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six7th
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Homework Statement
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 μ is related to the angular momentum L by the relation:
\textbf{μ} = \frac{5q}{6mc}\textbf{L}
Homework Equations
μ = IA
I = \frac{2}{5}MR^2
σ = \frac{q}{4πR^2}
\frac{1}{T} = \frac{ω}{2π}
The Attempt at a Solution
Split sphere into thin segments, each one has magnetic moment:
dμ = dI \cdot dA
where dA is the area enclosed by the segment:
dA = 2π r dr
The current dI on each segment is:
dI = \frac{dq}{T} = \frac{ω}{2π} dq
The charge on each segment is:
dq = σ da
Where da is the surface area of the segment:
da = r^2\sinθ dθ d\phi
This results in
dq = σ \cdot r^2\sinθ dθ d\phi = \frac{q}{4πR^2} r^2\sinθ dθ d\phi
We now have an expression for dI:
dI = \frac{ωqr^2\sinθ}{8π^2R^2} dθ d\phi
Now we can calculate the moment of each segment:
dμ = \frac{ωqr^2\sinθ}{8π^2R^2} 2πr dr d\phi dθ
Integrating this over the whole sphere gives
μ = \frac{ωq}{4πR^2} \int^{R}_{0} r^3 dr \int^{π}_{0} \sinθ dθ \int^{2π}_{0} d\phi
μ = \frac{wqR^2}{8}
Obviously this is not correct, the answer I should be getting is
μ = \frac{wqR^2}{3}
Where am I going wrong?
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