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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 `mu = q/(2 m)l`
Solution
Given:
Radius of the ring = r
Mass of the ring = m
Total charge of the ring = q
Angular speed, w = `(2pi)/T ⇒ T = (2pi)/w`
Current in the ring, `i = q/t = (qw)/(2pi)`
For the ring of area A with current i, magnetic moment,
`mu = niA =ia [n = 1]`
= `(qw)/(2pi)xx pir^2 = (qwr^2)/(2)`
Angular momentum, `l = Iw`,
where I is the moment of inertia of the ring about its axis of rotation.
`I = mr^2`
`so, l =mr^2w`
⇒ `wr^2 = 1/m=`
Putting this value in equation (i), we get:
`mu = (ql)/(2m)`
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