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Fall 2011 Homework 18 Solutions

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

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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 . 715A? (a) In the class lecture we applied Faraday’s Law to a solenoid with N turns. The self-induced emf is E = - N d Φ B dt , and we defined the self-inductance L by E = - L di dt Here we aren’t 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 . 256H (b) For this part we must notice that since the flux through one loop of a solenoid of length l is given by Φ B = BA = μ 0 N l iA, we can rewrite the expression for the inductance L given on the equation sheet in terms of the flux, L = μ 0 N 2 l A = 0 N l A = N Φ B i . Note that the relation above, L = N Φ 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 Φ B . Φ B = Li N = (0 . 256H)(0 . 715A) 405 = 4 . 52 × 10 4 Wb 30-19 An inductor with an inductance of 2.50 H and a resistance of 8 . 00Ω 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.
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