# 12 - kavo (ak22862) HW12 Janow (12121) 1 This print-out...

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Unformatted text preview: kavo (ak22862) HW12 Janow (12121) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A 59 mA current is carried by a uniformly wound air-core solenoid with 489 turns as shown in the figure below. The permeability of free space is 1 . 25664 10 6 N / A 2 . 5 9 m A 7 . 17 mm 18 . 4 cm Compute the magnetic field inside the solenoid. Correct answer: 0 . 000197039 T. Explanation: Let : N = 489 , = 18 . 4 cm , I = 59 mA , and = 1 . 25664 10 6 N / A 2 . I d The magnetic field inside the solenoid is B = n I = parenleftbigg N parenrightbigg I = (1 . 25664 10 6 N / A 2 ) parenleftbigg 489 . 184 m parenrightbigg (0 . 059 A) = . 000197039 T . 002 (part 2 of 3) 10.0 points Compute the magnetic flux through each turn. Correct answer: 7 . 95575 10 9 Wb. Explanation: Let : B = 0 . 000197039 T , and d = 7 . 17 mm = 0 . 00717 m . The magnetic flux through each turn is = B A = B parenleftBig 4 d 2 parenrightBig = (0 . 000197039 T) 4 (0 . 00717 m) 2 = 7 . 95575 10 9 Wb . 003 (part 3 of 3) 10.0 points Compute the inductance of the solenoid. Correct answer: 0 . 0659383 mH. Explanation: The inductance of the solenoid is L = N I = (489) (7 . 95575 10 9 Wb) . 059 A = . 0659383 mH . 004 (part 1 of 5) 15.0 points A circuit is set up as shown in the figure. L R 1 R 2 E S I 1 I 2 I kavo (ak22862) HW12 Janow (12121) 2 The switch is closed at t = 0. The current I through the inductor takes the form I = E R x parenleftBig 1- e t/ x parenrightBig where R x and x are to be determined. Find I immediately after the circuit is closed. 1. I = 0 correct 2. I = E R 1 3. I = E R 1 + R 2 4. I = E R 2 Explanation: Before the circuit is closed, no current is flowing. When we have just closed the circuit we are at t = 0 + , a mathematical nota- tion meaning a very short time after t = 0. (Nothing happens in the circuit at t = 0, only immediately after when the switch is, indeed, closed. However, this is just a mathematical detail.) There are two loops in the prob- lem, one with E , R 1 , R 2 and one with E , R 1 , L . So at t = 0 + , the battery wants to drive a current through both loops. The first loop presents no problem; since there is no in- ductance working against us, a current will immediately be set up. The second loop, how- ever, has an inductor which tries to prevent any change in the current going through it, and so goes up smoothly from I = 0, as can be seen in the given solution (just put t = 0 to find I = 0). Therefore, at this instant, the inductor L carries no current, and we can ne- glect it when we find the current through R 2 ....
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## 12 - kavo (ak22862) HW12 Janow (12121) 1 This print-out...

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