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hw8sol - Flores Michael Homework 8 Due 4:00 am Inst Sonia...

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Flores, Michael – Homework 8 – Due: Mar 23 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points A current I = 9 . 84 Amps flows through a wire perpendicular to the paper and towards the reader at A and back in the opposite direction at C . Consider the wires below the plane at A and C to be semi-infinite. In the figure, L 1 = 3 m, R = 2 . 2 m, and L 2 = 2 m and there is an external uniform magnetic field B = 7 . 4 T, which is directed into the paper. Being ’external’, this field is NOT 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. L 1 I I L 2 C A R O B What is the magnitude of the magnetic field at the center of the arc O due to the current in the wire? Correct answer: 7 . 64401 × 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 d~s × ˆ r = 0 on these segments. The contribution from the curved part of the wire is easy to find using the Biot-Savart law B = Z μ 0 I d~s × ˆ r 4 π r 2 = μ 0 I π 2 · 4 π R = μ 0 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 0 I 4 π ( L 1 + R ) ˆ ı , while for the wire below we have B 2 = μ 0 I 4 π ( L 2 + R ) ˆ  . These three must be added as vectors, giving the magnitude of the field | B O | = " μ 0 I 8 R 2 + μ 0 I 4 π ( R + L 1 ) 2 + μ 0 I 4 π ( R + L 2 ) 2 # 1 / 2 = £ (1 . 89231 × 10 - 7 T) 2 +(2 . 34286 × 10 - 7 T) 2 +(7 . 02574 × 10 - 7 T) 2 / 1 2 = 7 . 64401 × 10 - 7 T . 002 (part 1 of 2) 10 points The loop in the figure carries a current 3 . 72 A. The semicircular portion has a radius 3 . 23 cm. A 3 . 23 cm 3 . 23 cm 2 × 3 . 23 cm 3 . 72 A Determine the magnitude of the magnetic field at A. Correct answer: 6 . 87569 × 10 - 5 T. Explanation: Let : I = 3 . 72 A , R = 3 . 23 cm
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Flores, Michael – Homework 8 – Due: Mar 23 2004, 4:00 am – Inst: Sonia Paban 2 A R R 2R I Basic Concepts: The Biot-Savart Law is given by d ~ B = μ 0 4 π I d ~ × ˆ r r 2 . (1) Solution: Note: The distance from a current element on a circle to the center is a constant, namely r , so we can pull this out of the inte- gral. Also, the current element, I d ~ l , is always perpendicular to ˆ r , so sin θ = 1. Hence, B full circle = μ 0 4 π I r 2 Z d‘ = μ 0 4 π I r 2 2 π r = μ 0 I 2 r , where a half-circular arc is one-half this value. Assume: Out of the page to be positive. The angle θ (between the vector ~ and the vector ~r ) is from the Biot-Savart Law, Eq.
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