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Unformatted text preview: toupal (rgt374) 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 0 . 9 m long wire weighing 0 . 072 N / m is suspended directly above an infinitely straight wire. The top wire carries a current of 47 A and the bottom wire carries a current of 82 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: 10 . 7056 mm. Explanation: Let : I 1 = 47 A , I 2 = 82 A , f = 0 . 072 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 ) (47 A) (82 A) 2 (0 . 072 N / m) = 0 . 0107056 m = 10 . 7056 mm . 002 10.0 points A rectangular loop with dimensions (hor- izontal = 0 . 11 m) (vertical= 0 . 253 m), is suspended by a string, and the lower hori- zontal section of the loop is immersed in a magnetic field. If a current of 6 A is maintained in the loop, what is the magnitude of the magnetic field required to produce a tension of 0 . 035 N in the supporting string? Assume: Gravitational force is negligible. Correct answer: 0 . 0530303 T. Explanation: Let : F = 0 . 035 N , I = 6 A , and l hor = 0 . 11 m . The magnitude of the force on the lower leg is | ( F | = | I ( ' hor ( B | . Therefore, the magnitude of the magnetic field due to this force is B = F I ' hor = . 035 N (6 A) (0 . 11 m) = 0 . 0530303 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. toupal (rgt374) Homework 08 Chiu (58295) 2 11 cm 13 cm 66cm 4 . 3A 15A x y Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Correct answer: 4 . 1925 10- 5 N. Explanation: Let : c = 0 . 11 m , a = 0 . 13 m , ' = 0 . 66 m , I 1 = 4 . 3 A , and I 2 = 15 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 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
- Spring '10