solution_pdf 18

# solution_pdf 18 - malsam(wgm329 – hw18 – turner...

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Unformatted text preview: malsam (wgm329) – 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 7 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 . 47 T into the pa- per (not including the field due to the current in the wire). 5m 7A 7 A 8 m 7 A 2 m A C O ⊗ B = 7 . 47 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: 638 . 281 N. Explanation: Let : R = 2 m , I = 7 A , L 1 = 5 m , L 2 = 8 m , and B = 7 . 47 T . The vectors from the origin to points A and B are vectorr 1 = ( R + L 1 )ˆ = (7 m)ˆ and vectorr 2 = ( R + L 2 )ˆ ı = (10 m)ˆ ı. By the Biot-Savart law, d vector B = μ 4 π I dvectors × 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. vector B = B ˆ k is a constant, so the force on the current-carrying wire from point A at vectorr 1 = r 1 ˆ to point C at vectorr 2 = r 2 ˆ ı in the uniform field is vector F = I integraldisplay vectorr 2 vectorr 1 ( dvectors × vector B ) = − I vector B × integraldisplay vectorr 2 vectorr 1 dvectors = − I vector B × ( vectorr 2 − vectorr 1 ) = − I B ˆ k × ( r 2 ˆ ı − r 1 ˆ ) = − I B bracketleftBig r 2 ( ˆ k × ˆ ı ) − r 1 ( ˆ k × ˆ ) bracketrightBig = − I B ( r 2 ˆ + r 1 ˆ ı ) , with a magnitude of F = I B radicalBig ( r 2 ) 2 + ( r 1 ) 2 = (7 A)(7 . 47 T) radicalBig (10 m) 2 + (7 m) 2 = 638 . 281 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: 5 . 63167 × 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 dvectors × ˆ r = 0 on these segments. Applying malsam (wgm329) – hw18 – turner – (56705) 2 the Biot-Savart law to the curved part of the wire, vector B = integraldisplay μ I dvectors × ˆ r 4 π R 2 = μ I parenleftBig π 2 parenrightBig 4 π R ( − ˆ k ) = − μ I 8 R ˆ k = − (1 . 25664 × 10 − 6 T · m / A)(7 A) 8 (2 m) ˆ k = − (5 . 49779 × 10 − 7 T) ˆ k ....
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## This note was uploaded on 11/22/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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solution_pdf 18 - malsam(wgm329 – hw18 – turner...

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