F5ch30

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Unformatted text preview: 800 CHAPTER 30 AMPERE'S LAW CHAPTER 30 Answer to Checkpoint Questions 1. 2. 3. 4. (a), (c), (b) b, c, a d, tie of a and c, then b (d), (a), tie of (b) and (c) (zero) Answer to Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. (c), (d), then (a) and (b) tie (a) into; (b) greater 2 and 4 (c), (a), (b) (a), (b), (c) (see Eq. 30-11) (c), (a), (b) (b), (d), (a) (zero) (a) 1, 3, 2; (b) less than (a) 1, +x; 2, y; (b) 1, +y; 2, +x a, b, c outward a, tie of b and d, then c c and d tie, then b, a b, a, d, c (zero) (d), then tie of (a) and (e), then (b), (c) (a) 2 opposite 4; (b) 2 and 4 opposite 6; (c) 1 and 5 opposite 3 and 6; (d) 1 and 5 opposite 2, 3, and 4 17. 0 (scalar product is zero) CHAPTER 30 AMPERE'S LAW 801 18. (a) 4a; (b) 3d Solutions to Exercises & Problems 1E Solve i from Eq. 30-6: 6 2 i = 2rB = 2(88:0 410 10m)(7:30 10 T) = 32:1 A : 7 T m/A 0 2E Use Eq. 30-6: 10 7 0 i B = 2r = (42(2:6 T m/A)(50 A) = 7:7 10 3 T : 10 3 m=2) 3E Put r = 20 ft = 6:10 m. Then 7 T m/A)(100 A) B = (4 102(6:10 m) = 3:3 10 6 T = 3:3 T : (a) The magnitude of the magnetic eld due to the current in the wire, at a point a distance r from the wire, is given by 0 i B = 2r : (b) This is about one-sixth the magnitude of the Earth's eld. It will aect the compass reading. 4E The current i due to the electron ow is i = ne = (5:6 1014 =s)(1:6 10 19 C) = 9:0 10 5 A. Thus 7 T m/A)(9 5 0 i B = 2r = (4 102(1:5 10 3:0 10 A) = 1:2 10 8 T: m) 802 CHAPTER 30 AMPERE'S LAW 5E Use B (x; y; z) = (0 =4)i s r=r3 , where s = s j and r = x i + y j + z k. Thus 0 i s j (xi + y j + zk) = 0 i s (z i x k) : B(x; y; z ) = 4 (x2 + y2 + z 2 )3 2 4(x2 + y + z 2 )3 2 = s = (a) B(0; 0; 5:0 m) = (4 10 7 T m/A)(2:0 A)(3:0 10 2 m)(5:0 m) i 4[02 + 02 + (5:0 m)2 ]3 2 = (2:4 10 10 i) T : = (b) B(0; 6:0 m; 0), since x = z = 0. (c) B(7:0 m; 7:0 m; 0) = (4 10 7 T m/A)(2:0 A)(3:0 10 2 m)( 7:0 m) k 4[(7:0 m)2 + (7:0 m)2 + 02 ]3 2 = (4:3 10 11 k) T : = (d) (4 10 7 T m/A)(2:0 A)(3:0 10 2 m)(3:0 m) k B( 3:0 m; 4:0 m; 0) = 4[( 3:0 m)2 + ( 4:0 m)2 + 02 ]3 2 = (1:4 10 10 k) T : = 6E The points must be along a line parallel to the wire and a distance r from it, where r satis es 0 i B wire = 2r = Bext ; or 10 6 T m/A)(100 0 r = 2Bi = (1:26 (5:0 10 3 T) A) = 4:0 10 3 m : 2 ext 7E (a) The eld due to the wire, at a point 8:0 cm from the wire, must be 39 T and must be directed due south. Since B = 0 i=2r, 080 m)(39 10 6 i = 2rB = 2(0: 10 7 Tm/A T) = 16 A : 4 0 CHAPTER 30 AMPERE'S LAW 803 (b) The current must be from west to east to produce a eld to the south at points below it. 8E Set up a coordinate system as shown to the right. The B- eld at the location of the charge q is z i B = 0 ( k) : 2d i current F = q v B = 0 ( k v) : d 2d charge (a) Now v = v( i) so x 0 iv k ( i) = 0 iqv ( j) ; F = 2d 2d i.e., F has a magnitude of 0 iqv=2d and is directed against the current direction. (b) Now the direction v is reversed so F = 0 iqv j=2d. q q q q Thus y 9E The straight segment of the wire produces no magnetic elds at C . The eld from the two semi-circular loop cancel at C . So B...
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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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