Model the coils of wire are wrapped around the center

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Model: The coils of wire are wrapped around the center of the cylinder, and are wrapped tightly enough so that the loops all have the same area. Visualize: Please refer to figure P33.71. Solve: (a) For equilibrium, the gravitational torque on the cylinder is balanced by the torque on the current loops exerted by the magnetic field. The magnetic torque is ( ) sin 10 2 sin . B B R LIB τ μ θ θ = = Note that θ is the angle between the magnetic dipole μ G of the loops and . B G This is balanced by the gravitational torque . G τ Using the figure above, ( ) 2 3 sin sin sin . G G F R R L gR R L g τ θ π ρ θ π ρ θ = = = Equilibrium requires 3 2 20 sin sin 20 B G RLIB R L g R g I B τ τ θ π ρ θ π ρ = = = (b) Evaluating, ( ) ( )( ) ( ) 2 3 2 0.025 m 2500 kg/m 9.8 m/s 9.6 A 20 0.25 T I π = = Assess: This is a reasonable current. Note that the angle of the incline does not matter since both the gravitational and magnetic torques have the same dependence on θ .
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33.72. Model: Assume that the 20 cm is large compared with the effective radius of the magnetic dipole (bar magnet). Solve: From Equation 33.9, the on-axis field of a magnetic dipole at a distance z from the dipole is 0 dipole 3 2 4 B z μ μ π = The magnetic dipole moment can be obtained from the magnitude of the torque. We have ( ) 2 0.075 Nm 0.15 Am sin 0.50 T sin90 B B τ τ μ μ θ = × = = = ° G G G Thus, the magnetic field produced by the magnet is ( ) ( ) ( ) 2 7 6 dipole 3 2 0.15 Am 10 T m/A 3.8 10 T 0.20 m B = = ×
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33.73. Model: A current loop produces a magnetic field. Visualize: Solve: The field at the center of a current loop is B loop = μ 0 I /2 R . The electron orbiting an atomic nucleus is, on average, a small current loop. Current is defined as I = Δ q / Δ t . During one orbital period T , the charge Δ q = e goes around the loop one time. Thus the average current is I avg = e / T . For circular motion, the period is ( ) 11 19 16 3 avg 6 16 2 5.3 10 m 2 1.60 10 C 1.514 10 s 1.057 10 A 2.2 10 m/s 1.514 10 s R T I v π π × × = = = × = = × × × Thus, the magnetic field at the center of the atom is ( )( ) ( ) 7 3 0 avg center 11 4 10 T m/A 1.057 10 A 12.5 T 2 2 5.3 10 m I B R π μ × × = = = ×
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33.74. Model: The superposition principle holds for forces and torques. Visualize: Solve: (a) A graph of B vs x is shown above. (b) The magnetic force on a segment of wire Δ x in length located at x i is ( ) i i F IB x x Δ = Δ The total force on the wire is ( ) net 0 0 0 1 2 L i i x F F IB x x IBdx IB dx ILB L = ΣΔ = Σ Δ = = The right-hand rule gives the direction of the force along the + y -axis, so net 0 1 ˆ . 2 F ILB j = G (c) Each segment of wire experiences a torque about the origin of ( ) sin90 . i i i i i r F x IB x x τ = Δ ° = Δ The total torque on the wire is thus ( ) 2 2 0 0 0 1 3 L i i i x x IB x x IB dx IL B L τ τ = Σ = Σ Δ = The direction of the torque is along the + z -axis by the right-hand rule, indicating that the magnetic field will exert a torque causing the wire to rotate counter clockwise in the xy -plane.
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33.75.
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