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Unformatted text preview: 0) 0 3 2 y 6 2 2 = (1:26 10 T m/A)(6:0 3 10 m A) j 2(5:0 m) 11 T) j : = (9:6 10 For (x; y; z) = (5:0 m; 0; 0), note that the line joining the end point of interest and the location of the dipole is perpendicular to the axis of the dipole. You can check easily that if an electric diploe is used, the eld wold be E (1=40 )(p=x3 ), which is half of the magnitude of E for a point on the y axis the same distance from the dipole. By analogy, in our case B is also half the value or B (0; 5:0 m; 0), i.e., 1 1 B (5:0 m; 0; 0) = 2 B (0; 5:0 m; 0) = 2 (9:6 10 11 T) = 4:8 10 11 T : CHAPTER 30 AMPERE'S LAW 827 Just like the electric dipole case, B(5:0 m; 0; 0) points in the negative y direction. (a) The magnitude of the magnetic eld on the axis of a circular loop, a distance z from the loop center, is given by Eq. 30{28: where R is the radius of the loop, N is the number of turns, and i is the current. Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense and the elds they produce are in the same direction in the region between them. Place the origin at the center of the lefthand loop and let x be the coordinate of a point on the axis between the loops. To calculate the eld of the lefthand loop set z = x in the equation above. The chosen point on the axis is a distance s x from the center of the righthand loop. To calculate the eld it produces put z = s x in the equation above. The total eld at the point is therefore 1 1 N0 iR2 + : B= 2 (R2 + x2 )3 2 (R2 + x2 2sx + s2 )3 2 Its derivative with respect to x is dB = N0 iR2 3x 3(x s) + : dx 2 (R2 + x2 )5 2 (R2 + x2 2sx + s2 )5 2 When this is evaluated for x = s=2 (the midpoint between the loops) the result is
= = = = 72P N0 2 B = 2(R2 + iR)3 2 ; z2
= dB = N0 iR2 3s=2 dx 2 2 (R2 + s2 =4)5 2 independently of the value of s. = s= 3s=2 = 0; (R2 + s2 =4 s2 + s2 )5 2 = (b) The second derivative is d2 B = N0 iR2 dx2 2 2 3 + 2 15x 2 7 2 (R2 + x2 )5 2 (R + x ) # 2 3 15(x s) 2 + x2 2sx + s2 )5 2 + (R2 + x2 2sx + s2 )7 2 : (R
= = = = " At x = s=2, 2 d2 B = N0 iR2 6 + 2 30s 2=4 7 2 dx2 2 2 (R2 + s2 =4)5 2 (R + s =4) N0 R2 6(R2 + s2 =4) + 30s2 =4 = 3N iR2 s2 R2 : = 2 0 (R2 + s2 =4)7 2 (R2 + s2 =4)7 2 = = s= = = 828 CHAPTER 30 AMPERE'S LAW Clearly, this is zero if s = R. (a) B from sum: 7:069 10 (b) B from sum: 1:043 10 (c) B from sum: 2:506 10
74 73 5 T; 4 T; 4 T; 0 in = 5:027 10 0 in = 1:005 10 0 in = 2:513 10 5 T; 40% dierence; 4 T; 4% dierence; 4 T; 0:3% dierence (a) each side contributes 3:1410 7 Tm; the total is 1:2610 6 Tm; 0 i = 1:2610 6 Tm; (b) the sides contribute 2:98 10 7 T m, 2:48 10 7 T m, 2:98 10 7 T m, and 4:12 10 7 T m; the total is 1:26 10 6 T m; (c) the sides contribute 2:52 10 7 T m, 2:03 10 7 T m, 2:52 10 7 T m, and 5:49 10 7 T m; the total is 1:26 10 6 T m; (d) the sides contribute 1:89 10 7 Tm, 1:71 10 7 Tm, 1:89 10 7 Tm, and 5:49 10 7 T m; the total is essentially zero (a) B = (0 =2)[i1 =(x a) + i2 =x] j; (b) B = (0 =2)(i1 =a)(1 + b=2) j
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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.
 Fall '07
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