Third Exam-solutions - Version 077 Third Exam Hoffmann...

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Unformatted text preview: Version 077 Third Exam Hoffmann (56745) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points No units are given for the answer choices. The units are milliVolts (mV). A conducting bar moves as shown near a long wire carrying a constant I = 93 A cur- rent. I a v L A B If a = 6 . 7 mm, L = 21 cm, and v = 14 m / s, what is the potential difference, V V A V B ? 1. 10.9414 2. 53.0526 3. 24.634 4. 13.9437 5. 2.07065 6. 8.16179 7. 22.8516 8. 36.7644 9. 18.0 10. 19.2 Correct answer: 8 . 16179 mV. Explanation: Given; E = d B dt = d dt ( B x ) = B dx dt = B v . From Amperes law, the strength of the magnetic field created by the long current- carrying wire at a distance a from the wire is B = I 2 a . Hence the potential difference is V = B L v = parenleftbigg I 2 a parenrightbigg ( L v ) = 8 . 16179 mV . 002 10.0 points No units are shown for the answer choices. The units are Amperes (A). The magnetic flux threading a metal ring varies with time t according to B = 3 a t 3 b t 2 , with a = 6 . 8 s- 3 m 2 T, and b = 7 s- 2 m 2 T. The resistance of the ring is 4 . 4 . Determine the maximum current induced in the ring during the interval from t 1 = 4 s to t 2 = 4 s. 1. 0.544006 2. 0.220627 3. 0.0635549 4. 0.4803 5. 0.0150816 6. 0.0743322 7. 0.181967 8. 0.0169214 9. 2.62812 10. 0.564972 Correct answer: 0 . 181967 A. Explanation: From Faradays law, the induced emf should be E = d B dt = (9 a t 2 2 b t ) , so the maximum E occurs when d E dt = 18 a t + 2 b = 0 t = b 9 a Version 077 Third Exam Hoffmann (56745) 2 and the maximum emf is E max = 9 a parenleftbigg b 9 a parenrightbigg 2 + 2 b parenleftbigg b 9 a parenrightbigg = b 2 9 a + 2 b 2 9 a = b 2 9 a . Thus the maximum current is I max = E max R = b 2 9 a R = (7 s- 2 m 2 T) 2 9 (6 . 8 s- 3 m 2 T) (4 . 4 ) = . 181967 A . 003 10.0 points An electron under the influence of some central force moves at speed v i in a counter- clockwise circular orbit of radius R . A uni- form magnetic field B perpendicular to the plane of the orbit is turned on (see figure). B increasing v electron circular path of electron radius R Suppose that the magnitude of the field changes at a given rate d B dt . What is the magnitude of the electric field induced at the radius of the electron orbit? 1. E = R d B dt 2. E = 2 2 R d B dt 3. E = R 2 d B dt 4. E = 2 R d B dt 5. E = 2 R 2 d B dt 6. E = R d B dt 7. E = 2 R d B dt 8. E = R 2 d B dt 9. E = R 2 d B dt correct 10. E = R 2 d B dt Explanation: Basic concepts: Faradays law states E = d dt ....
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This note was uploaded on 01/27/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Third Exam-solutions - Version 077 Third Exam Hoffmann...

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