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Unformatted text preview: Version One Homework 8 Savrasov 39821 May 14, 2007 1 This printout should have 7 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. BiotSavart Law 30:01, trigonometry, numeric, > 1 min, nor mal. 001 (part 1 of 1) 2 points The segment of wire in the figure carries a cur rent of 5 A, where the radius of the circular arc is 3 cm. The permeability of free space is 1 . 25664 10 6 T m / A . 5 A 3 c m O Determine the magnitude of the magnetic field at point O , the origin of the arc. Correct answer: 26 . 1799 T. Explanation: Let : I = 5 A and R = 3 cm = 0 . 03 m . For the straight sections d ~s r = 0. The quarter circle makes onefourth the field of a full loop B = I 8 R into the paper. Or, you can use the equation B = o I 4 , where = 2 . Thus the magnetic field is B = I 8 R , = (1 . 25664 10 6 T m / A)(5 A) 8(0 . 03 m) = 26 . 1799 T , into the paper . keywords: Magnetic Field from a Segment 01 30:01, calculus, numeric, > 1 min, wording variable. 002 (part 1 of 2) 1 points Consider two radial legs extending to infinity and a line segment carrying a current I as shown below. r I r I 5 7 I O What is the magnitude of the magnetic field B at the origin O due to the current through this path? 1. I 2 r tan 5 14 correct 2. I 2 r tan 9 14 3. I 2 r tan 5 7 4. I 2 r sin 5 14 5. I 2 r sin 9 14 6. I 2 r sin 5 7 7. I 2 r cos 5 14 8. I 2 r cos 9 14 9. I 2 r cos 5 7 Version One Homework 8 Savrasov 39821 May 14, 2007 2 10. None of these Explanation: Let : = 5 7 , and = 2 2 = 1 2 5 14 = 9 14 , and 2 = 2 = 5 14 . By the BiotSavart law, dB = 4 I d~s r r 2 . Consider a thin, straight wire carring a con stant current I along the xaxis with the y axis pointing towards the vertex point O , as in the following figure. y x O r P x I r ds a 2 Figure: Is not drawn to scale. Let us calculate the total magnetic field at the point P located at a distance a from the wire. An element d~s is at a distance r from P . The direction of the field at P due to this element is out of the paper, since d~s r is out of the paper. In fact, all elements give a contribution directly out of the paper at P . Therefore, we have only to determine the magnitude of the field at P . In fact, taking the origin at O and letting P be along the positive y axis, with k being a unit vector pointing out of the paper, we see that d~s r = k  d~s r  = k ( dx sin ) . Substituting into BiotSavart law gives d ~ B = k dB , with dB = I 4 dx sin r 2 . (1) In order to integrate this expression, we must relate the variables , x , and r . One approach is to express x and r in terms of . From the geometry in the figure and some simple differ entiation, we obtain the following relationship...
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 Spring '10
 Dr.AndyGavrin

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