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Unformatted text preview: Anderson, Matthew Homework 11 Due: Apr 15 2008, 9:00 am Inst: Vitaly 1 This printout should have 11 questions. Multiplechoice 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) 15 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 = aB t 2. E = aAB e 2 at 3. E = aAB e at correct 4. E = AB e at 5. E = aB e at 6. E = aAB Explanation: Basic Concepts: Faradays Law: E I 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 = AB 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 = AB d dt e a t = aAB e a t That is, the induced emf decays exponentially in time. Note: The maximum emf occurs at t = 0 , where E = aAB . B = B e at B ~ t The plot of E versus t is similar to the B versus t curve shown in the figure above. keywords: 002 (part 1 of 1) 15 points The plane of a rectangular coil, 7 . 2 cm by 11 . 4 cm, is perpendicular to the direction of a uniform magnetic field B . If the coil has 38 turns and a total resistance of 11 . 1 , at what rate must the magnitude of B change to induce a current of 0 . 14 A in the windings of the coil? Correct answer: 4 . 9823 T / s. Explanation: Given : x = 7 . 2 cm = 0 . 072 m , y = 11 . 4 cm = 0 . 114 m , N = 38 turns , r = 11 . 1 , and I = 0 . 14 A . The induced emf is E = I R = N d dt = N d ( B A ) dt = N A dB dt , so dB dt = I R N ( xy ) = (0 . 14 A)(11 . 1 ) 38(0 . 072 m)(0 . 114 m) = 4 . 9823 T / s . Anderson, Matthew Homework 11 Due: Apr 15 2008, 9:00 am Inst: Vitaly 2 keywords: 003 (part 1 of 1) 15 points A conducting bar moves as shown near a long wire carrying a constant I = 30 A current. I a v L A B If a = 2 . 4 mm, L = 140 cm, and v = 26 m / s, what is the potential difference, V V A V B ?...
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This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.
 Spring '08
 Opyrchal
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