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Third Exam-solutions

Third Exam-solutions - Version 077 – Third Exam –...

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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, Δ 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 Φ B dt = − d dt ( B ℓ x ) = − B ℓ dx dt = − B ℓ v . From Ampere’s law, the strength of the magnetic field created by the long current- carrying wire at a distance a from the wire is B = μ I 2 π a . Hence the potential difference is Δ V = B L v = parenleftbigg μ I 2 π 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 Φ 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 Ω. 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 Faraday’s law, the induced emf should be E = − d Φ 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 Ω) = . 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 π 2 R d B dt 3. E = π R 2 d B dt 4. E = 2 π R d B dt 5. E = π 2 R 2 d B dt 6. E = π R d B dt 7. E = π 2 R d B dt 8. E = π R 2 d B dt 9. E = R 2 d B dt correct 10. E = R 2 d B dt Explanation: Basic concepts: Faraday’s law states E = − d Φ dt ....
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Third Exam-solutions - Version 077 – Third Exam –...

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