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\Tau_{\mu\ni}
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Homework Statement
So, we can presumably write that m = g L, where L is the angular momentum, g the ratio wanted, and m magnetic dipole moment of an axially symmetric body. Total mass is M, total charge Q, mass density [itex]\rho_m(r)=\frac{M}{Q}\rho_e(r)[/itex], where [itex]\rho_e(r)[/itex] is charge density.
Homework Equations
Moment of inertia can also be written [itex]I_\omega=\int \rho_m d^2 d\tau[/itex] where d is the distance from the axis of symmetry.
The Attempt at a Solution
I guess the dipole moment is in the direction of the axis of the symmetry, as is the angular velocity and it can be written:
m = g I[itex]_\omega [/itex] ω
RHS:
= g ω [itex]\hat{z}[/itex] I[itex]_\omega [/itex]
LHS:
m = I [itex] \int d\vec{a} [/itex]
Here is where I guess I'm having conceptual problems. I is the total current, part of which should be a current flowing through a differential circular loop inside the body, at distance d, that is to say:
I = [itex]\int \left| v \right| \rho_e d\tau [/itex] , where [itex]\left|v\right|[/itex] is the module of radial speed of a differential volume element (at distance d) that can be written: [itex]\left|v\right|=\left|\omega\right|\left|r\right|[/itex]sin[itex]\theta [/itex]= ωd. [itex] \int d\vec{a} [/itex] of a differential loop in question is simply [itex] d^2 \pi[/itex]. So, I am getting:
g ω [itex]\int \rho_m d^2 d\tau[/itex] =ω [itex]\int d^3 \pi \frac{M}{Q} \rho_m d\tau [/itex]
Where did I go wrong, how to find g?