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45o (b) i d = 2i n = 2(20:0 10 6:300 A turns/cm) = 4:77 cm : A)(10 B = 2B Bw P = 2(4 10 7 T m/A)(20:0 10 3 A)(10 turns=0:0100 m) = 3:55 10 5 Tm : p p s 61P Let and solve for i: mv r = mv = e ni eB 0 mv i = e nr 0 (9:11 10 31 kg)(0:0460)(3 00 108 m/s) = (1:60 10 19 C)(4 10 7 T m/A)(100:=0:0100 m)(2:30 10 2 m) = 0:272 A : 62E The magnitude of the dipole moment is given by = NiA, where N is the number of turns, i is the current, and A is the area. Use A = R2 , where R is the radius. Thus = (200)(0:30 A)(0:050 m)2 = 0:47 A m2 : 63E (a) Set z = 0 in Eq. 3028: B (0) / i=R. Since case b has two loops, B = 2i=R = 2R = 4 : B b a i=R b a R a b 824 CHAPTER 30 AMPERE'S LAW (b) = 2iA = 2R2 = 2 1 2 = 1 : iA R2 2 2
b b b a a a 64E Use Eq. 3028 and note that the contribute to B from the two coils are the same. Thus
P 2N p B = 2[R22+0(iR 2)2 ]3 2 = 80 Ni : R= 5 5R
P = BP is in the positive x direction. (a) Similar to 62E, the magnitude of the dipole moment is given by = NiA, where N is the number of turns, i is the current, and A is the area. Use A = R2 , where R is the radius. Thus = NiR2 = (300)(4:0 A)(0:025 m)2 = 2:4 A m2 : (b) The magnetic eld on the axis of a magnetic dipole, a distance z away, is given by Eq. 30{29: 0 B = 2 z3 : Solve for z : 0 1 3 = (4 10 7 T m/A)(2:36 A m2 ) 1 3 = 46 cm : z = 2 B 2(5:0 10 6 T) = = 65E 66E (a) For x a the result of 21P reduces to
2 0 (ia2 B (x) (4x420 ia 2 )1 2 = 4x3 ) ; )(4x
= indeed the B eld of a magnetic diploe (see Eq. 3029). (b) = ia2 , by comparison between Eq. 3029 and the result above. (a) The two straight segments of the wire do not contribute to B . For the larger semicircular loop B 1 = 0 i=4b (see 12P), and for the smaller one B 2 = 0 i=4a. Thus
P P P 67P 0 1 B = B 1 + B 2 = 4 i a + 1 : b p P P CHAPTER 30 AMPERE'S LAW 825 B points into the page. (b) = A loop i = 1 (a2 + b2 )i, pointing into the page. 2
P 68P Use Eq. 3028 to obtain
2 B (x) = 0 iR 2
" R2 + (x R=2)2 1 32
= 3 2 # 1 + R2 + (x + R=2)2 : = The plot of B (x) is as follows. Note that B (x) is almost a constant in the vicinity of x = 0.
0.9 0.8 B (mT) 0.7 0.6 0.5 5 4 3 2 1 0 1 2 3 4 5 x (cm) 69P (b) Denote the large and small loops with subscripts 1 and 2, respectively. (a) 7 0 B1 = Ri1 = (4 102(0T m/A)(15 A) = 7:9 10 5 T : 2 :12 m)
1 2 = j 2 B1 j = B1 sin 90 = N2 i2 A2 B1 = N2 i2 r2 B1 = (50)(1:3 A)(0:82 10 2 m)2 (7:9 10 5 T) = 1:1 10 6 N m : 70P (a) Contribution to B from the straight segment of the wire:
C B 1 = 20 i : R
C 826 CHAPTER 30 AMPERE'S LAW Contribution from the circular loop:
0 B 2 = Ri : 2
C Thus
BC points out of the page. (b) Now B 1 ? B 2 so
C C 1 0 B = B 1 + B 2 = Ri 1 + : 2 C C C and B C 1 0 B = B 2 1 + B 2 2 = Ri 1 + 2 : 2 points at an angle out of the page, where 1 B 1 = tan 1 1 = 18 : = tan B 2
q
C C C C C r 71P (a) By imagining that both bg and cf have a pair of currents of the same magnitude (= i) and opposite direction, you can justify the superposition. (b) = + + = (ia2 )( j i + i) = ia2 j = (6:0 A)(0:10 m)2 j = (6:0 10 2 m2 A) j :
bcf gb abgha cdef c (c) Since both points are far from the cube we can use the dipole approximation. For (x; y; z) = (0; 5:0 m; 0) B(0; 5:0 m;...
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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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