# A spherical shell of radius R and uniformly charged with charge Q is rotating about its axis with frequency f. Find the magnetic moment of the sphere?

##### 1 Answer
Jan 4, 2018

Uniformly charged spherical shell of radius $R$ carries a total charge $= Q$
Hence it has surface charge density

sigma=Q/(4πR^2)

It rotates about its axis with frequency$= f$
$\therefore$ Its angular velocity $\omega = 2 \pi f$
Suppose the angular velocity $\vec{\omega} = \omega \hat{z}$

To find the magnetic moment of spinning shell we can divide
it into infinitesimal charges.
Using spherical polar coordinates $\left(\rho , \phi , \theta\right)$ with origin at the center of the spherical shell, we consider infinitesimal area $\mathrm{dS}$ of a circular ring of infinitesimal width and of radius $r$ located at a distance $R$ from the origin on the surface of sphere.
Here $\vec{r} = R \sin \theta \hat{r}$ and coordinate $\rho = R$, is constant for each elemental area.

$\mathrm{dq} = \sigma \mathrm{dS} = \sigma {R}^{2} \sin \theta d \theta \mathrm{dp} h i$

Current in the ring is given by

$\mathrm{dI} = \left(\mathrm{dq}\right) \times f = \left(\sigma {R}^{2} \sin \theta d \theta \mathrm{dp} h i\right) \times \frac{\omega}{2 \pi}$
$\implies \mathrm{dI} = \frac{\omega \sigma {R}^{2} \sin \theta d \theta \mathrm{dp} h i}{2 \pi}$

Now the magnetic dipole moment of ring is given by the expression in terms of position vector $\vec{r}$ and current density $\vec{J}$ as

$\mathrm{dv} e c m = \frac{1}{2} {\int}_{\text{ring}} \vec{r} \times \vec{J}$
$\implies \mathrm{dv} e c m = \frac{1}{2} \mathrm{dI} {\int}_{\text{ring}} \vec{r} \times \mathrm{dv} e c l$
where $\mathrm{dv} e c l$ is element length of the ring.

Line integral becomes equal to the circumference of ring$= 2 \pi R \sin \theta$

$\therefore \mathrm{dv} e c m = \mathrm{dI} \left(\pi {R}^{2} {\sin}^{2} \theta\right) \hat{z}$

Inserting value of $\mathrm{dI}$ we get

$\mathrm{dv} e c m = \frac{\omega \sigma {R}^{2} \sin \theta d \theta \mathrm{dp} h i}{2 \pi} \left(\pi {R}^{2} {\sin}^{2} \theta\right) \hat{z}$
$\implies \mathrm{dv} e c m = \frac{\vec{\omega} \sigma {R}^{4} {\sin}^{3} \theta d \theta \mathrm{dp} h i}{2}$

Total magnetic moment is integral with respect to both variables within respective limits

$\vec{m} = \frac{\vec{\omega} \sigma {R}^{4}}{2} {\int}_{0}^{\pi} {\sin}^{3} \theta d \theta {\int}_{0}^{2 \pi} \mathrm{dp} h i$
$\implies \vec{m} = \frac{\vec{\omega} \sigma {R}^{4}}{2} \times \frac{4}{3} \times 2 \pi$
$\implies \vec{m} = \frac{4}{3} \pi \sigma {R}^{4} \vec{\omega}$

Rewriting in terms of charge $Q$

vecm=4/3pi(Q/(4πR^2))R^4vecomega
$\vec{m} = \frac{Q}{3} {R}^{2} \vec{\omega}$

.-.-.-.-.-.-.-.-

Reference figure for $\mathrm{dS}$

Using online integral calculator

${\int}_{0}^{\pi} {\sin}^{3} \theta d \theta = | {\cos}^{3} \frac{\theta}{3} - \cos \theta {|}_{0}^{\pi} = \frac{4}{3}$