Discussion Forum

 A particle of charge q and mass m moves in a circular orbit of radius r with angular speed ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

Answer:

Explanation:

 The angular momentum L of the particle is given by

 $L=mr^{2}\omega$   where  $\omega =2\pi n$

 $\therefore$  Freequency n= $\frac{\omega}{2\pi}$

 Further  $i= q\times n= \frac{\omega q}{2\pi}$

 Magnetic momemt ,  $M=i A= \frac{\omega q}{2\pi}\times \pi r^{2}$

 $\therefore$    $M= \frac{\omega q r^{2}}{2}$

So,  $\frac{M}{L}= \frac{\omega q r^{2}}{2mr^{2}\omega}=\frac{q}{2m}$