20. Lenz' and Faraday's laws-solutions

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

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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(part1of3)10.0points 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 B b a n turns The direction of the induced current in the loop when ω t = π 2 , after the magnetic field begins to oscillate is 1. clockwise. correct 2. undetermined, since the current is zero. 3. counter-clockwise. Explanation: When ω t = π 2 , B = B 0 cos π 2 = 0; i.e. , the field has been decreasing, and is about to change direction. The induced current will be in a direction to oppose this change; i.e. , clockwise. 002(part2of3)10.0points 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 , E = d Φ dt (negative sign not required) = a b ω B 0 sin ω t . Using Ohm’s Law I = E R = a b ω B 0 R sin ω t . 003(part3of3)10.0points Select a sketch a graph of the induced cur- rent I vs ω t , taking clockwise current to be positive. 1. ω t I 0 π 2 π 3 π 2 2 π 5 π 2 3 π
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gilvin (jg47854) – 20. Lenz’ and Faraday’s laws – meyers – (21235) 2 2. ω t I 0 π 2 π 3 π 2 2 π 5 π 2 3 π correct 3. ω t I 0 π 2 π 3 π 2 2 π 5 π 2 3 π 4. ω t I 0 π 2 π 3 π 2 2 π 5 π 2 3 π 5. ω t I 0 π 2 π 3 π 2 2 π 5 π 2 3 π 6. ω t I 0 π 2 π 3 π 2 2 π 5 π 2 3 π Explanation: The graph is a sine wave with period 2 π . ω t I 0 π 2 π 3 π 2 2 π 5 π 2 3 π 004 10.0points 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 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. Explanation: BasicConcepts: E = N d Φ B dt Φ B integraldisplay vector B · d vector A = B · A Solution: E = N d Φ B dt = N A Δ B Δ t = N π r 2 ( B 2 B 1 ) Δ t = (382) π (62 cm) 2 × (60 mT) (24 mT) 42 ms = 395 . 412 V |E| = 395 . 412 V .
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