Q. A charged particle (charge q) is moving in a circle of radius $R$ with uniform speed $v.$ The associated magnetic moment $\mu$ is given by
A) $\frac{qvR}{2}$
B) $qvR^{2}$
C) $\frac{qvR^{2}}{2}$
D) $qvR$
Explanation
(a) As revolving charge is equivalent to a current, so
$I = q\ f = q \times \frac{\omega}{2\pi}$
But $\omega = \frac{v}{R}$
Where $R$ is radius of circle and $v$ is uniform speed of charged particle.
Therefore, $I = \frac{qv}{2\pi R}$
Now, magnetic moment associated with charged particle is given by
$\mu = IA = I \times \pi R^{2}$
or $\mu = \frac{qv}{2\pi R}\ \times \pi R^{2} = \frac{1}{2}\ qvR$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.