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solution12_pdf

# solution12_pdf - maini(nm7637 – hw12 – Shneidman...

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Unformatted text preview: maini (nm7637) – hw12 – Shneidman – (12108) 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 62 . 1 mA current is carried by a uniformly wound air-core solenoid with 238 turns as shown in the figure below. The permeability of free space is 1 . 25664 × 10 − 6 N / A 2 . 6 2 . 1 m A 13 . 4 mm 8 . 63 cm Compute the magnetic field inside the solenoid. Correct answer: 0 . 000215213 T. Explanation: Let : N = 238 , ℓ = 8 . 63 cm , I = 62 . 1 mA , and μ = 1 . 25664 × 10 − 6 N / A 2 . I d ℓ The magnetic field inside the solenoid is B = μ n I = μ parenleftbigg N ℓ parenrightbigg I = (1 . 25664 × 10 − 6 N / A 2 ) parenleftbigg 238 . 0863 m parenrightbigg × (0 . 0621 A) = . 000215213 T . 002 (part 2 of 3) 10.0 points Compute the magnetic flux through each turn. Correct answer: 3 . 03506 × 10 − 8 Wb. Explanation: Let : B = 0 . 000215213 T , and d = 13 . 4 mm = 0 . 0134 m . The magnetic flux through each turn is Φ = B A = B parenleftBig π 4 d 2 parenrightBig = (0 . 000215213 T) π 4 (0 . 0134 m) 2 = 3 . 03506 × 10 − 8 Wb . 003 (part 3 of 3) 10.0 points Compute the inductance of the solenoid. Correct answer: 0 . 116319 mH. Explanation: The inductance of the solenoid is L = N Φ I = (238) (3 . 03506 × 10 − 8 Wb) . 0621 A = . 116319 mH . 004 (part 1 of 5) 10.0 points A circuit is set up as shown in the figure. L R 1 R 2 E S I 1 I 2 I maini (nm7637) – hw12 – Shneidman – (12108) 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/τ x parenrightBig where R x and τ x are to be determined. Find I immediately after the circuit is closed. 1. I = E R 1 2. I = E R 2 3. I = E R 1 + R 2 4. I = 0 correct 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 ǫ 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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solution12_pdf - maini(nm7637 – hw12 – Shneidman...

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