Physics - Magnetism Question with Solution | TestHub

As revolving charge is equivalent to a current, so $\begin{array}{rl}& I=qf=q\times \frac{\omega }{2\pi }\\ But& \omega =\frac{v}{R}\end{array}$ where $R$ is radius of circle and $v$ is uniform speed of charged particle. Therefore, $I=\frac{qv}{2\pi R}$ Now, magnetic moment associated with charged particle is given by or $\begin{array}{rl}\mu & =IA=I\times \pi R^2\\ or\mu & =\frac{qv}{2\pi R}\times \pi R^2\\ & =\frac{1}{2}qvR\end{array}$