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midterm 2

# midterm 2 - Version 052 midterm 02 chiu(56565 This...

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Version 052 – midterm 02 – chiu – (56565) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)5.0points 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. perpendicular to and out of the page cor- rect 2. 45 counterclockwise from the + x -axis 3. along the negative y -axis 4. 225 counterclockwise from the + x -axis 5. along the positive y -axis 6. along the negative x -axis 7. 135 counterclockwise from the + x -axis 8. 315 counterclockwise from the + x -axis 9. along the positive + x -axis 10. perpendicular to and into the page Explanation: From the Biot-Savart 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(part2of2)5.0points 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. 3.5708e-06 2. 6.35602e-06 3. 6.2846e-06 4. 1.6724e-06 5. 1.07613e-06 6. 2.61433e-06 7. 5.55457e-06 8. 4.20094e-06 9. 1.28945e-06 10. 2.42814e-06 Correct answer: 1 . 07613 × 10 6 T. Explanation: Let : I = 2 . 2 A , and a = 0 . 73 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 = μ 0 4 π I integraldisplay π/ 2 0 a dθ a 2 = μ 0 4 π I π 2 a = μ 0 I 8 a . For the straight sections, we apply the formu- las derived from the figure below, θ 1 θ 2 a I

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Version 052 – midterm 02 – chiu – (56565) 2 vector B = ˆ z μ 0 I 4 π a integraldisplay θ 2 θ 1 sin θ = ˆ z μ 0 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 y-directed current, θ 2 = π 2 and θ 1 = 0. For the horizontal x-directed 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 0 I 4 π a (1 + 1) = ˆ z μ 0 I 2 π a , /noindent the same as in a long straight wire. Adding the contributions from the straight sections and the arc, B tot = μ 0 I 8 a + parenleftbigg μ 0 I 2 π a parenrightbigg = μ 0 I 2 a parenleftbigg 1 4 + 1 π parenrightbigg = (1 . 25664 × 10 6 T · m / A) (2 . 2 A) 2 (0 . 73 m) × parenleftbigg 1 4 + 1 π parenrightbigg = 1 . 07613 × 10 6 T .
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