hw 19 - yang(ey942 – ohw19 – turner –(56705 This...

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yang (ey942) – ohw19 – turner – (56705) 1 This print-out should have 18 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 Consider two long parallel wires which are perpendicular to the plane of the paper ( i.e. , the x - y plane). Both wires carry the same current, I . Wire #1 intersects the plane a distance a above point O and wire #2 intersects the plane a distance a below point O . Point C is equidistant from both wires and is a distance a from point O . a a a a O C D wire #2 wire #1 45 45 x y I II III IV O What is the direction of the magnetic field at C ? 1. in the negative y direction 2. in quadrant III 3. into the plane 4. in the positive x direction 5. in quadrant IV 6. in quadrant II 7. in the negative x direction 8. in the positive y direction correct 9. out of the plane 10. in quadrant I Explanation: B 2 B 1 O C wire #2 wire #1 From this figure, we can see by symmetry that the y components of the magnetic fields cancel, leaving only the component in the positive x direction. 002 (part 2 of 2) 10.0 points What is the magnitude of the magnetic field at C ? 1. B C = 1 2 μ 0 I a 2. B C = 1 4 π μ 0 I a 3. B C = 1 2 π μ 0 I a 4. B C = 1 2 2 μ 0 I a 5. B C = 1 4 μ 0 I a 6. B C = 1 2 μ 0 I a 7. B C = 0 8. B C = 1 2 2 π μ 0 I a 9. B C = 1 π μ 0 I a 10. B C = 1 2 π μ 0 I a correct Explanation: Note: The distance r from each wire to C is r = a sin 45 = 2 a . The magnitude of the x component of the
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yang (ey942) – ohw19 – turner – (56705) 2 magnetic field due to each wire at C is given by B 1 = B 2 = μ 0 I 2 π 2 a cos 45 = μ 0 I 2 π 2 a 1 2 = μ 0 I 4 π a . Hence the contribution from both wires at C is just twice this value: B C = 2 B 1 = μ 0 I 2 π a . 003 10.0 points A circular loop of wire of radius R = 4 . 2 m lies in the xy -plane, centered about the origin. The loop is carrying a current of I = 5 . 88 A flowing in counterclockwise direction. Consider two l = 1 . 73 mm segments of the loop: one centered about the positive x -axis, the other centered about the positive y -axis. Hint: Use Biot-Savart law. The permeability of free space is 1 . 25664 × 10 6 Tm / A . R y x I r l =ds 1 I 2 1 1 l =ds 2 2 What is the magnitude of the force the first exerts on the second? Correct answer: 2 . 07397 × 10 13 N. Explanation: Let : μ 0 = 1 . 25664 × 10 6 Tm / A , l = 1 . 73 mm = 0 . 00173 m , I = 5 . 88 A , and R = 4 . 2 m . Since the radius, R , of the current loop is much bigger than the length, l , of both small segments, we can regard the segments as a small straight line. From the Biot-Savart law d vector B = μ 0 I 4 π dvectors × ˆ r r 2 . The field vector B produced by the first segment at the position of the second is vector B = μ 0 I 4 π integraldisplay dvectors × ˆ r r 2 = μ 0 I 4 π l ˆ y × ˆ r ( 2 R ) 2 = μ 0 I 4 π l sin 45 ˆ z 2 R 2 .
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