- #1

\Tau_{\mu\ni}

- 3

- 0

## 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*?