CH 20 - Induced Voltages-solutions

CH 20 - Induced Voltages-solutions - This print-out should...

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Unformatted text preview: This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A rod slides along a metal frame. The earth’s magnetic field is 3 . 5 × 10- 5 T at right angles to the plane of the frame. The rod is of length 0 . 32 m inside the frame and moves at . 62 m / s. What is the motional emf produced by the rod? Correct answer: 6 . 944 × 10- 6 V. Explanation: Let : B e = 3 . 5 × 10- 5 T , = 0 . 32 m , and v = 0 . 62 m / s . The motional emf is given by E = B e v = (3 . 5 × 10- 5 T) (0 . 62 m / s) × (0 . 32 m) = 6 . 944 × 10- 6 V . 002 (part 2 of 3) 10.0 points If the resistance of the frame and rod com- bined is 8 . 7 Ω , what is the resulting current? Correct answer: 7 . 98161 × 10- 7 A. Explanation: Let : r = 8 . 7 Ω . From Ohm’s law, I = E r = (6 . 944 × 10- 6 V) (8 . 7 Ω ) = 7 . 98161 × 10- 7 A . 003 (part 3 of 3) 10.0 points What is the magnitude of the force on the rod? Correct answer: 8 . 9394 × 10- 12 N. Explanation: The force on a current element due to a magnetic field is given by F = B I . In our case, F = B e i = (3 . 5 × 10- 5 T) (7 . 98161 × 10- 7 A) × (0 . 32 m) = 8 . 9394 × 10- 12 N . The direction of the force may be found di- rectly, via the right-hand rule, or by energy conservation: the energy needed to create the current induced in the circuit must come from the kinetic energy of the wire, so the force op- poses the wire’s motion. keywords: 004 10.0 points A 604-turn circular-loop coil 19 . 6 cm in di- ameter is initially aligned so that its axis is parallel to Earth’s magnetic field. In 1 . 76 ms the coil is flipped so that its axis is perpendic- ular to Earth’s magnetic field....
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This note was uploaded on 10/03/2010 for the course PH 102 taught by Professor Staff during the Fall '08 term at Alabama.

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CH 20 - Induced Voltages-solutions - This print-out should...

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