Physics - Magnetism Question with Solution | TestHub

Magnetic moment of the revolving charge is given by $M=IA$

Current $I=\frac{q}{T}=\frac{q\omega }{2\pi }$

Area $A=\pi r^2$

$M=\left(\frac{q\omega }{2\pi }\right)(\pi r^2)=\frac{1}{2}q\omega r^2$

Angular momentum of the particle is $L=mvr=m(\omega r)r=m\omega r^2$

The ratio of the magnitude of magnetic moment to that of angular momentum is $\frac{M}{L}=\frac{\frac{1}{2}q\omega r^2}{m\omega r^2}=\frac{q}{2m}$

This ratio depends only on $q$ and $m$.

Answer: q and m