A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity `omega` with respect to normal axis then the magnetic moment of the loop is :

Options

A. `3/2 q omega r^(2)`
B. `1/2 q omega r^(2)`
C. `q omega r^(2)`
D. `4/3 q omega r^(2)`

Answer

b

Explanation

Current in the loop `I = q/T = (q omega)/(2 pi)`
Area of loop ` A = pi r^(2)`
Magnetic moment (M) of loop is defined as multiplication of current and area of the loop.
`M = IA = (q omega)/(2pi) pi r^(2) = (qomega r^(2))/2 `