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# Zconsider 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 - Physics

Sum

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 =mr
so ,  I =mr2ω
⇒ ωr^2 = 1/m
Putting this value in equation (i), we get:
mu =(ql)/(2m)

Concept: Force on a Moving Charge in Uniform Magnetic and Electric Fields
Is there an error in this question or solution?
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 12 Magnetic Field
Q 59 | Page 234
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