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Unformatted text preview: Version 052 – midterm 02 – chiu – (56565) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 5.0 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 xaxis. a I O O y x What is the direction of the magnetic field at O due to this current configuration? 1. perpendicular to and out of the page cor rect 2. 45 ◦ counterclockwise from the + xaxis 3. along the negative yaxis 4. 225 ◦ counterclockwise from the + xaxis 5. along the positive yaxis 6. along the negative xaxis 7. 135 ◦ counterclockwise from the + xaxis 8. 315 ◦ counterclockwise from the + xaxis 9. along the positive + xaxis 10. perpendicular to and into the page Explanation: From the BiotSavart law we know that d vector B ∝ I d vector l × ˆ 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) 5.0 points Let I = 2 . 2 A and a = 0 . 73 m. What is the magnitude of the magnetic field at O due to the current configuration? 1. 3.5708e06 2. 6.35602e06 3. 6.2846e06 4. 1.6724e06 5. 1.07613e06 6. 2.61433e06 7. 5.55457e06 8. 4.20094e06 9. 1.28945e06 10. 2.42814e06 Correct answer: 1 . 07613 × 10 − 6 T. Explanation: Let : I = 2 . 2 A , and a = 0 . 73 m . Consider first the onequarter of a circular arc. Since each current element is perpendic ular to the unit vector pointing to O , we can write B arc = μ 4 π I integraldisplay π/ 2 a dθ a 2 = μ 4 π I π 2 a = μ I 8 a . For the straight sections, we apply the formu las derived from the figure below, θ 1 θ 2 a I Version 052 – midterm 02 – chiu – (56565) 2 vector B = ˆ z μ I 4 π a integraldisplay θ 2 θ 1 dθ sin θ = ˆ z μ I 4 π a (cos θ 1 − cos θ 2 ) where ˆ z is the unit vector perpendicular to the plane of the paper that points to the reader. For the downward ydirected current, θ 2 = π 2 and θ 1 = 0. For the horizontal xdirected current θ 2 = π and θ 1 = π 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 vector B = ˆ z μ I 4 π a (1 + 1) = ˆ z μ I 2 π a , /noindent the same as in a long straight wire....
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This note was uploaded on 12/01/2011 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Current

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