A charged particle (charge q ) is moving in a circle of radius R...

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