# HW 24 - terry (ect328) – homework 24 – Turner –...

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Unformatted text preview: 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 – 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 B t 2. E = a B e- at 3. E = a A B e- at correct 4. E = a A B e- 2 at 5. E = a A B 6. E = A B e- at Explanation: Basic Concepts: Faraday’s Law: E ≡ contintegraldisplay E · ds =- d Φ B dt Solution: Since B is perpendicular to the plane of the loop, the magnetic flux through the loop at time t > 0 is Φ B = B A = A B e- a t Also, since the coefficient AB and the pa- rameter a are constants, and Faraday’s Law says E =- d Φ B dt the induced emf can be calculated the from Equation above: E =- d Φ 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 Φ B = 3 a t 3- b t 2 , with a = 4 . 1 s- 3 · m 2 · T, and b = 7 . 5 s- 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 =- 3 s to t 2 = 2 s. Correct answer: 0 . 8967 A. Explanation: From Faraday’s law, the induced emf should be E =- d Φ 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 . terry (ect328) – homework 24 – Turner – (59130) 2 Thus the maximum current is I max = E max R = b 2 9 a R = (7 . 5 s- 2 · m 2 · T) 2 9 (4 . 1 s- 3 · m 2 · T) (1 . 7 Ω) = . 8967 A ....
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## This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas.

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HW 24 - terry (ect328) – homework 24 – Turner –...

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