Question

Consider a non-conducting plate of radius r and mass m that has a charge q distributed uniformly over it. The plate is rotated about its axis with an angular speed ω. Show that the magnetic moment µ and the angular momentum l of the plate are related as $\mathrm{\mu }=\frac{q}{2m}l$.

Answer 1

Given:
Radius of the ring = r
Mass of the ring = m
Total charge of the ring = q
Angular speed, $\omega =\frac{2\pi }{T}\Rightarrow T=\frac{2\pi }{\omega }$
Current in the ring, $i=\frac{q}{T}=\frac{q\omega }{2\pi }$
For the ring of area A with current i, magnetic moment,
µ = niA = ia [n = 1]
$=\frac{q\omega }{2\mathrm{\pi }}\times \mathrm{\pi }{r}^2\mathit{=}\frac{q\omega r^2}{2}$ ...(i)
Angular momentum, l = $I\omega$,
where I is the moment of inertia of the ring about its axis of rotation.
$I=m{r}^2$
So, $l=m{r}^2\omega$
$\Rightarrow \omega r^2=\frac{l}{m}$
Putting this value in equation (i), we get:
$\mu =\frac{ql}{2m}$