# HW8 - Version One – Homework 8 – Savrasov – 39819 –...

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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 π 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 µ I r correct 2. B = 5 23 µ I r 2 3. B = 5 23 µ I π r 4. B = 5 23 µ I π r 2 5. B = 5 23 µ I π 6. B = 2 23 µ I r 7. B = 2 23 µ I r 2 8. B = 2 23 µ I π r 9. B = 2 23 µ I π r 2 10. B = 2 23 µ I π Explanation: Using the Biot-Savart law, the magnetic field due to the two radial legs is zero since d~s × ˆ r = 0 . However, around the arc | d~s × ˆ r | = ds = r dθ . The magnetic field at the center of an arc with a current I is B = µ I 4 π Z d~s × ˆ r r 2 = µ I 4 π r 2 Z ds = µ I 4 π r 2 Z r dθ = µ I 4 π r Z 20 23 π dθ = µ I 4 π r θ ¯ ¯ ¯ ¯ 20 23 π = µ I 4 π r µ 20 23 π − ¶ = 5 23 µ 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 B is out of the page. correct 2. b B is into the page. 3. b B is to the right. 4. b B is to the left. 5. b B is down the page. 6. b B is up the page. 7. b 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. k ~ B k = µ I 2 π ` √ 2 correct 2. k ~ B k = µ I 3 π ` √ 2 3. k ~ B k = µ I 2 π ` 4. k ~ B k = µ I π ` √ 2 5. k ~ B k = µ I √ 2 ` 6. k ~ B k = µ I 2 √ 2 π ` 7. k ~ B k = µ I π ` 2 √ 2 8. k ~ B k = 2 µ I π ` 9. k ~ B k = µ I ` √ 2 3 10. k ~ B k = µ I 2 ` Explanation: By the Biot-Savart law, dB = µ 4 π I d~s × ˆ r r 2 ....
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## This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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HW8 - Version One – Homework 8 – Savrasov – 39819 –...

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