Phys121-10sol

Phys121-10sol - Anderson, Matthew Homework 10 Due: Apr 8...

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Unformatted text preview: Anderson, Matthew Homework 10 Due: Apr 8 2008, noon Inst: Vitaly 1 This print-out should have 14 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) 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 current-carrying 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 Biot-Savart 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 Biot-Savart 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 Biot-Savart 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 x-direction, j is in y-direction, and k direction is perpendicular to paper to- wards reader. Determine the direction of the magnetic field ~ B at O due to the current-carrying 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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Phys121-10sol - Anderson, Matthew Homework 10 Due: Apr 8...

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