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Unformatted text preview: maini (nm7637) hw10 Shneidman (12108) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points The wire is carrying a current I . x y I I I 180 O R Find the magnitude of the magnetic field vector 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 4 r correct 2. B = I 3 r 3. B = I 2 r 4. B = I 7 r 5. B = I 4 r 6. B = I 8 r 7. B = I r 8. B = I 2 r 9. B = I r 10. B = I 3 r Explanation: By the Biot-Savart Law, vector B = I 4 integraldisplay dvectors r r 2 . Consider the left straight part of the wire. The line element dvectors at this part, if we come in from , points towards O; i.e. , in the x- direction. We need to find dvectors r to use the Biot-Savart Law. However, in this part of the wire, r is pointing towards O as well, so dvectors and r are parallel meaning dvectors r = 0 for this part of the wire. It is now easy to see that the right part, having a dvectors antiparallel to r , also gives no contribution to vector B at O . Let us go through the semicircle C. The element dvectors , which is along the wire, will now be perpendicular to r , which is pointing along the radius towards O . Therefore | dvectors 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 integraldisplay C ds r 2 Since the distance r to the element dvectors is con- stant everywhere on the semicircle C, we will be able to pull it out of the integral. The integral is integraldisplay C ds r 2 = 1 r 2 integraldisplay 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) 10.0 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 vector B at O due to the current-carrying wire. 1. hatwide B = 1 2 parenleftBig k + parenrightBig 2. hatwide B = + 3. hatwide B = + k maini (nm7637) hw10 Shneidman (12108) 2 4. hatwide B =- 5. hatwide B = + 6. hatwide B = 1 2 parenleftBig k- parenrightBig 7. hatwide B =- 8. hatwide B = 1 2 parenleftBig i- parenrightBig 9....
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This note was uploaded on 01/31/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.
- Spring '08