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PS7 Solutions - Physics 217 Problem Set 7 Solutions by...

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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 Lenz’s 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 it’s 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. Self-inductance is always positive, in accordance with Lenz’s law. 2 Problem 2: Purcell 7.13 This is a simple LR circuit, and the time constant τ = L/R . The current takes the form of exponential I ( t ) = I 0 (1 - e - t/τ ) where I 0 is the terminal current, and is simply the voltage divided by the total resistance in the circuit—when it all settles down, I no longer changes, and the inductor functions simply as a wire. More mathematically, it is the solution to the first-order, linear o.d.e. in I E 0 = L dI dt + RI with appropriate initial condition. 1
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The time at which it reaches 90% of its final value is when e - t/τ = 10% = t = - τ ln(1 / 10) = τ ln(10) = 2 . 30 τ Given that L = 0 . 50 millihenry, and R = 0 . 01 ohm, so τ = 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 (0 . 9) 2 = 1 2 (0 . 0005) 12 0 . 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 0 ) = Z t 0 0 I ( t ) 2 R dt = Z t 0 0 I 2 0 R 1 - e - t/τ · 2 dt = I 2 0 R t - 2 e - t/τ - 1 + e - 2 t/τ - 2 t 0 0 = E 2 R t 0 + 2 τ ( e - t 0 - 1) - 1 2 τ ( e - 2 t 0 - 1) Remember t 0 = τ ln(10) is the time that e - t 0 = 0 . 1, so we have Q = E 2 R τ ln(10) - 2 × 0 . 9 τ + 1 2 0 . 99 τ = 12 2 0 . 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 0 0 E I ( t ) dt = E 2 R Z t 0 0 1 - e - t/τ · dt = E 2 R t - e - t/τ - 1 t 0 0 = E 2 R h t 0 + τ ( e - t 0 - 1) i = 12 2 0 . 01 watt (0 . 05 second)(2 . 30 - 0 . 9) = 1008 joules 3 Problem 3: Purcell 7.14 Call the horizontal direction the ˆ x direction, with the right being positive. Assume the magnetic field is pointing into the page.
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