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 `pi = (q)/(2m)` *l,* where *l* is the angular momentum of the ring about its axis of rotation.

#### Solution

Given:

Radius of the ring = r

Mass of the ring = m

Total charge of the ring = q

(a) Angular speed, ω = `(2pi)/T` ⇒ T = `(2pi)/(ω)`

Current in the ring i ,= `q/T = (qw)/(12pi)`

(b) For a ring of area A with current i, magnetic moment,

`(qw)/(2pi)xxpir^2= (qwr^2)/2`....(i)

(c) Angular momentum, l = Iω

where *I* is moment of inertia of the ring about its axis of rotation.

I =mr^{2 }

so , I =mr^{2}ω

`⇒ ωr^2 = 1/m`

Putting this value in equation (i), we get:

`mu =(ql)/(2m)`