solutions 18 - Patel, Kinal – Homework 18 – Due: Apr 8...

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Unformatted text preview: Patel, Kinal – Homework 18 – Due: Apr 8 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 10 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 4) 10 points A cylindrical filament has radius 1 . 63 mm. Its resistance is 2 . 17 Ω and its length is 1 . 26 m. The filament carries a uniform, steady current of 0 . 806 A. The permeability of free space is 4 π × 10- 7 N / A 2 . Find the electric field inside the filament. Correct answer: 1 . 38811 N / C. Explanation: Let : ‘ = 1 . 26 m , R = 2 . 17 Ω , and I = 0 . 806 A . The electric field inside the filament has to point in the direction of the current; the field causes the current to flow. To find the field, first consider the voltage drop across the ends of the wire, V = I R . The electric field inside the filament is E = V ‘ = I R ‘ = (2 . 17 Ω)(0 . 806 A) 1 . 26 m = 1 . 38811 N / C . 002 (part 2 of 4) 10 points Find B at the surface of the filament. Correct answer: 9 . 88957 × 10- 5 T. Explanation: Let : r = 1 . 63 mm = 0 . 00163 m , and μ = 4 π × 10- 7 N / A 2 . Using Ampere’s Law I ~ B · ds = μ I , the magnetic field from a long straight wire is given by B = μ I 2 π r (considering a circular loop around the wire). B = μ I 2 π r = (4 π × 10- 7 N / A 2 )(0 . 806 A) 2 π (0 . 00163 m) = 9 . 88957 × 10- 5 T ....
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This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.

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solutions 18 - Patel, Kinal – Homework 18 – Due: Apr 8...

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