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Unformatted text preview: Physics 217: Problem Set 7 Solutions by Jolyon Bloomfield and Yao Liu Due Monday, March 30th, 2009 1 Problem 1: Purcell 7.11 Looking at the figure, the current I 2 produces a ~ B field upward, so increasing I 2 will make ~ B 2 increase. By Lenzs law, the induced emf in the first coil should be in such a direction that oppose it, i.e., the ~ B field of the induced current should point down. If the first coil is as in the figure, the induced current should be opposite to the one indicated, so its a back emf E 1 = L 1 dI 1 dt M dI 2 dt The same analysis is applied to the induced emf of the second coil: increasing I 1 produces an increasing upward ~ B field, so in order to oppose it, the induced emf in the second coil is negative: E 2 = L 2 dI 2 dt M dI 1 dt If the two coils are connected as in (b), it is the same as setting I 1 = I 2 , and the total induced emf is the sum of the two E = E 1 + E 2 = ( L 1 + L 2 + 2 M ) dI 1 dt and if they are connected as in (c), I 1 = I 2 instead, and the total emf is the difference E = E 1 E 2 = ( L 1 + L 2 2 M ) dI 1 dt In particular, L 1 + L 2 > 2 M in this case, and as you can imagine it holds for any system of two coils. Selfinductance is always positive, in accordance with Lenzs law. 2 Problem 2: Purcell 7.13 This is a simple LR circuit, and the time constant = L/R . The current takes the form of exponential I ( t ) = I (1 e t/ ) where I is the terminal current, and is simply the voltage divided by the total resistance in the circuitwhen it all settles down, I no longer changes, and the inductor functions simply as a wire. More mathematically, it is the solution to the firstorder, linear o.d.e. in I E = L dI dt + RI with appropriate initial condition. 1 The time at which it reaches 90% of its final value is when e t/ = 10% = t = ln(1 / 10) = ln(10) = 2 . 30 Given that L = 0 . 50 millihenry, and R = 0 . 01 ohm, so = L/R = 50 milliseconds. The time t = 115 milliseconds = 0.115 seconds. The energy stored in the inductor is given by U = 1 2 LI 2 = 1 2 LI 2 (0 . 9) 2 = 1 2 (0 . 0005) 12 . 01 2 (0 . 9) 2 = 292 joules which is the energy transfered from the battery to the inductor. The energy lost to joule heat is Q ( t ) = Z t I ( t ) 2 Rdt = Z t I 2 R 1 e t/ 2 dt = I 2 R t 2 e t/ 1 / + e 2 t/ 2 / t = E 2 R t + 2 ( e t / 1) 1 2 ( e 2 t / 1) Remember t = ln(10) is the time that e t / = 0 . 1, so we have Q = E 2 R ln(10) 2 . 9 + 1 2 . 99 = 12 2 . 01 watt (0 . 05 second)(2 . 30 1 . 8 + 0 . 495) = 716 joules The total energy left the battery is 1008 joules, as one can calculate directly W = Z t E I ( t ) dt = E 2 R Z t 1 e t/ dt = E 2...
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This note was uploaded on 06/10/2011 for the course PHYS 2217 taught by Professor Leclair, a during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 LECLAIR, A
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