Right 1 b jb1 b2 j 2b1 cos 0 i d2 2 2r r 0 id

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Unformatted text preview: )(30 cm + 8A)(8:0 cm)(30 10 m) 2(1:0 :0 cm) 3 N; = 3:2 10 and F points toward the wire. Consider a segment of the projectile between x and x + dx. Use Eq. 30-13 to nd the magnetic force on the segment: 39P wire 1 O R i y dF = dF1 + dF2 = i(dx i) B1 (x) + i(dx i) B2 (x) = i[B1 (x) + B2 (x)] j dx 0 i + 0 i = i 4x 4(2R + w x) j dx ; where j is a unit vector pointing to the right. Here we used the expression for the magnetic eld of a semi-in nite wire. Thus j W projectile i R wire 2 i x CHAPTER 30 AMPERE'S LAW 815 i2 Z + 1 + 1 i2 w F = dF = j dx = 0 ln 1 + 4 x 2R + w x 2 R R (b) Use K = 1 mv2 = Wext = F ds for the projectile: 2 Z R w R j: f " #1 2 2Wext 1 2 = 2 Z 0 i2 ln 1 + w dy v = m m 0 2 R 2(4 10 7 T m/A)(450 103 A)2 (4:0 m) ln(1 + 1:2 cm=6:7 cm) 1 2 = 2(10 10 3 kg) = 2:3 103 m/s : = L = f = 40E (a) Two of the currents are out of the page and one is into the page, so the net current enclosed by the path is 2:0 A, out of the page. Since the path is traversed in the clockwise sense a current into the page is positive and a current out of the page is negative, as indicated by the right-hand rule associated with Ampere's law. Thus I B ds = 0 i = (2:0 A)(4 10 7 T m/A) = 2:5 10 6 T m : (b) The net current enclosed by the path is zero (two currents are out of the page and two H are into the page), so B ds = 0 i = 0. A close look at the path reveals that only currents no.1 and no.6 are enclosed. Thus I 41E B ds = 0 (6i0 i0 ) = 50 i : 42E The area enclosed by the loop L is A = 1 (4d)(3d) = 6d2 . Thus 2 I c B ds = 0 i = 0 jA = (4 10 7 T m/A)(15 A/m2 )(6)(0:20 m)2 = 4:5 10 6 T m : 43E Use Eq. 30-22 for the B - eld inside the wire and 30-19 for that ouside the wire. The plot is shown in the next page. 816 CHAPTER 30 AMPERE'S LAW B (mT) inside 0.8 outside 0 1 2 3 4 5 6 r (cm) Use Ampere's law: B ds = 0 i, where the integral is around a closed loop and i is the net current through the loop. For path 1 H I 44P 1 B ds = 0 ( 5:0 A + 3:0 A) = ( 2:0 A)(4 10 7 T m/A) = 2:5 10 6 T m ; 5:0 A 5:0 A 3:0 A) = ( 13:0 A)(4 10 7 T m/A) and for path 2 I 2 B ds = 0 ( = 1:6 10 5 T m : R 45P Use Ampere's law. For the dotted loop shown on the diagram i = 0. The integral B ds is zero along the bottom, right, and top sides of the loop. Along the right side the eld is zero, along the top and bottom sides the eld is perpendicular to ds. If ` is the length of H the left edge then direct integration yields B ds = B`, where B is the magnitude of the eld at the left side of the loop. Since neither B nor ` is zero, Ampere's law is contradicted. We conclude that the geometry shown for the magnetic eld lines is in error. The lines actually bulge outward and their density decreases gradually, not precipitously as shown. (a) For the circular path L of radius r concentric with the conductor I (r2 b2 ) B ds = 2rB (r) = 0 i enclosed = 0 i 2 2 : (a b ) L 46P CHAPTER 30 AMPERE'S LAW 817 Thus (b) At r = a; 0 i r2 b2 : B (r) = 2(a2 b2 ) r a2 b2 = 0 i : 0 i B (a) = 2(a2 b2 ) a 2a 2 b2 = 0. For b = 0 At r = b; B (b) / r...
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