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Unformatted text preview: song (shs546) Homework 13 weathers (22202) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points An infinite wire is bent as shown in the figure. The current is I . It consists of 3 segments AB , BDE , and EF . AB is vertical with A extending to infinity. EF is horizontal with F extending to infinity. BDE is onequarter of a circular arc. We use the Roman numerals I, II, III and IV to denote the direction of a vector in the xy plane, which begins at the origin and is pointing into the quadrants I, II, III and IV respectively. B O D C E F A O y x I II III IV The direction of the magnetic vector at O due to the current segment BD , that due to DE and that due to EF are given by respectively (in, out mean into, or out ofthe xy plane) 1. I, II, B ED = in 2. B BD = in, B DE = out, B ED = out 3. B BD = in, B DE = in, B ED = in 4. IV, IV, B ED = out 5. B BD = out, B DE = in, B ED = out 6. B BD = out, B DE = out, B ED = in 7. B BD = in, B DE = in, B ED = out 8. B BD = in, B DE = out, B ED = in 9. B BD = out, B DE = out, B ED = out correct 10. B BD = out, B DE = in, B ED = in Explanation: The BiotSavart Law is given by d vector B = 4 I d vector l r r 2 . Since both d vector l and r lie in the xy plane, their cross product must be perpendicular to this plane. Apply the BiotSavart Law to each line element of the wire, d vector B is found always to direct out the xy plane at O . 002 (part 2 of 3) 10.0 points What is the magnitude of vector B at the center of the arc due to the arc BDE alone? 1. B = I 2 r 2. B = I r 3. B = I 8 r 4. B = I 4 r 5. B = I r 6. B = 2 I r 7. B = I 8 r correct 8. B = I 4 r 9. B = I r 10. B = I 2 r Explanation: Note: The distance r from a current el ement of BDE to O is a constant and the current element I d vector l is always perpendicular to r . Hence, the magnitude of the vector B at the center of the arc due to the arc BDC alone is song (shs546) Homework 13 weathers (22202) 2 B BDE = 4 I r 2 integraldisplay dl = 4 I r 2 2 r = I 8 r . 003 (part 3 of 3) 10.0 points Let r = 0 . 14 m and I = 0 . 29 A. The permeability of free space is 1 . 25664 10 6 N / A 2 . Find the magnitude of vector B at O due to the entire wire ABDEF ? Correct answer: 7 . 39665 10 7 T. Explanation: Let : = 1 . 25664 10 6 N / A 2 , I = 0 . 29 A , and r = 0 . 14 m . First calculate the magnetic field of a straight wire carrying a current(See figure be low). I r s y y O B From the BiotSavart Law, the contribution to the magnetic field at O due to a current element I y at y is given by B = 4 I y s 2 sin = I 4 r sin ....
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
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