Question

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

Answer 1

it seems your question is ----> A charge q is spread uniformly over an insulated loop of radius r. If it is rotatedwith an angular velocity ω with respect to normal axis then the magnetic moment of the loop is :

Let linear charge density of loop is $\lambda$
cut an elementary length of length dl,
so, charge on dq= $\lambda dl$

or, $dq=\frac{q}{2πr}dl$

now, magnetic moment, M = dq/dt × A
= dq × n × A
where n is frequency e.g., $n=\frac{\omega}{2\pi}$

so, M = q/2πr dl × ω/2π × πr²

M = ωqr/4π $\int dl$

M = ωqr/4π × 2πr

M = 1/2 ωqr²