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# The eld points upward in the diagram if the current

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Unformatted text preview: ion in a cross-sectional plane of the cylinder. P is a point within the hole, A is y B1 on the axis of the cylinder, and C is on the axis of the B2 hole. The magnetic eld due to the cylinder without 1 2 the hole, carrying a uniform current out of the page, is x P labeled B1 and the magnetic eld of the cylinder that lls the hole, carrying a uniform current into the page, is labeled B2 . The line from A to P makes the angle r2 r1 1 with the line that joins the centers of the cylinders and the line from C to P makes the angle 2 with that line, as shown. B1 is perpendicular to the line from A to P and so makes the angle 1 with the vertical. 2 1 Similarly, B2 is perpendicular to the line from C to P C A d and so makes the angle 2 with the vertical. The x component of the total eld is 0 id B = 2a2 : B = B2 sin 2 B1 sin 1 = 2(0 ir2 b2 ) sin 2 2(0 ir1 b2 ) sin 1 a2 a2 = 2(a0 i b2 ) [r2 sin 2 r1 sin 1 ] : 2 x 820 CHAPTER 30 AMPERE'S LAW As the diagram shows r2 sin 2 = r1 sin 1 , so B = 0. The y component is given by x B = B2 cos 2 + B1 cos 1 = 2(0 ir2 b2 ) cos 2 + 2(0 ir1 b2 ) cos 1 a2 a2 = 2(a0 i b2 ) [r2 cos 2 + r1 cos 1 ] : 2 y The diagram shows that r2 cos 2 + r1 cos 1 = d, so 0 B = 2(a2 id b2 ) : y This is identical to the result found in part (a) for the eld on the axis of the hole. It is independent of r1 , r2 , 1 , and 2 , showing that the eld is uniform in the hole. 51P (a) Suppose the eld is not parallel to the sheet, as shown in the upper diagram. Reverse the direction of the current. According to the Biot-Savart law the eld reverses, so it will be as in the second diagram. Now rotate the sheet by 180 about a line that perpendicular to the sheet. The eld, of course, will rotate with it and end up in the direction shown in the bottom diagram. The current distribution is now exactly as it was originally, so the eld must also be as it was originally. But it is not. Only if the eld is parallel to the sheet will be nal direction of the eld be the same as the original direction. If the current is out of the page any in nitesimal portion of the sheet, in the form of a long straight wire, produces a eld that is to B the left above the sheet and to the right below the sheet. The eld must be as drawn in Fig. 30{65. Integrate the tangential component of the magnetic eld B around the rectangular loop shown with dotted lines. The L upper and lower edges are the same distance from the current sheet and each has length L. This means the eld has the same magnitude along these edges. It points to the left along the upper edge and to the right along the lower. If the integration is carried out in the counterclockwise sense the contribution of the upper edge is BL, the contribution of the lower edge is also BL, and the contribution of each of H the sides is zero because the eld is perpendicular to the sides. Thus B ds = 2BL. The total current through the loop is L. Ampere's law yields 2BL = 0 L, so B = 0 =2. CHAPTER 30 AMPERE'S LAW 821 52P* (a) I L B ds = Z = Z 1 d + + + 2 3 Z Z...
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## This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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