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. $\frac{q vR }{2}$
• B. $q v R ^{2}$
• C. $\frac{q vR ^{2}}{2}$
• D. $qvR$

Answer 1

Answer: (A)

$\frac{q vR }{2}$

Answer 2

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