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Unformatted text preview: maini (nm7637) – hw12 – Shneidman – (12108) 1 This printout should have 17 questions. Multiplechoice 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 aircore 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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 Spring '08
 moro
 Current, Inductance, Inductor, Electrical resistance, RL circuit

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