A nonconducting ring of radius r and mass m carries a total charge q that is uniformly distributed over it. The ring is rotated about its axis with an angular speed . Determine the magnetic moment of the ring.

Step-by-step solution

The rotating charge on the ring constitutes an equivalent electric current. The equivalent current is given by i = q T , where T is the time period of rotation. Since the angular speed is , the time period is T = 2 . Therefore, the current is i = q 2 . The magnetic moment of a current loop is the product of the current and the area of the loop. The area of the ring is A = r^2 . Substituting the values, we get = i A = ( q 2 ) ( r^2) = q r^2 2 .