# HW 24 - terry(ect328 homework 24 Turner(59130 This...

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terry (ect328) – homework 24 – Turner – (59130) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A plane loop oF wire oF area A is placed in a region where the magnetic feld is perpendicu- lar to the plane. The magnitude oF B varies in time according to the expression B = B 0 e - at . That is, at t = 0 the feld is B 0 , and For t> 0, the feld decreases exponentially in time. ±ind the induced emF, E , in the loop as a Function oF time. 1. E = aB 0 t 2. E = 0 e - at 3. E = a A B 0 e - at correct 4. E = a A B 0 e - 2 at 5. E = a A B 0 6. E = AB 0 e - at Explanation: Basic Concepts: ±araday’s Law: E ≡ ± E · ds = - d Φ B dt Solution: Since B is perpendicular to the plane oF the loop, the magnetic ²ux through the loop at time 0 is Φ B = BA = 0 e - at Also, since the coe³cient AB 0 and the pa- rameter a are constants, and ±araday’s Law says E = - d Φ B dt the induced emF can be calculated the From Equation above: E = - d Φ B dt = - 0 d dt e - = a A B 0 e - That is, the induced emF decays exponentially in time. Note: The maximum emF occurs at t =0 , where E = a A B 0 . B = B 0 e - at B 0 0 0 ± t The plot oF E versus t is similar to the B versus t curve shown in the fgure above. 002 10.0 points The magnetic ²ux threading a metal ring varies with time t according to Φ B =3 3 - bt 2 , with a =4 . 1s - 3 · m 2 · T, and b = 7 . 5s - 2 · m 2 · T. The resistance oF the ring is 1 . 7 Ω. Determine the maximum current induced in the ring during the interval From t 1 = - 3s to t 2 = 2 s. Correct answer: 0 . 8967 A. Explanation: ±rom ±araday’s law, the induced emf should be E = - d Φ B dt = - (9 2 - 2 ) , so the maximum E occurs when d E dt = - 18 +2 b t = b 9 a and the maximum emf is E max = - 9 a ² b 9 a ³ 2 b ² b 9 a ³ = - b 2 9 a + 2 b 2 9 a = b 2 9 a .

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terry (ect328) – homework 24 – Turner – (59130) 2 Thus the maximum current is I max = E max R = b 2 9 aR = (7 . 5s - 2 · m 2 · T) 2 9 (4 . 1s - 3 · m 2 · T) (1 . 7 Ω) = 0 . 8967 A . 003 (part 1 of 2) 10.0 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 ± .
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## This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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HW 24 - terry(ect328 homework 24 Turner(59130 This...

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