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121homework 11

# 121homework 11 - Felder Jacob Homework 12 Due Dec 5 2006...

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Felder, Jacob – Homework 12 – Due: Dec 5 2006, 9:00 pm – Inst: Vitaly 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points A 38 mA current is carried by a uniformly wound air-core solenoid with 426 turns as shown in the figure below. The permeability of free space is 1 . 25664 × 10 - 6 N / A 2 . 38 mA 8 . 81 mm 8 . 02 cm Compute the magnetic field inside the solenoid. Correct answer: 0 . 000253646 T. Explanation: Let : N = 426 , = 8 . 02 cm , I = 38 mA , and μ 0 = 1 . 25664 × 10 - 6 N / A 2 . I d The magnetic field inside the solenoid is B = μ 0 n I = μ 0 N I = (1 . 25664 × 10 - 6 N / A 2 ) 426 0 . 0802 m × (0 . 038 A) = 0 . 000253646 T . 002 (part 2 of 3) 10 points Compute the magnetic flux through each turn. Correct answer: 1 . 54622 × 10 - 8 Wb. Explanation: Let : B = 0 . 000253646 T , and d = 8 . 81 mm = 0 . 00881 m . The magnetic flux through each turn is Φ = B A = B π 4 d 2 · = (0 . 000253646 T) π 4 (0 . 00881 m) 2 = 1 . 54622 × 10 - 8 Wb . 003 (part 3 of 3) 10 points Compute the inductance of the solenoid. Correct answer: 0 . 173339 mH. Explanation: The inductance of the solenoid is L = N Φ I = (426) (1 . 54622 × 10 - 8 Wb) 0 . 038 A = 0 . 173339 mH . keywords: 004 (part 1 of 5) 10 points A circuit is set up as shown in the figure. L R 1 R 2 E S I 1 I 2 I

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Felder, Jacob – Homework 12 – Due: Dec 5 2006, 9:00 pm – Inst: Vitaly 2 The switch is closed at t = 0. The current I through the inductor takes the form I = E R x 1 - e - t/τ x · where R x and τ x are to be determined. Find I immediately after the circuit is closed. 1. I = E R 1 + R 2 2. I = E R 1 3. I = 0 correct 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 ² 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 . The equivalent resistance is R eq = R 1 + R 2 so, from E = R eq I we find I 2 = E R 1 + R 2 005 (part 2 of 5) 10 points Find I 2 immediately after the circuit is closed.
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121homework 11 - Felder Jacob Homework 12 Due Dec 5 2006...

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