Sum

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`

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#### 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)`

Concept: Force on a Current - Carrying Conductor in a Uniform Magnetic Field

Is there an error in this question or solution?

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