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Unformatted text preview: Kapoor (mk9499) – homework21 – Turner – (60230) 1 This print-out should have 12 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 A current I = 5 A flows through a wire perpendicular to the paper and towards the reader at A and back in the opposite direc- tion at C . Consider the wires below the plane at A and C to be semi-infinite. In the figure, L 1 = 4 m, R = 8 m, and L 2 = 8 m and there is a B = 6 . 82 T magnetic field into the paper (not including the field due to the current in the wire). Caution: 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. 4m 5A 5 A 8 m 5 A 8 m A C O ⊗ B = 6 . 82 T What is the magnitude of the force on the wire due to the external magnetic field B ? Correct answer: 682 N. Explanation: Let : R = 8 m , I = 5 A , L 1 = 4 m , L 2 = 8 m , and B = 6 . 82 T . 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. The force on a current carrying wire from point A, at vectorr 1 , to point C, at vector r 2 , in a uniform field is vector F = I integraldisplay vectorr 2 vectorr 1 ( dvectors × vector B ) . Since vector B is a constant, it can be taken out of the integral and we can write (recalling that if we change the order of the cross product, we need to change the overall sign) vector F = − I vector B × integraldisplay vectorr 2 vectorr 1 dvectors = − I vector B × ( vectorr 2 − vectorr 1 ) . Now vectorr 2 = ( R + L 2 ) ˆ i vectorr 1 = ( R + L 1 ) ˆ j vector B = B ˆ k . Thus, the total force is vector F = − I B [( R + L 2 ) ( ˆ k × ˆ i ) − ( R + L 1 ) ( ˆ k × ˆ j )] = − I B [( R + L 2 ) j + ( R + L 1 ) i ] . Thus the magnitude of the force is F = I B radicalBig ( R + L 2 ) 2 + ( R + L 1 ) 2 = 682 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: 1 . 11135 × 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. Therefore dvectors × ˆ r = 0 on these segments. The Kapoor (mk9499) – homework21 – Turner – (60230) 2 contribution from the curved part of the wire is easy to find using the Biot-Savart law B = integraldisplay μ I dvectors × ˆ r 4 π r 2 = μ I parenleftBig π 2 parenrightBig 4 π R = μ I 8 R ( − ˆ k ) , but one must not forget to take into account the contribution from the long straight wire which runs into and out of the page. For the wire above this is B 1 = μ I 4 π ( L 1 + R ) ˆ ı , while for the wire below we have B 2 = μ I 4 π ( L 2 + R ) ˆ ....
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