# 18-4 - brown(twb493 Ch18-h4 chiu(56565 This print-out...

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brown (twb493) – Ch18-h4 – chiu – (56565) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of6)5.0points Conventional current flows through the ring shown in the figure below in such a way that if you stand at location A on the + x axis and look towards the ring, current appears to be flowing clockwise. The ring lies in the yz plane, encircling the x axis. x z y A B C D E F What is the direction of the magnetic field due to the ring at A? 1. ˆ B A = ( 0 , 1 , 0 ) 2. ˆ B A = ( 0 , 0 , 1 ) 3. ˆ B A = ( 0 , 0 , 1 ) 4. ˆ B A = ( 0 , 1 , 0 ) 5. ˆ B A = (− 1 , 0 , 0 ) correct 6. ˆ B A = ( 1 , 0 , 0 ) Explanation: Given the direction in which the current is traveling around the ring, we can use the right hand rule to find that the magnetic field must point along the x axis. Therefore the correct choice for point A is ˆ B A = (− 1 , 0 , 0 ) . 002(part2of6)5.0points At B? 1. ˆ B B = ( 1 , 0 , 0 ) correct 2. ˆ B B = ( 0 , 1 , 0 ) 3. ˆ B B = ( 0 , 1 , 0 ) 4. ˆ B B = ( 0 , 0 , 1 ) 5. ˆ B B = (− 1 , 0 , 0 ) 6. ˆ B B = ( 0 , 0 , 1 ) Explanation: At any point in the plane of the ring and inside the ring, the magnetic field is pointing through the ring along the x axis. At any point in the plane of the ring but outside of the ring, like B, D, E, and F, the magnetic field lines will be wrapping back around toward the positive x axis to travel back through the ring again. So at these points the correct choice for the direction of the magnetic field is ˆ B = ( 1 , 0 , 0 ) . 003(part3of6)5.0points At C? 1. ˆ B C = ( 0 , 0 , 1 ) 2. ˆ B C = ( 0 , 1 , 0 ) 3. ˆ B C = ( 0 , 1 , 0 ) 4. ˆ B C = ( 1 , 0 , 0 ) 5. ˆ B C = ( 0 , 0 , 1 ) 6. ˆ B C = (− 1 , 0 , 0 ) correct Explanation: See the explanation for part 1. 004(part4of6)5.0points At D? 1. ˆ B D = ( 0 , 1 , 0 ) 2. ˆ B D = ( 0 , 1 , 0 ) 3. ˆ B D = ( 0 , 0 , 1 )

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brown (twb493) – Ch18-h4 – chiu – (56565) 2 4. ˆ B D = ( 1 , 0 , 0 ) correct 5. ˆ B D = (− 1 , 0 , 0 ) 6. ˆ B D = ( 0 , 0 , 1 ) Explanation: See the explanation for part 2. 005(part5of6)5.0points At E? 1. ˆ B E = ( 0 , 0 , 1 ) 2. ˆ B E = ( 0 , 1 , 0 ) 3. ˆ B E = ( 1 , 0 , 0 ) correct 4. ˆ B E = ( 0 , 1 , 0 ) 5. ˆ B E = (− 1 , 0 , 0 ) 6. ˆ B E = ( 0 , 0 , 1 ) Explanation: See the explanation for part 2. 006(part6of6)5.0points At F? 1. ˆ B F = ( 0 , 0 , 1 ) 2. ˆ B F = ( 0 , 0 , 1 ) 3. ˆ B F = ( 1 , 0 , 0 ) correct 4. ˆ B F = ( 0 , 1 , 0 ) 5. ˆ B F = ( 0 , 1 , 0 ) 6. ˆ B F = (− 1 , 0 , 0 ) Explanation: See the explanation for part 2. 007(part1of2)10.0points A loop of wire carries a conventional current of 0 . 9 A. The radius of the loop is 0 . 08 m. Calculate the magnitude of the magnetic field at a distance of 0 . 32 m from the center of the loop, along the axis of the loop. Use μ 0 4 π = 1 × 10 7 T · m / A . Correct answer: 1 . 00846 × 10 7 T. Explanation: Here we use the formula for the magnetic field due to the loop of current: B = μ 0 4 π 2 I A ( r 2 + R 2 ) 3 / 2 , where A is the area of the loop, R is the radius of the loop, and r is the distance of the observation point from the center of the loop. We have B = 2(1 × 10 7 T · m / A)(0 . 9 A) π (0 . 08 m) 2 ((0 . 32 m) 2 + (0 . 08 m) 2 ) 3 / 2 = 1 . 00846 × 10 7 T . 008(part2of2)10.0points What would the magnitude of the magnetic field be at the same location if there were 250 loops of wire in a coil instead of one loop?
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