- #1
six7th
- 14
- 0
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?
Last edited:
