Question
A particle of charge $$q$$ and mass $$m$$ moves in a circular orbit of radius $$r$$ with angular speed $$\omega .$$ The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on
A.
$$\omega $$ and $$q$$
B.
$$\omega ,$$ $$q$$ and $$m$$
C.
$$q$$ and $$m$$
D.
$$\omega $$ and $$m$$
Answer :
$$q$$ and $$m$$
Solution :
The angular momentum $$L$$ of the particle is given by $$L = m{r^2}\omega \,{\text{where}}\,\,\omega = 2\pi n.$$
$$\eqalign{
& \therefore {\text{Frequency}}\,n = \frac{\omega }{{2\pi }};{\text{Further}}\,i = q \times n = \frac{{\omega q}}{{2\pi }} \cr
& {\text{Magnetic}}\,{\text{moment,}}\,M = iA = \frac{{\omega q}}{{2\pi }} \times \pi {r^2}; \cr
& \therefore M = \frac{{\omega q{r^2}}}{2}\,{\text{so,}}\,\frac{M}{L} = \frac{{\omega q{r^2}}}{{2m{r^2}\omega }} = \frac{q}{{2m}} \cr} $$