If the integration is carried out in the

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Unformatted text preview: Z Z 4 B ds y y 0 [3:0 i + 8:0(x2 =d2 ) j] i dxj =0 d + + + 0 Z Z d [3:0 i + 8:0(x2 =d2 ) j] j dyj = x 3 (d, d, 0) d 0 0 [3:0 i + 8:0(x2 =d2 ) j] i dxj = y x d 4 2 path L [3:0 i + 8:0(x2 =d2 ) j] j dyj =0 = (3:0d + 8:0d 3:0d) mT = (8:0d) mT ; d 1 (d, 0, 0) x where d isH in meters. (b) From B ds = (8:0d) mT = 0 i enclosed , we get L (c) Since 53E H L B ds > 0; i enclosed > 0, so the current is in the positive k direction. T) d : i enclosed = (8:0) mT = (80)(0:50)(10m/A = 3:2 103 A: 7 T 4 10 0 3 The magnetic eld inside an ideal solenoid is given by Eq. 30{25. The number of turns per unit length is n = (200 turns)=(0:25 m) = 800 turns/m. Thus B = 0 ni0 = (4 10 7 T m/A)(800 m 1 )(0:30 A) = 3:0 10 4 T : 54E B = 0 i0 n = (4 10 7 T m/A)(3:60 A)(1200=0:950 m) = 5:71 10 3 T : 55E Solve N , the number of turns of the solenoid, from B = 0 in = 0 iN=L: N = BL=0 i: Thus the length of wire is l = 2rN = 2rBL 10 2 m)(23 10 3 = 2(2:60 26 10 6 T :0m/A)(18T)(1:30 m) = 108 m : 2(1: :0 A) 0 i 822 CHAPTER 30 AMPERE'S LAW 56E (a) Use Eq. 30-26. The inner radius is r = 15:0 cm so the eld there is 7 T m/A)(0: iN B = 20r = (4 10 2(0:150 m)800 A)(500) = 5:33 10 4 T : (b) The outer radius is r = 20:0 cm. The eld there is 7 T m/A)(0: iN B = 20r = (4 10 2(0:200 m)800 A)(500) = 4:00 10 4 T : 57E In this case L = 2r is roughly the length of the toroid so B = 0 i0 2N = 0 ni0 : r This result is expected, since from the perspective of a point inside the toroid the portion of the toroid in the vicinity of the point resembles part of a long solenoid. 58P Use B = 0 i0 n and note that ni0 = . Thus B = 0 : Also from 51P we have B = 1 0 ( 1 0 ) = 0 as we move through an in nite plane current sheet of current 2 2 density . 59P Consider a circle of radius r, inside the toroid and concentric with it. The current that passes through the region between the circle and the outer rim of the toroid is Ni, where N is the number of turns and i is the current. The current per unit length of circle is = Ni=2r and 0 is 0 Ni=2r, the magnitude of the magnetic eld at the circle. Since the eld is zero outside a toroid, this is also the change in the magnitude of the eld encountered as you move from the circle to the outside. The equality is not really surprising in light of Ampere's law. You are moving perpendicularly to the magnetic eld lines. Consider an extremely narrow loop, with the narrow sides along eld lines and the two long sides perpendicular H the eld lines. If B1 is the to eld at one end and B2 is the eld at the other end then B ds = (B2 B1 )w, where w is the width of the loop. The current through the loop is w, so Ampere's law yields (B2 B1 )w = 0 w and B2 B1 = 0 . 60P (a) Denote the B- elds at point P on the axis due to the solenoid and the wire as B s CHAPTER 30 AMPERE'S LAW 823 B = 0 i n = B = 0 i ; 2d which gives the separation d to point P on the axis: w s s w w s and B , respectively. Since B is along the axis of the solenoid and B is perpendicular to it, B ? B , respectively. For the net eld B to be at 45 with the axis we then must have B = B . Thus...
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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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