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

The correct option is A (1/2)qωr²

Let us take an element at an angle θ subtending an angle dθ.
The charge dq on the element can be written as dq = q/(2π)dθ
We know that i = dq/dt = q/(2π) × dθ/dt
The time dt can be written as dt = dθ/ω
Hence i = qω/(2π)
Magnetic moment = iA
Hence Magnetic moment M = (qω/(2π)) × πr² = (1/2)qωr²