × 10 - 31 kg J=joule m=Meter “milli”=10 - 3 “micro”=10 - 6 n=“nano”=10 - 9 “pico”=10 - 12 C=coulomb g = 9 . 8 m/s 2 1. A magnetic field CANNOT: (1) change the kinetic energy of a charged particle. (2) exert a force on a charged particle. (3) change the velocity of a charged particle. (4) change the momentum of a charged particle. (5) change the trajectory of a charged particle. 2. A proton (charge +e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge +2e) which is also traveling perpendicular to the same field. The ratio of their speeds, v proton /v alpha , is: (1) 2 (2) 0.5 (3) 1 (4) 4 (5) 8 3. An infinitely long straight wire and a single circular current loop with radius R are arranged as shown in the figure. If both the straight wire and the loop are carrying the same current I, what is the magnitude of the magnetic field at the center of the loop? R I I I (1) ( π + 1) μ 0 I/ (2 πR ) (2) ( π - 1) μ 0 I/ (2 πR ) (3) μ 0 I/ (2 R ) (4) μ 0 I/ (2 πR ) (5) 2 πμ 0 I/R 4. An electron (charge q = - 1 . 6 × 10 - 19 C) undergoing uniform circular motion with a radius R = 2 × 10 - 8 m acts like a current loop. If the speed of the electron is 20 km/s, what is the corresponding current flow around the circle (in Amps)? (1) 2 . 5 × 10 - 8 (2) 5 . 0 × 10 - 8 (3) 1 . 2 × 10 - 8 (4) 2 . 5 × 10 - 7 (5) 5 . 0 × 10 - 7 5. Two isotopes of helium start from rest and pass through a potential difference Δ V = 860 V and then move out of a slit at the point S into a 0.2 T uniform magnetic field pointing into the paper as shown in the figure. Helium-4 consists
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This note was uploaded on 10/13/2011 for the course PHY 2054 taught by Professor Avery during the Spring '08 term at University of Florida.

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PHY 2054 Exam 2 Spring 2011 Key - 77777 Instructor(s):...

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