Question

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 $

Answer 1

Answer: (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, $I=\frac{q v}{2 \pi R}$
Now, magnetic moment associated with charged particle is given by
$\mu =I A=I \times \pi R^{2}$
or $\mu =\frac{q v}{2 \pi R} \times \pi R^{2}$
$=\frac{1}{2} q v R$