This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 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 = 1 . 25664 10- 6 N / A 2 . Correct answer: 180 . 96 A. Explanation: Solution: This problem can be solved with Amperes law, I ~ B d~s = I Choose the integration path to be a loop around the surface. This gives us B ( d ) = I where d is the diameter. Or, I = B d = (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 N. Explanation: The central wire creates field ~ B = I 1 2 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 I 1 2 R = I 1 I 2 L R to the right k ~ F k = (59 . 6 A)(25 . 2 A)(0 . 332 m) (12 . 6 cm) = 1582 . 98 N . 003 (part 1 of 2) 10 points A superconducting solenoid is to generate a magneticfieldof4 . 51T. Thesolenoidwinding 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 = N I l = nI , then I = B l N Husain, Zeena Homework 9 Due: Mar 29 2004, 4:00 am Inst: Sonia Paban 2 = (4 . 51 T) (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 ....
View Full Document
This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
- Spring '08