# oldhw19 - oldhomewk 19 LEE BENJAMIN Due 4:00 am Question 1...

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oldhomewk 19 – LEE, BENJAMIN – Due: Oct 16 2007, 4:00 am 1 Question 1, chap 29, sect 5. part 1 of 2 10 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 = μ 0 I 3 π r 2. B = μ 0 I 7 π r 3. B = μ 0 I 2 π r 4. B = μ 0 I 3 r 5. B = μ 0 I 4 π r 6. B = μ 0 I 8 π r 7. B = μ 0 I π r 8. B = μ 0 I 2 r 9. B = μ 0 I r 10. B = μ 0 I 4 r correct Explanation: By the Biot-Savart Law, vector B = μ 0 I 4 π integraldisplay dvectors × ˆ 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 × ˆ r to use the Biot-Savart Law. However, in this part of the wire, ˆ r is pointing towards O as well, so dvectors and ˆ r are parallel meaning dvectors × ˆ r = 0 for this part of the wire. It is now easy to see that the right part, having a dvectors antiparallel to ˆ 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 ˆ r , which is pointing along the radius towards O . Therefore | dvectors × ˆ r | = ds using the fact that ˆ r is a unit vector. So the Biot-Savart Law gives for the magnitude B of the magnetic field at O B = μ 0 I 4 π 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 = π r is the length of the semicircle. Thus the magnitude of the magnetic field is B = μ 0 I 4 π 1 r 2 π r = μ 0 I 4 r . Question 2, chap 29, sect 5. part 2 of 2 10 points Note: ˆ i is in x-direction, ˆ j is in y-direction, and ˆ 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 = ˆ k correct 2. hatwide B = +ˆ 3. hatwide B = ˆ ı

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oldhomewk 19 – LEE, BENJAMIN – Due: Oct 16 2007, 4:00 am 2 4. hatwide B = 1 2 parenleftBig ˆ i ˆ parenrightBig 5. hatwide B = 1 2 parenleftBig ˆ k + ˆ parenrightBig 6. hatwide B = + ˆ k 7. hatwide B = +ˆ ı 8. hatwide B = ˆ 9. hatwide B = 1 2 parenleftBig ˆ k ˆ parenrightBig 10. hatwide B = 1 2 parenleftBig ˆ i + ˆ parenrightBig 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 dvectors × ˆ r for one typi- cal segment dvectors . If we go along the semicircle from left to right, and we know that ˆ 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.
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