Question

Consider a non-conducting ring of radius r and mass m that has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speed ω. (a) Find the equivalent electric current in the ring. (b) Find the magnetic moment µ of the ring. (c) Show that $\mathrm{\mu }=\frac{q}{2m}$ l, where l is the angular momentum of the ring about its axis of rotation.

Answer 1

Given:
Radius of the ring = r
Mass of the ring = m
Total charge of the ring = q
(a) Angular speed, $\omega =\frac{2\mathrm{\pi }}{T}\Rightarrow T=\frac{2\mathrm{\pi }}{\omega }$
Current in the ring, $i=\frac{q}{T}=\frac{q\omega }{2\mathrm{\pi }}$

(b) For a ring of area A with current i, magnetic moment,
µ = niA = ia [n = 1]
$=\frac{q\omega }{2\mathrm{\pi }}\times \mathrm{\pi }{r}^2\mathit{}\mathit{=}\mathit{}\frac{q\omega r^2}{2}$ ...(i)

(c) Angular momentum, l = $I\omega$,
where I is moment of inertia of the ring about its axis of rotation.
$I=m{r}^2$
So, $I=m{r}^2\omega$
$\Rightarrow \omega r^2=\frac{l}{m}$
Putting this value in equation (i), we get:
$\mu =\frac{ql}{2m}$