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Unformatted text preview: Anderson, Matthew Homework 10 Due: Apr 8 2008, noon Inst: Vitaly 1 This printout should have 14 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) 15 points The wire is carrying a current I . x y I I I 180 O R Find the magnitude of the magnetic field ~ B at O due to a currentcarrying wire shown in the figure, where the semicircle has radius r , and the straight parts to the left and to the right extend to infinity. 1. B = I 2 r 2. B = I r 3. B = I 4 r 4. B = I 2 r 5. B = I 7 r 6. B = I 8 r 7. B = I 4 r correct 8. B = I 3 r 9. B = I 3 r 10. B = I r Explanation: By the BiotSavart Law, ~ B = I 4 Z d~s r r 2 . Consider the left straight part of the wire. The line element d~s at this part, if we come in from , points towards O; i.e. , in the x direction. We need to find d~s r to use the BiotSavart Law. However, in this part of the wire, r is pointing towards O as well, so d~s and r are parallel meaning d~s r = 0 for this part of the wire. It is now easy to see that the right part, having a d~s antiparallel to r , also gives no contribution to ~ B at O . Let us go through the semicircle C. The element d~s , which is along the wire, will now be perpendicular to r , which is pointing along the radius towards O . Therefore  d~s r  = ds using the fact that r is a unit vector. So the BiotSavart Law gives for the magnitude B of the magnetic field at O B = I 4 Z C ds r 2 Since the distance r to the element d~s is con stant everywhere on the semicircle C, we will be able to pull it out of the integral. The integral is Z C ds r 2 = 1 r 2 Z C ds = 1 r 2 L C , where L C = r is the length of the semicircle. Thus the magnitude of the magnetic field is B = I 4 1 r 2 r = I 4 r . 002 (part 2 of 2) 15 points Note: i is in xdirection, j is in ydirection, and k direction is perpendicular to paper to wards reader. Determine the direction of the magnetic field ~ B at O due to the currentcarrying wire. 1. b B = + 2. b B = Anderson, Matthew Homework 10 Due: Apr 8 2008, noon Inst: Vitaly 2 3. b B = 1 2 k 4. b B = 5. b B = + 6. b B = 1 2 k + 7. b B = k correct 8. b B = 1 2 i 9. b B = + k 10. b B = 1 2 i + Explanation: We know from Part 1 that the only contri bution to the magnetic field at O comes from the semicircle. Furthermore, we need only consider the direction of d~s r for one typi cal segment d~s . If we go along the semicircle from left to right, and we know that r is point ing in towards O , the right hand rule tells us that the field resulting from this element is into the paper. Since this holds for every el ement on the semicircle, the total field is also pointing into the paper.pointing into the paper....
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