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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:

[itex]\textbf{μ} = \frac{5q}{6mc}\textbf{L}[/itex]

## Homework Equations

[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]

## The Attempt at a Solution

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?

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