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.
$$qv{R^2}$$
C.
$$\frac{{qv{R^2}}}{2}$$
D.
$$qvR$$
Answer :
$$\frac{{qvR}}{2}$$
Solution :
As revolving charge is equivalent to a current, so $$ = qf = q \times \frac{\omega }{{2\pi }}$$
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
$$\mu = iA = i \times \pi {R^2}\,\,{\text{or}}\,\,\mu = \frac{{qv}}{{2\pi R}} \times \pi {R^2} = \frac{1}{2}qvR$$
Alternative
Current produced due to circular motion of charge $$q$$ is given by
$$i = \frac{q}{T}\,\,\left[ {T = {\text{Time period}}\,{\text{of revolution}}} \right]$$
Now, $$T = \frac{{2\pi R}}{v}$$
So, $$i = \frac{q}{{2\pi R}} \times v$$
Now, magnetic moment $$\left( \mu \right)$$ is given by
$$\mu = iA$$
So, $$\mu = \frac{{qv}}{{2\pi R}} \times \pi {R^2}$$
$$ \Rightarrow \mu = \frac{{qvR}}{2}$$