Q106 / 120 · VITEEE 2016 · Physics
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
- A$\omega$ and $q$
- B$\omega, q$ and $m$
- Correct answerC$q$ and $m$
- D$\omega$ and $m$
Solution
The angular momentum $\mathrm{L}$ of the particle is given by
$L=m r^{2} \omega$ where $\omega=2 \pi n$
$\therefore$ Frequency $n=\frac{\omega}{2 \pi}$;
Further $i=q \times n=\frac{\omega q}{2 \pi}$
Magnetic moment, $M=i A=\frac{\omega q}{2 \pi} \times \pi r^{2}$;
$\therefore M=\frac{\omega q r^{2}}{2}$
So, $\frac{M}{L}=\frac{\omega q r^{2}}{2 m r^{2} \omega}=\frac{q}{2 m}$
Hence, the correct answer is option: 'C'.