Physics - Magnetism Question with Solution | TestHub

As revolving charge is equivalent to a current, so

$I=q f=q\times \frac{\omega }{2\mathrm{\pi }}$

But$\omega =\frac{v}{R}$

Where$R$is radius of circle and$v$is uniform speed of charged particle.

Therefore,$I=\frac{qv}{2\mathrm{\pi }R}$

Now, magnetic moment associated with charged particle is given by

$\mathrm{\mu }=IA=I\times \mathrm{\pi }{R}^2$

or$\mathrm{\mu }=\frac{qv}{2\mathrm{\pi }R} \times \mathrm{\pi }{R}^2=\frac{1}{2} qvR$