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Question
Consider a non-conducting ring of radius r and mass m that has a total charge qdistributed 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.
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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 =mr2
so , I =mr2ω
`⇒ ωr^2 = 1/m`
Putting this value in equation (i), we get:
`mu =(ql)/(2m)`
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