Homework 9 Solutions

5 t b the torque has magnitude equal to 2 b1

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Unformatted text preview: CHAPTER 29 56. We use Eq. 29-26 and note that the contributions to BP from the two coils are the same. Thus, BP = 2μ 0iR 2 N 2 2 ⎡ R2 + ( R 2) ⎤ ⎣ ⎦ 32 −7 8μ 0 Ni 8 ( 4π×10 T ⋅ m/A ) (200) ( 0.0122A ) = = = 8.78 ×10−6 T. 5 5R 5 5 ( 0.25m ) BP is in the positive x direction. 57. (a) The magnitude of the magnetic dipole moment is given by μ = NiA, where N is the number of turns, i is the current, and A is the area. We use A = πR2, where R is the radius. Thus, 2 μ = NiπR 2 = 300 4.0 A π 0.025 m = 2.4 A ⋅ m2 . b gb gb g (b) The magnetic field on the axis of a magnetic dipole, a distance z away, is given by Eq. 29-27: B= μ0 μ 2π z 3 . We solve for z: 13 ⎛ ( 4π ×10−7 T ⋅ m A )( 2.36 A ⋅ m 2 ) ⎞ ⎛ μ0 μ ⎞ ⎟ = 46 cm . z =⎜ ⎟ =⎜ ⎜ ⎟ 2π ( 5.0 ×10−6 T ) ⎝ 2π B ⎠ ⎝ ⎠ 13 58. (a) We set z = 0 in Eq. 29-26 (which is equivalent using to Eq. 29-10 multiplied by the number of loops). Thus, B(0) ∝ i/R. Since case b has two loops, Bb 2i Rb 2 Ra = = = 4.0 . Ba i Ra Rb (b) The ratio of their magnetic dipole moments is 2 μb 2iAb 2 Rb2 ⎛1⎞ 1 = = 2 = 2 ⎜ ⎟ = = 0.50. μ a iAa Ra ⎝ 2⎠ 2 59. The magnitude of the magnetic dipole moment is given by μ = NiA, where N is the number of turns, i is the current, and A is the area. We use A = πR2, where R is the radius. Thus, 2 μ = 200 0.30 A π 0.050 m = 0.47 A ⋅ m2 . b gb gb g 60. Using Eq. 29-26, we find that the net y-component field is 1147 By = μ0i1 R 2 2π ( R 2 + z12 )3/ 2 − μ0i2 R 2 2 2π ( R 2 + z2 )3/ 2 , where z12 = L2 (see Fig. 29-73(a)) and z22 = y2 (because the central axis here is denoted y instead of z). The fact that there is a minus sign between the two terms, above, is due to the observation that the datum in Fig. 29-73(b) corresponding to By = 0 would be impossible without it (physically, this means that one of the currents is clockwise and the other is counterclockwise). (a) As y → ∞, only the first term contributes and (with By = 7.2...
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