solutions 13

solutions 13 - Patel Kinal Homework 13 Due Mar 7 2008 7:00...

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Unformatted text preview: Patel, Kinal Homework 13 Due: Mar 7 2008, 7:00 pm Inst: Weathers 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Consider a current configuration shown be- low. A long (effectively infinite) wire segment is connected to a quarter of a circular arc with radius a . The other end of the arc is connected to another long horizontal wire segment. The current is flowing from the top coming down vertically and flows to the right along the pos- itive x-axis. a I O O y x What is the direction of the magnetic field at O due to this current configuration? 1. along the positive + x-axis 2. 225 counterclockwise from the + x-axis 3. along the positive y-axis 4. 135 counterclockwise from the + x-axis 5. along the negative y-axis 6. 45 counterclockwise from the + x-axis 7. 315 counterclockwise from the + x-axis 8. perpendicular to and into the page 9. along the negative x-axis 10. perpendicular to and out of the page cor- rect Explanation: From the Biot-Savart law we know that d ~ B I d ~ l r . One should first verify that the magnetic field at O contributed by any infinitesimal current element along the current configuration given is perpendicular to the page, which is coming out of the page. Therefore the resulting mag- netic field must also be pointing out of the page. 002 (part 2 of 2) 10 points Let I = 5 . 6 A and a = 0 . 96 m. What is the magnitude of the magnetic field at O due to the current configuration? Correct answer: 2 . 08296 10- 6 T. Explanation: Let : I = 5 . 6 A , and a = 0 . 96 m . Consider first the one-quarter of a circular arc. Since each current element is perpendic- ular to the unit vector pointing to O , we can write B arc = 4 I Z / 2 ad a 2 = 4 I 2 a = I 8 a . For the straight sections, we apply the formu- las derived from the figure below, 1 2 a I ~ B = z I 4 a Z 2 1 d sin = z I 4 a (cos 1- cos 2 ) where z is the unit vector perpendicular to the plane of the paper that points to the reader. Patel, Kinal Homework 13 Due: Mar 7 2008, 7:00 pm Inst: Weathers 2 For the downward y-directed current, 2 = 2 and 1 = 0. For the horizontal x-directed current 2 = and 1 = 2 . Thus we find the contribution at the point O for the magnetic field from the long vertical wire is same as in the long horizontal wire, and the sum is equal to ~ B = z I 4 a (1 + 1) = z I 2 a . (Note that the sum of the two is the same as a long straight wire) Adding the contributions from the straight sections and the arc, B tot = I 8 a + I 2 a = I 2 a 1 4 + 1 = (1 . 25664 10- 6 T m / A) (5 . 6 A) 2 (0 . 96 m) 1 4 + 1 = 2 . 08296 10- 6 T ....
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solutions 13 - Patel Kinal Homework 13 Due Mar 7 2008 7:00...

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