Question
A charged particle of charge $$q$$ and mass $$m$$ enters perpendicularly in a magnetic field $$B.$$ Kinetic energy of the particle is $$E,$$ then frequency of rotation is
A.
$$\frac{{qB}}{{m\pi }}$$
B.
$$\frac{{qB}}{{2\pi m}}$$
C.
$$\frac{{qBE}}{{2\pi m}}$$
D.
$$\frac{{qB}}{{2\pi E}}$$
Answer :
$$\frac{{qB}}{{2\pi m}}$$
Solution :
Magnetic force = centripetal force
i.e. $$qvB = \frac{{m{v^2}}}{r}$$
or $$qvB = mr{\omega ^2}\,\,\left( {v = r\omega } \right)$$
or $${\omega ^2} = \frac{{qvB}}{{mr}} = \frac{{q\left( {r\omega } \right)B}}{{mr}}$$
Angular frequency, $$\omega = \frac{{qB}}{m}$$
If $$\nu $$ is the frequency of rotation, then
$$\eqalign{
& \omega = 2\pi \nu \Rightarrow \nu = \frac{\omega }{{2\pi }} \cr
& \therefore \nu = \frac{{qB}}{{2\pi m}} \cr} $$
NOTE
In the resultant expression $$\frac{q}{m}$$ is known as specific charge. It is sometimes denoted by $$\alpha .$$ So, in terms of $$\alpha ,$$ the above formula can be written as
$$\omega = B\alpha \,\,{\text{and}}\,\,\nu = \frac{{B\alpha }}{{2\pi }}$$