# Model assume that the 10 cm distance is much larger

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Model: Assume that the 10 cm distance is much larger than the size of the small bar magnet. Solve: (a) From Equation 33.9, the on-axis field of a magnetic dipole is ( ) ( ) ( ) 3 6 3 2 0 3 7 0 5.0 10 T 0.10 m 2 4 0.025 A m 4 2 2 10 T m/A Bz B z μ μ π μ π μ × = = = = (b) The on-axis field strength 15 cm from the magnet is ( ) ( ) ( ) 2 7 6 0 3 3 2 0.025 A m 2 10 T m/A 1.48 10 T 1.48 T 4 0.15 m B z μ μ μ π = = = × =

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33.16. Solve: (a) The magnetic dipole moment of the superconducting ring is ( ) ( ) ( ) 2 2 3 4 2 1.0 10 m 100 A 3.1 10 A m R I μ π π = = × = × (b) From Example 33.5, the on-axis magnetic field of the superconducting ring is ( ) ( ) ( ) ( ) ( ) ( ) 2 7 3 2 7 0 ring 3/ 2 3 2 2 2 2 2 2 10 T m/A 100 A 1.0 10 m 5.0 10 T 2 0.050 m 0.0010 m IR B z R π μ × = = = × + +
33.17. Model: The size of the loop is much smaller than 50 cm, so that the magnetic dipole moment of the loop is . AI μ = Solve: The magnitude of the on-axis magnetic field of the loop is ( ) ( ) ( ) ( ) ( ) 3 9 9 4 2 0 3 7 0.50 m 7.5 10 T 2 7.5 10 T A 1.88 10 m 4 2 25 A 10 Tm/A AI B z μ π × = = × = = × Let L be the edge-length of the equilateral triangle. Thus 4 2 1 1.88 10 m sin60 0.021 m 2.1 cm. 2 L L L × = × ° ⇒ = = Assess: The 2.1 cm edge-length is much smaller than 50 cm, as we assumed.

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33.18. Model: The radius of the earth is much larger than the size of the current loop. Solve: (a) From Equation 33.9, the magnetic field strength at the surface of the earth at the earth’s north pole is ( )( ) ( ) 7 22 2 5 0 3 3 6 2 10 T m/A 8.0 10 A m 6.2 10 T 2 6.38 10 m B z μ μ π × × = = = × × This value is close to the value of 5 × 10 5 T given in Table 33.1. (b) The current required to produce a dipole moment like that on the earth is ( ) 2 earth AI R I μ π = = 8.0 × 10 22 A m 2 = π (6.38 × 10 6 m) 2 I I = 6.3 × 10 8 A Assess: This is an extremely large current to run through a wire around the equator.
33.19. Visualize: Please refer to Figure EX33.19. Solve: Because B G is in the same direction as the integration path s G from i to f, the dot product of B G and d s G is simply Bds . Hence the line integral ( ) ( ) ( ) ( ) ( ) f f f 2 2 i i i 0.50 m 0.50 m 0.10 T 2 0.50 m 0.071 T m B d s Bds B ds B = = = + = = G G

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33.20. Visualize: Please refer to Figure EX33.20. Solve: The line integral of B G between points i and f is f i B d s G G Because B G is perpendicular to the integration path from i to f, the dot product is zero at all points and the line integral is zero.
33.21. Model: The magnetic field is that of the three currents enclosed by the loop. Visualize: Please refer to Figure EX33.21. Solve: Ampere’s law gives the line integral of the magnetic field around the closed path: 6 0 through 3.77 10 T m B d s I μ = = × G G ú ( ) ( ) ( ) 7 0 1 2 3 3 4 10 T m/A 6.0 A 4.0 A I I I I μ π = + = × + ( ) 6 3 7 3.77 10 T m 2.0 A 4 10 T m/A I π × + = × I 3 = 1.0 A Assess: The right-hand rule was used above to assign positive signs to I 1 and I 3 and a negative sign to I 2 .

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33.22.
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