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HW24Sol - Platt David Homework 24 Due Nov 4 2005 4:00 am...

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Platt, David – Homework 24 – Due: Nov 4 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 A long cylindrical wire of radius 18 cm carries a current of 84 A uniformly distributed over its cross-sectional area. The permeability of free space is 4 π × 10 - 7 T · m / A. Find the magnetic energy per unit length within the wire. Correct answer: 0 . 0001764 J / m. Explanation: r a dr Let : a = 18 cm , I = 84 A , and μ 0 = 4 π × 10 - 7 T · m / A . Use Ampere’s law, the magnetic field inside the wire at a distance r < a from its center is B = μ 0 2 π r I C = μ 0 2 π r πr 2 πa 2 I = μ 0 r I 2 πa 2 . The magnetic energy within the cylindrical annulus is d U m = B 2 2 μ 0 V annulus = B 2 2 μ 0 2 π r ‘ dr = B 2 μ 0 π r ‘ dr = μ 0 r I 2 π a 2 2 π r ‘ dr μ 0 = μ 0 4 π I 2 r 3 ‘ dr a 4 . Thus the magnetic energy per unit length within the wire is U m = Z a 0 d U m = μ 0 4 π I 2 a 4 Z a 0 r 3 dr = μ 0 4 π I 2 a 4 a 4 4 = μ 0 4 π I 2 4 = (1 × 10 - 7 T · m / A) (84 A) 2 4 = 0 . 0001764 J / m .
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