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HW22Sol - Platt David – Homework 22 – Due 4:00 am –...

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Unformatted text preview: Platt, David – Homework 22 – Due: Oct 31 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A(n) 0 . 58 m length of wire moves across a 2 . 14 T magnetic field. At what speed would it have to move to induce an EMF of 13 . 1 V? Correct answer: 10 . 5543 m / s. Explanation: Let : ‘ = 0 . 58 m , B = 2 . 14 T , and EMF = 13 . 1 V . The induced EMF is V = B l v Thus v = V B l = 13 . 1 V (2 . 14 T)(0 . 58 m) = 10 . 5543 m / s . 002 (part 1 of 1) 10 points A135-turn5cmsquarewirecoilrotatesabout a vertical axis at 1210 rev / min. The horizon- tal component of the Earth’s magnetic field at the location of the loop is 2 . 8 × 10- 5 T. B x = 2 . 8 × 10- 5 T B x = 2 . 8 × 10- 5 T 5cm 5 c m 1210 rev / min Calculate the maximum emf induced in the coil by Earth’s field. Correct answer: 1 . 19742 mV. Explanation: Let : ‘ = 5 cm = 0 . 05 m , A = ‘ 2 = 25 cm 2 = 0 . 0025 m 2 , ω = 1210 rev / min , N = 135 turns , and B = 2 . 8 × 10- 5 T . The maximum emf is E max = N AB ω = (135) ( . 0025 m 2 ) ( 2 . 8 × 10- 5 T ) × (1210 rev / min) · 2 π rad 1rev · 1min 60sec × 10 3 mV 1 V = 1 . 19742 mV . 003 (part 1 of 2) 10 points The resistance of the rectangular current loop is R , and the metal rod is sliding to the left. The length of the rod is d , while the width of the rails is ‘ . Note: a and b are the contact points where the rod touches the rails, and d > ‘. v m R a b B B ‘ d What is the magnitude of the induced cur- rent around the loop?...
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HW22Sol - Platt David – Homework 22 – Due 4:00 am –...

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