Phys121-12sol

Phys121-12sol - Anderson Matthew Homework 12 Due noon Inst...

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Anderson, Matthew – Homework 12 – Due: Apr 24 2008, noon – Inst: Vitaly 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 3) 15 points A 20 . 8 mA current is carried by a uniFormly wound air-core solenoid with 233 turns as shown in the fgure below. The permeability oF Free space is 1 . 25664 × 10 - 6 N / A 2 . 20 . 8 mA 23 . 2 mm 14 . 9 cm Compute the magnetic feld inside the solenoid. Correct answer: 4 . 08736 × 10 - 5 T. Explanation: Let : N = 233 , = 14 . 9 cm , I = 20 . 8 mA , and μ 0 = 1 . 25664 × 10 - 6 N / A 2 . I d The magnetic feld inside the solenoid is B = μ 0 n I = μ 0 µ N I = (1 . 25664 × 10 - 6 N / A 2 ) µ 233 0 . 149 m × (0 . 0208 A) = 4 . 08736 × 10 - 5 T . 002 (part 2 oF 3) 15 points Compute the magnetic ±ux through each turn. Correct answer: 1 . 72786 × 10 - 8 Wb. Explanation: Let : B = 4 . 08736 × 10 - 5 T , and d = 23 . 2 mm = 0 . 0232 m . The magnetic ±ux through each turn is Φ = B A = B π 4 d 2 · = (4 . 08736 × 10 - 5 T) π 4 (0 . 0232 m) 2 = 1 . 72786 × 10 - 8 Wb . 003 (part 3 oF 3) 15 points Compute the inductance oF the solenoid. Correct answer: 0 . 193554 mH. Explanation: The inductance oF the solenoid is L = N Φ I = (233) (1 . 72786 × 10 - 8 Wb) 0 . 0208 A = 0 . 193554 mH . keywords: 004 (part 1 oF 5) 15 points A circuit is set up as shown in the fgure. L R 1 R 2 E S I 1 I 2 I
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Anderson, Matthew – Homework 12 – Due: Apr 24 2008, noon – 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 2 3. I = 0 correct 4. I = E R 1 Explanation: Before the circuit is closed, no current is ±owing. 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 ²rst 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 ²nd I = 0). Therefore, at this instant, the inductor L carries no current, and we can ne- glect it when we ²nd the current through R 2 . The equivalent resistance is R eq = R 1 + R 2 so, from E = R eq I we ²nd I 2 = E R 1 + R 2 005 (part 2 of 5) 15 points Find I 2 immediately after the circuit is closed.
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Phys121-12sol - Anderson Matthew Homework 12 Due noon Inst...

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