hw 24 - nguyen (jmn727) homework 24 Turner (59070) 1 This...

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Unformatted text preview: nguyen (jmn727) homework 24 Turner (59070) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A plane loop of wire of area A is placed in a region where the magnetic field is perpendicu- lar to the plane. The magnitude of B varies in time according to the expression B = B e- at . That is, at t = 0 the field is B , and for t > 0, the field decreases exponentially in time. Find the induced emf, E , in the loop as a function of time. 1. E = a A B e- 2 at 2. E = a A B 3. E = A B e- at 4. E = a B e- at 5. E = a A B e- at correct 6. E = a B t Explanation: Basic Concepts: Faradays Law: E contintegraldisplay E ds =- d B dt Solution: Since B is perpendicular to the plane of the loop, the magnetic flux through the loop at time t > 0 is B = B A = A B e- a t Also, since the coefficient AB and the pa- rameter a are constants, and Faradays Law says E =- d B dt the induced emf can be calculated the from Equation above: E =- d B dt =- A B d dt e- a t = a A B e- a t That is, the induced emf decays exponentially in time. Note: The maximum emf occurs at t = 0 , where E = a A B . B = B e- at B vector t The plot of E versus t is similar to the B versus t curve shown in the figure above. 002 10.0 points The magnetic flux threading a metal ring varies with time t according to B = 3 a t 3- b t 2 , with a = 4 . 7 s- 3 m 2 T, and b = 3 . 1 s- 2 m 2 T. The resistance of the ring is 2 . 5 . Determine the maximum current induced in the ring during the interval from t 1 =- 4 s to t 2 = 3 s. Correct answer: 0 . 0908747 A. Explanation: From Faradays law, the induced emf should be E =- d B dt =- (9 a t 2- 2 b t ) , so the maximum E occurs when d E dt =- 18 a t + 2 b = 0 t = b 9 a and the maximum emf is E max =- 9 a parenleftbigg b 9 a parenrightbigg 2 + 2 b parenleftbigg b 9 a parenrightbigg =- b 2 9 a + 2 b 2 9 a = b 2 9 a . nguyen (jmn727) homework 24 Turner (59070) 2 Thus the maximum current is I max = E max R = b 2 9 a R = (3 . 1 s- 2 m 2 T) 2 9 (4 . 7 s- 3 m 2 T) (2 . 5 ) = . 0908747 A ....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw 24 - nguyen (jmn727) homework 24 Turner (59070) 1 This...

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