# HW20Sol - Platt David – Homework 20 – Due 4:00 am –...

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Unformatted text preview: Platt, David – Homework 20 – Due: Oct 26 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 5 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 1) 10 points An air-core solenoid has a length of 55 . 9 cm, a cross-sectionalareaof0 . 0885m 2 , andcontains 660 turns. The current through the solenoid increases by 9 . 55 A. The permeability of free space is 1 . 25664 × 10- 6 N / A 2 . By how much does the magnetic flux through the solenoid change? Correct answer: 1253 . 97 μ Wb. Explanation: Let : ‘ = 55 . 9 cm = 0 . 559 m , A = 0 . 0885 m 2 , N = 660 turns , I = 9 . 55 A , and μ = 1 . 25664 × 10- 6 N / A 2 . The magnetic field inside the solenoid is B = μ nI = (1 . 25664 × 10- 6 N / A 2 ) × µ 660 turns . 559 m ¶ (9 . 55 A) = 0 . 0141692 T . ΔΦ B = B A = (0 . 0141692 T)(0 . 0885 m 2 ) = 0 . 00125397 Wb = 1253 . 97 μ Wb . 002 (part 1 of 1) 10 points An open hemispherical surface of radius . 141 m is in a magnetic field of 0 . 121 T....
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HW20Sol - Platt David – Homework 20 – Due 4:00 am –...

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