PHY303LHW8S

# PHY303LHW8S - rosen (arr956) Homework 08 Chiu (58295) 1...

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Unformatted text preview: rosen (arr956) Homework 08 Chiu (58295) 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 10.0 points A 3 m long wire weighing 0 . 087 N / m is sus- pended directly above an infinitely straight wire. The top wire carries a current of 14 A and the bottom wire carries a current of 93 A . The permeablity of free space is 1 . 25664 10 6 N / A 2 . Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. Correct answer: 2 . 9931 mm. Explanation: Let : I 1 = 14 A , I 2 = 93 A , f = 0 . 087 N / m , and = 1 . 25664 10 6 N / A 2 . In order for the system to be in equilibrium, the magnetic force per unit length F on the top wire must be equal to its weight per unit length. Thus F = F L = I 1 I 2 2 d so d = I 1 I 2 2 F = (1 . 25664 10 6 N / A 2 ) (14 A) (93 A) 2 (0 . 087 N / m) = 0 . 0029931 m = 2 . 9931 mm . 002 10.0 points A rectangular loop with dimensions (hori- zontal = 0 . 4 m) (vertical= 0 . 92 m), is sus- pended by a string, and the lower horizontal section of the loop is immersed in a magnetic field. If a current of 5 . 1 A is maintained in the loop, what is the magnitude of the magnetic field required to produce a tension of 0 . 054 N in the supporting string? Assume: Gravita- tional force is negligible. Correct answer: 0 . 0264706 T. Explanation: Let : F = 0 . 054 N , I = 5 . 1 A , and l hor = 0 . 4 m . The magnitude of the force on the lower leg is | vector F | = | I vector hor vector B | . Therefore, the magnitude of the magnetic field due to this force is B = F I hor = . 054 N (5 . 1 A) (0 . 4 m) = 0 . 0264706 T . 003 (part 1 of 2) 10.0 points Consider a current in the long, straight wire which lies in the plane of the rectangular loop, that also carries a current, as shown. rosen (arr956) Homework 08 Chiu (58295) 2 16 cm 6 . 5 cm 72cm 5 . 2A 12A x y Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Correct answer: 1 . 6224 10 5 N. Explanation: Let : c = 0 . 16 m , a = 0 . 065 m , = 0 . 72 m , I 1 = 5 . 2 A , and I 2 = 12 A . c a I 1 I 2 x y The magnetic forces on the top and the bottom segments of the rectangle cancel. We now concern ourselves with the seg- ments of the rectangle which are parallel to the long wire. Using Amperes law, the mag- netic fields from the long wire at distances c and ( c + a ) away are B c = + I 1 2 c k B ca = + I 1 2 ( c + a ) k . The forces on the left and right vertical segments of the rectangle are, respectively, F c = I 2 B c and F ca = I 2 B ca . The forces are oppositely directed since the current I 2 has a direction in the loop at distances c op- posite from the current I 2 in the loop at a distance c + a , away from the long wire....
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## This note was uploaded on 04/13/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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PHY303LHW8S - rosen (arr956) Homework 08 Chiu (58295) 1...

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