# Lenz' and Faraday's laws-solutions - gilvin (jg47854) - 20....

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The direction of the induced current in the gilvin (jg47854) - 20. Lenz’ and Faraday’s laws - meyers - (21235) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page - find all choices before answering. This homework is due Tuesday, April 26, at midnight Tucson time. 001 (part 1 of 3) 10.0 points A rectangular loop of copper wire of resis- tance R has width a and length b. The loop is stationary in a constant, the magnetic field B at time t = 0 seconds is directed into the page as shown below. The uniform magnetic field varies with time t according to the rela- tionship B = B 0 cos ω t, where ω and B 0 are positive constants and B is positive when the field is directed into the page. B b B n turns B B π be in a direction to oppose this change; i.e., clockwise. 002 (part 2 of 3) 10.0 points What is the expression for the magnitude of the induced current in the loop as a function of time in terms of a, b, B 0 , ω, R, t, and fundamental constants. 1. I = a b B 0 ω R sin ω t 2. I = a b B 0 R sin ω t 3. I = R B 0 a b sin ω t 4. I = a b ω B 0 R sin ω t 5. I = a b ω B 0 R sin ω t correct 6. I = R ω B 0 a b sin ω t Explanation: Calculating the flux, Φ = a b B 0 cos ω t . Calculating the emf, loop when ω t = 2 , after the magnetic field begins to oscillate is 1. clockwise. correct 2. undetermined, since the current is zero. E = −d Φ (negative sign not required) dt = a b ω B 0 sin ω t . Using Ohm’s Law I = E t . 3. counter-clockwise. Explanation: When ω t = π 2 , B = B 0 cos 2 R = a b R B 0 sin ω 003 (part 3 of 3) 10.0 points Select a sketch a graph of the induced cur- rent I vs ω t, taking clockwise current to be positive. I 1. ω t 0 π π = 0; i.e., 2 2 2 the field has been decreasing, and is about to change direction. The induced current will

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gilvin (jg47854) - I 2. 20. Lenz’ and Faraday’s laws - meyers - (21235) 2 I ω t ω t 0 π π 0 π π 2 correct I 3. 2 2 ω t 2 2 2 004 10.0 points A coil is wrapped with 382 turns of wire on the perimeter of a circular frame (of radius 62 cm). Each turn has the same area, equal 0 π π 2 I 4. 2 2 ω t to that of the frame. A uniform magnetic field is directed perpendicular to the plane of the coil. This field changes at a constant rate from 24 mT to 60 mT in 42 ms. What is the magnitude of the induced E in the coil at the instant the magnetic field has a magnitude of 49 mT? Correct answer: 395.412 V. 0 π π 2 2 2 Explanation: Basic Concepts: E = −N d Φ B dt I Φ B B · dA = B · A Solution: 5. ω t 0 π π 2 2 I 6. 0 π π 2 2 Explanation: 2 ω t 2 E = −N d Φ B dt = −N A ΔB Δt = −N π r 2 (B 2 − B 1 ) Δt = −(382) π (62 cm) 2 × (60 mT) − (24 mT) 42 ms = −395.412 V |E | = 395.412 V . 005
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## This note was uploaded on 04/29/2011 for the course PHYS 202 taught by Professor Doc during the Spring '11 term at North Texas.

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Lenz' and Faraday's laws-solutions - gilvin (jg47854) - 20....

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