HW8 - Version One Homework 8 Savrasov 39819 May 15, 2006 1...

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Unformatted text preview: Version One Homework 8 Savrasov 39819 May 15, 2006 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Magnetic Field from an Arc 01 30:01, calculus, multiple choice, > 1 min, wording-variable. 001 (part 1 of 2) 1 points Consider two radial legs extending to infin- ity and a circular arc carrying a current I as shown below. x y I I I I 20 23 O r What is the magnitude of the magnetic field B at the origin O due to the current through this path? 1. B = 5 23 I r correct 2. B = 5 23 I r 2 3. B = 5 23 I r 4. B = 5 23 I r 2 5. B = 5 23 I 6. B = 2 23 I r 7. B = 2 23 I r 2 8. B = 2 23 I r 9. B = 2 23 I r 2 10. B = 2 23 I Explanation: Using the Biot-Savart law, the magnetic field due to the two radial legs is zero since d s r = 0 . However, around the arc | d s r | = ds = r d . The magnetic field at the center of an arc with a current I is B = I 4 d s r r 2 = I 4 r 2 ds = I 4 r 2 r d = I 4 r 20 23 d = I 4 r 20 23 = I 4 r 20 23 - = 5 23 I r . 002 (part 2 of 2) 1 points What is the direction of the magnetic field B at point O due to the current through the path? 1. B is out of the page. correct 2. B is into the page. 3. B is to the right. 4. B is to the left. 5. B is down the page. 6. B is up the page. 7. B is zero. Explanation: The right-hand rule dictates that magnetic field is out of the page. Version One Homework 8 Savrasov 39819 May 15, 2006 2 Square Loop of Wire 01 30:01, calculus, multiple choice, > 1 min, wording-variable. 003 (part 1 of 3) 1 points A conductor in the shape of a square of edge length carries a counter-clockwise current I as shown in the figure below. I P What is the magnitude of the magnetic field at point P (at the center of the square loop) due to the current in only one of the sides of the wire? 1. B = I 2 2 correct 2. B = I 3 2 3. B = I 2 4. B = I 2 5. B = I 2 6. B = I 2 2 7. B = I 2 2 8. B = 2 I 9. B = I 2 3 10. B = I 2 Explanation: By the Biot-Savart law, dB = 4 I d s r r 2 ....
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This note was uploaded on 11/03/2010 for the course PHYSICS 303 taught by Professor Shih during the Spring '10 term at University of Texas at Austin.

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HW8 - Version One Homework 8 Savrasov 39819 May 15, 2006 1...

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