13 - roofner (bar784) Homework 13 Weathers (22202) 1 This...

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Unformatted text preview: roofner (bar784) Homework 13 Weathers (22202) 1 This print-out should have 11 questions. Multiple-choice 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 one-quarter 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. B BD = in, B DE = out, B ED = out 2. I, II, B ED = in 3. IV, IV, B ED = out 4. B BD = in, B DE = out, B ED = in 5. B BD = out, B DE = out, B ED = out correct 6. B BD = out, B DE = out, B ED = in 7. B BD = in, B DE = in, B ED = in 8. B BD = out, B DE = in, B ED = in 9. B BD = in, B DE = in, B ED = out 10. B BD = out, B DE = in, B ED = out Explanation: The Biot-Savart 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 Biot-Savart 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 8 r 2. B = 2 I r 3. B = I 8 r correct 4. B = I 4 r 5. B = I 2 r 6. B = I 2 r 7. B = I r 8. B = I 4 r 9. B = I r 10. B = I 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 roofner (bar784) 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 . 063 m and I = 1 . 1 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: 6 . 23472 10 6 T. Explanation: Let : = 1 . 25664 10 6 N / A 2 , I = 1 . 1 A , and r = 0 . 063 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 Biot-Savart 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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13 - roofner (bar784) Homework 13 Weathers (22202) 1 This...

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