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Unformatted text preview: Nguyen, Thanh – Homework 13 – Due: Mar 7 2008, 7:00 pm – Inst: Weathers 1 This printout should have 15 questions. Multiplechoice 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 xaxis. 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 + xaxis 2. 135 ◦ counterclockwise from the + xaxis 3. 45 ◦ counterclockwise from the + xaxis 4. 315 ◦ counterclockwise from the + xaxis 5. perpendicular to and into the page 6. perpendicular to and out of the page cor rect 7. along the negative yaxis 8. along the positive yaxis 9. 225 ◦ counterclockwise from the + xaxis 10. along the negative xaxis Explanation: From the BiotSavart law we know that d ~ B ∝ I d ~ 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) 10 points Let I = 6 . 9 A and a = 0 . 79 m. What is the magnitude of the magnetic field at O due to the current configuration? Correct answer: 3 . 1188 × 10 6 T. Explanation: Let : I = 6 . 9 A , and a = 0 . 79 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 Z π/ 2 adθ 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 ~ B = ˆ z μ I 4 π a Z 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. Nguyen, Thanh – Homework 13 – Due: Mar 7 2008, 7:00 pm – Inst: Weathers 2 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 ~ B = ˆ z μ I 4 π a (1 + 1) = ˆ z μ I 2 π 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 = μ I 8 a + µ μ I 2 π a ¶ = μ I 2 a µ 1 4 + 1 π ¶ = (1 . 25664 × 10 6 T · m / A) (6 . 9 A) 2 (0 . 79 m) × µ 1 4 + 1 π ¶ = 3 . 1188 × 10 6 T ....
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This note was uploaded on 10/24/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
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