Fall 2011 Homework 18 Solutions

Fall 2011 Homework 18 Solutions - Physics 21 Fall, 2011...

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Unformatted text preview: Physics 21 Fall, 2011 Solution to HW-18 30-7 At the instant when the current in an inductor is in- creasing at a rate of 6 . 45 10 2 A / s, the magnitude of the self-induced emf is 1 . 65 10 2 V. (a) What is the induc- tance of the inductor? (b) If the inductor is a solenoid with 405 turns, what is the average magnetic flux through each turn when the current is 0 . 715 A? (a) In the class lecture we applied Faradays Law to a solenoid with N turns. The self-induced emf is E =- N d B dt , and we defined the self-inductance L by E =- L di dt Here we arent concerned with the direction of the emf, only the magnitude. Therefore we can substitute the given values to obtain L = E di/dt = 1 . 65 10 2 V 6 . 45 10 2 A / s = 0 . 256 H (b) For this part we must notice that since the flux through one loop of a solenoid of length l is given by B = BA = N l iA, we can rewrite the expression for the inductance L given on the equation sheet in terms of the flux, L = N 2 l A = N N l A = N B i . Note that the relation above, L = N B i , is in fact the definition used in the book for L [Eq. (30.6)]. We can use it to answer this problem by solving for B . B = Li N = (0 . 256 H) (0 . 715 A) 405 = 4 . 52 10 4 Wb 30-19 An inductor with an inductance of 2.50 H and a resistance of 8 . 00 is connected to the terminals of a battery with an emf of 6.00 V and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is 0.500 A; (c) the current 0.250 s after the circuit is closed; (d) the final steady-state current.(d) the final steady-state current....
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This note was uploaded on 12/18/2011 for the course PHYS 021 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Fall 2011 Homework 18 Solutions - Physics 21 Fall, 2011...

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