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Introductory Physics Homework Help
Magnetic moment of a solid, uniformly charged ball
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[QUOTE="six7th, post: 4498409, member: 351050"] [h2]Homework Statement [/h2] Show for a solid spherical ball of mass m rotating through its centre with a charge q uniformly distributed on the surface of the ball that the magnetic moment [B]μ[/B] is related to the angular momentum [B]L[/B] by the relation: [itex]\textbf{μ} = \frac{5q}{6mc}\textbf{L}[/itex][h2]Homework Equations[/h2] [itex]μ = IA [/itex] [itex]I = \frac{2}{5}MR^2[/itex] [itex]σ = \frac{q}{4πR^2}[/itex] [itex]\frac{1}{T} = \frac{ω}{2π}[/itex][h2]The Attempt at a Solution[/h2] Split sphere into thin segments, each one has magnetic moment: [itex]dμ = dI \cdot dA [/itex] where dA is the area enclosed by the segment: [itex]dA = 2π r dr[/itex] The current dI on each segment is: [itex]dI = \frac{dq}{T} = \frac{ω}{2π} dq[/itex] The charge on each segment is: [itex]dq = σ da[/itex] Where da is the surface area of the segment: [itex]da = r^2\sinθ dθ d\phi[/itex] This results in [itex]dq = σ \cdot r^2\sinθ dθ d\phi = \frac{q}{4πR^2} r^2\sinθ dθ d\phi[/itex] We now have an expression for dI: [itex]dI = \frac{ωqr^2\sinθ}{8π^2R^2} dθ d\phi[/itex] Now we can calculate the moment of each segment: [itex]dμ = \frac{ωqr^2\sinθ}{8π^2R^2} 2πr dr d\phi dθ[/itex] Integrating this over the whole sphere gives [itex]μ = \frac{ωq}{4πR^2} \int^{R}_{0} r^3 dr \int^{π}_{0} \sinθ dθ \int^{2π}_{0} d\phi [/itex] [itex]μ = \frac{wqR^2}{8}[/itex] Obviously this is not correct, the answer I should be getting is [itex]μ = \frac{wqR^2}{3}[/itex] Where am I going wrong? [/QUOTE]
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Magnetic moment of a solid, uniformly charged ball
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