A particle of charge ' $\mathrm{q}$ ' and mass ' $\mathrm{m}$ ' moves in a circular orbit of radius ' $r$ ' with angular speed ' $\omega$ '. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

Question

Aicharya · Accepted Answer

Angular momentum $(\mathrm{L})=\mathrm{I} \omega=\mathrm{mr}^2 \omega$ Magnetic moment $(M)=\pi r^2 q f=\pi r^2 q \frac{\omega}{2 \pi}=\frac{1}{2} q \omega r^2$ $	herefore \frac{\mathrm{M}}{\mathrm{L}}=\frac{\mathrm{q}}{2 \mathrm{~m}}$

A particle of charge ' $\mathrm{q}$ ' and mass ' $\mathrm{m}$ ' moves in a circular orbit of radius ' $r$ '…

Solution