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Unformatted text preview: 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 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 = μ I 3 π r 2. B = μ I 7 π r 3. B = μ I 2 π r 4. B = μ I 3 r 5. B = μ I 4 π r 6. B = μ I 8 π r 7. B = μ I π r 8. B = μ I 2 r 9. B = μ I r 10. B = μ I 4 r correct Explanation: By the BiotSavart Law, vector B = μ 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 BiotSavart 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 BiotSavart Law gives for the magnitude B of the magnetic field at O B = μ 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 I 4 π 1 r 2 π r = μ I 4 r . Question 2, chap 29, sect 5. part 2 of 2 10 points Note: ˆ i is in xdirection, ˆ j is in ydirection, 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 currentcarrying wire. 1. hatwide B = − ˆ k correct 2. hatwide B = +ˆ 3. hatwide B = − ˆ ı 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...
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This note was uploaded on 04/07/2008 for the course PHY 303K taught by Professor Turner during the Fall '07 term at A.T. Still University.
 Fall '07
 Turner
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