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11homework09

# 11homework09 - Husain Zeena Homework 9 Due 4:00 am Inst...

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Husain, Zeena – Homework 9 – Due: Mar 29 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 21 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 The recent discovery of “high temperature” superconductors with negligible resistance has raised many possibilities for technological applications (magnetically levitated trains, zero-dissipation power transmission lines, in- credibly strong refrigerator magnets, etc.) However, these materials lose their supercon- ductivity if they are subjected to excessive magnetic fields. This limits the amount of current that a superconductor can carry, since the current results in a magnetic field at the surface of the wire. Calculate this “critical current” for a d = 2 . 08 mm diameter superconducting wire if the superconductivity is destroyed when the surface magnetic field exceeds B = 0 . 0348 T. Given μ 0 = 1 . 25664 × 10 - 6 N / A 2 . Correct answer: 180 . 96 A. Explanation: Solution: This problem can be solved with Ampere’s law, I ~ B · d~s = μ 0 I Choose the integration path to be a loop around the surface. This gives us B ( π d ) = μ 0 I where d is the diameter. Or, I = π B d μ 0 = π (0 . 0348 T) (0 . 00208 m) 1 . 25664 × 10 - 6 N / A 2 = 180 . 96 A 002 (part 1 of 1) 10 points An infinitely long straight wire carrying a current I 1 = 59 . 6 A is partially surrounded by a loop as in figure. The loop has a length L = 33 . 2 cm, a radius R = 12 . 6 cm, and carries a current I 2 = 25 . 2 A. The axis of the loop coincides with the wire. R L I 1 I 2 Calculate the force exerted on the loop. Correct answer: 1582 . 98 μ N. Explanation: The central wire creates field ~ B = μ 0 I 1 2 π R counterclockwise . The curved portions of the loop feels zero force since ~ l × ~ B = 0 there. The straight portions both feel I 2 ~ l × ~ B forces to the right, amounting to ~ F = I 2 2 L μ 0 I 1 2 π R = μ 0 I 1 I 2 L π R to the right k ~ F k = μ 0 (59 . 6 A) (25 . 2 A) (0 . 332 m) π (12 . 6 cm) = 1582 . 98 μ N . 003 (part 1 of 2) 10 points A superconducting solenoid is to generate a magnetic field of 4 . 51 T. The solenoid winding has 2290 turns per meter. What is the required current? Correct answer: 1567 . 22 A. Explanation: Magnetic field at the center of solenoid is B = μ 0 N I l = μ 0 n I , then I = B l μ 0 N

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Husain, Zeena – Homework 9 – Due: Mar 29 2004, 4:00 am – Inst: Sonia Paban 2 = (4 . 51 T) μ 0 (2290) = 1567 . 22 A . 004 (part 2 of 2) 0 points What force per unit length is exerted on the windings by this magnetic field? Correct answer: 7068 . 18 N / m. Explanation: Force per unit length is F l = I B = (1567 . 22 A) (4 . 51 T) = (7068 . 18 N / m) radially outward . Although the force per unit length is non- zero (in an outward direction) the net force per unit length will be zero. 005 (part 1 of 1) 10 points A sinusoidal voltage is applied directly across a 11 . 4 μ F capacitor. The frequency of the source is 4 . 81 kHz, and the voltage amplitude is 51 . 2 V.
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11homework09 - Husain Zeena Homework 9 Due 4:00 am Inst...

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