18 - Shie, Gary – Homework 18 – Due: Oct 15 2004, 4:00...

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Unformatted text preview: Shie, Gary – Homework 18 – Due: Oct 15 2004, 4:00 am – Inst: Turner 1 This print-out should have 12 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 the application of the Biot-Savart law Δ ~ B = μ 4 π I Δ ~ ‘ × ˆ r r 2 . Consider the magnetic field at O due to the current segments B B +semicircle+ CC . The length of the linear segments, B B = CC = d . The semicircle has a radius R . x y I I I 180 ◦ O R B B C C A The magnetic field B O at the origin O due to the current through this path is 1. out of the page. correct 2. into the page. 3. undetermined, since B = 0 . Explanation: Basic Concepts: Biot-Savart Law Solution: ˆ r is pointing from A to O ; i.e. , from the location of the source element to the location of the field of concern. So- ˆ is the direction. 002 (part 2 of 2) 10 points At O , the magnitude of the magnetic field due to the current segments described above is given by 1. B O = μ I 4 R correct 2. B O = μ I 4 π R 3. B O = μ I π 2 R 4. B O = μ I π R 5. B O = μ I π R 6. B O = μ I 2 π R 7. B O = μ I π 4 R 8. B O = μ I 2 R 9. B O = μ I R 10. B O = 0 Explanation: Since the Biot-Savart equation has the cross product ˆ ı × b R in the numerator, this term implies that the contribution of B B and CC to the field at O is zero. The magnetic field at at the center of an arc with a current I is B = μ I 4 π Z d~s × b 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 π dθ = μ I 4 π R θ fl fl fl fl π = μ I 4 R . 003 (part 1 of 1) 10 points Given: The set up as shown in the figure, where 39 A is flowing in the wire segments, AB = CD = 39 m, and the wire segment arc has a radius 18 m subtending an angle of 90 ◦ . Shie, Gary – Homework 18 – Due: Oct 15 2004, 4:00 am – Inst: Turner 2 39m 39A 3 9 A 39 m 39 A 1 8 m A B C D O Find the magnitude of the magnetic field at O due to the current segment ABCD . Correct answer: 3 . 40339 × 10- 7 T. Explanation: Let : I = 39 A , x = 39 m , y = 39 m , and a = 18 m . Along CD d~s and ˆ r are antiparallel, so again d~s × ˆ r = 0 . Therefore that segment of the current also creates no magnetic field at O . Along BC d~s is perpendicular to ˆ r so | d~s × ˆ r | = ds = adθ . Also d~s × ˆ r is in the same direction for all d~s along BC , while r = a , so the magnitude of the magnetic field at O due to ABCD is B = μ 4 π I Z π/ 2 adθ a 2 = μ 4 π a I θ fl fl fl π/ 2 = μ I 8 a . 004 (part 1 of 2) 10 points Consider two radial legs (extending to in- finity) and a connecting 20 23 π circular arc car- rying a current I as shown below....
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This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas.

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18 - Shie, Gary – Homework 18 – Due: Oct 15 2004, 4:00...

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