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Unformatted text preview: aldawsari (ma28683) – HW15 – Betancourt – (16856) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 30.130.6 001 (part 1 of 3) 10.0 points A 23 . 2 mA current is carried by a uniformly wound aircore solenoid with 669 turns as shown in the figure below. The permeability of free space is 1 . 25664 × 10 6 N / A 2 . 2 3 . 2 m A 8 . 85 mm 6 . 63 cm Compute the magnetic field inside the solenoid. Correct answer: 0 . 000294178 T. Explanation: Let : N = 669 , ℓ = 6 . 63 cm , I = 23 . 2 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 669 . 0663 m parenrightbigg × (0 . 0232 A) = . 000294178 T . 002 (part 2 of 3) 10.0 points Compute the magnetic flux through each turn. Correct answer: 1 . 80962 × 10 8 Wb. Explanation: Let : B = 0 . 000294178 T , and d = 8 . 85 mm = 0 . 00885 m . The magnetic flux through each turn is Φ = B A = B parenleftBig π 4 d 2 parenrightBig = (0 . 000294178 T) π 4 (0 . 00885 m) 2 = 1 . 80962 × 10 8 Wb . 003 (part 3 of 3) 10.0 points Compute the inductance of the solenoid. Correct answer: 0 . 521825 mH. Explanation: The inductance of the solenoid is L = N Φ I = (669) (1 . 80962 × 10 8 Wb) . 0232 A = . 521825 mH . 004 (part 1 of 2) 10.0 points A solenoid has 135 turns of wire uniformly wrapped around an airfilled core, which has a diameter of 13 mm and a length of 6 . 2 cm. The permeability of free space is 1 . 25664 × 10 6 N / A 2 . Calculate the selfinductance of the solenoid....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue.
 Spring '10
 Gavrine
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