# 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 - Physics

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

#### 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
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 12 Magnetic Field
Q 60 | Page 234