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hw18-solutions - howard(cah3459 hw18 turner(56705 This...

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howard (cah3459) – hw18 – turner – (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Assume the wire visible in the plane is bent perpendicularly at A and at C, with the re- mainder of the wire behind the paper. A current of 5 A flows toward you at A , through the visible portion, and back in the opposite direction at C . Consider the wires below the plane at A and C to be semi-infinite. There is a magnetic field of strength 7 . 36 T into the pa- per (not including the field due to the current in the wire). 3 m 5 A 5 A 8 m 5 A 2 m A C O B = 7 . 36 T What is the magnitude of the force on the wire due to the external magnetic field B ? It may be necessary to take into account the contribution from the long straight wire which runs up to and down from the underneath side of the page. The permeability of free space is 1 . 25664 × 10 - 6 T · m / A. Correct answer: 411 . 437 N. Explanation: Let : R = 2 m , I = 5 A , L 1 = 3 m , L 2 = 8 m , and B = 7 . 36 T . The vectors from the origin to points A and B are r 1 = ( R + L 1 = (5 m)ˆ and r 2 = ( R + L 2 ı = (10 m)ˆ ı . By the Biot-Savart law, d B = μ 0 4 π I d s × ˆ r r 2 . The contribution from the long straight wire which runs into and out of the page is zero since the external field and the current are parallel. B = B ˆ k is a constant, so the force on the current-carrying wire from point A at r 1 = r 1 ˆ to point C at r 2 = r 2 ˆ ı in the uniform field is F = I r 2 r 1 ( d s × B ) = - I B × r 2 r 1 d s = - I B × ( r 2 - r 1 ) = - I B ˆ k × ( r 2 ˆ ı - r 1 ˆ ) = - I B r 2 ( ˆ k × ˆ ı ) - r 1 ( ˆ k × ˆ ) = - I B ( r 2 ˆ + r 1 ˆ ı ) , with a magnitude of F = I B ( r 2 ) 2 + ( r 1 ) 2 = (5 A)(7 . 36 T) (10 m) 2 + (5 m) 2 = 411 . 437 N . 002 (part 2 of 2) 10.0 points What is the magnitude of the magnetic field at the center of the arc O due to the current in the wire? Correct answer: 4 . 08305 × 10 - 7 T. Explanation: The two straight current segments within the plane of the paper do not contribute to the magnetic field at point O because they are parallel to the radius vector from that point, so d s × ˆ r = 0 on these segments. Applying
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howard (cah3459) – hw18 – turner – (56705) 2 the Biot-Savart law to the curved part of the wire, B = μ 0 I d s × ˆ r 4 π R 2 = μ 0 I π 2 4 π R ( - ˆ k ) = - μ 0 I 8 R ˆ k = - (1 . 25664 × 10 - 6 T · m / A)(5 A) 8 (2 m) ˆ k = - (3 . 92699 × 10 - 7 T) ˆ k . The contribution from the long straight wire which runs into the plane at A is B A = μ 0 I 4 π r 1 ˆ ı = (1 . 25664 × 10 - 6 T · m / A) (5 A) 4 π (5 m) ˆ ı = (1 × 10 - 7 T)ˆ ı , while for the wire running out of the plane at B is B B = μ 0 I 4 π r 2 ˆ = (1 . 25664 × 10 - 6 T · m / A) (5 A) 4 π (10 m) ˆ = (5 × 10 - 8 T)ˆ . Thus the magnitude of the field is | B O | = B 2 + B 2 A + B 2 B = ( - 3 . 92699 × 10 - 7 T) 2 + (1 × 10 - 7 T) 2 +(5 × 10 - 8 T) 2 1 / 2 = 4 . 08305 × 10 - 7 T . 003 10.0 points Consider two radial legs extending to infinity and a circular arc carrying a current I as shown below.
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