Model the spinning charged disk forms a series of

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Model: The spinning charged disk forms a series of concentric current loops. Solve: The magnetic field at the center of a current loop of radius r is taken from Example 33.5 with z = 0: 0 loop 2 I B r μ = Consider a ring of charge on the disk at radius r i with width Δ r . The charge on the ring is 2 2 i r r Q Q R π π Δ Δ = . The time for one revolution of the ring is 2 t π ω Δ = , so the ring can be considered a current loop with 2 2 2 2 i i i Q Qr r Q I r r t R R ω ω π π Δ Δ ⎛ = = = Δ Δ Each ring contributes a field ( B loop ) i , so by superposition ( ) 0 0 center loop 2 0 0 2 0 2 2 2 2 i i i R I Q B B r r R Q Q dr R R μ μ ω π μ ω μ ω π π = = = Δ =
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33.80. Model: The net force is the sum of the forces on individual segments of the wire. Visualize: Please refer to figure CP33.80. Solve: The magnetic field due to the wire carrying the current I 1 = 10 A at a distance r from the wire is ( ) 0 1 2 I B r r μ π = The force on a segment of the horizontal wire located at x i of length Δ x and a distance r i from the 10 A current is ( ) ( ) 0 1 2 2 2 sin sin 2 i i i i i i I I F I xB r I xB r x r μ θ θ π Δ = Δ = Δ = Δ Note that only the perpendicular component of the magnetic field contributes to the force on the wire. From the figure, we can identify ( ) 2 2 sin and 0.010 m i i i i i x r x r θ = = + . The total force is the sum of the forces on each segment: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 0.050 m 0 1 2 0 1 2 2 2 2 0 m 2 2 0.050 m 2 2 0 1 2 0 1 2 2 0 m 7 5 2 2 0.010 m 0.050 m 0.010 m 1 ln 0.010 m ln 2 2 2 0.010 m 10 Tm/A 10 A 5.0 A ln 26 1.63 10 N i i i I I x I I xdx F F x r x I I I I x μ μ π π μ μ π π = Δ = Δ + + = + = = = × Assess: This is a fairly small force, but can be understood with the realization that at the closest points between the wires, where B is strongest, its direction is nearly parallel to the horizontal wire, so the force it exerts on the wire is nearly zero there.
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33.81. Model: The magnetic field is that of a conducting wire that has a nonuniform current density. Visualize: Solve: (a) Consider a small circular disk of width dr at a distance r from the center. The current through this disk is ( ) 2 0 0 2 2 r J r dr di JdA J r dr R R π π = = = Integrating this expression, we get 3 2 2 0 0 0 0 2 0 0 2 2 2 3 3 3 2 R R J J r J R I I di r dr J R R R π π π π = = = = = (b) Applying 0 through B d s I μ = G G ú to the circular path of integration, we note that the wire has perfect cylindrical symmetry with all the charges moving parallel to the wire. So, the magnetic field must be tangent to circles that are concentric with the wire. The enclosed current is the current within radius r . Thus, 2 0 0 0 0 0 2 r r J Bds di r dr R π μ μ = = ( ) 3 3 2 0 0 0 0 2 3 2 2 3 2 3 2 3 2 J r I r Ir B r B R R R R μ π μ π μ π π π ⎞⎛ = = = ⎟⎜ ⎠⎝ (c) At r = R , 2 0 0 3 2 2 IR I B R R μ μ π π = = This is the same result as obtained in Example 33.3 for the magnetic field of a long straight wire.
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33.82.
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