(N/A) When a charged particle having charge $q$ and velocity $\vec{v}$ enters a uniform magnetic field $\vec{B}$,it experiences a magnetic Lorentz force given by:
$\vec{F} = q(\vec{v} \times \vec{B})$
Since the velocity $\vec{v}$ is perpendicular to the magnetic field $\vec{B}$,the force $\vec{F}$ acts perpendicular to both $\vec{v}$ and $\vec{B}$. This force acts as a centripetal force,causing the particle to move in a uniform circular path.
For uniform circular motion,the magnetic force provides the necessary centripetal force:
$\text{Centripetal force} = \text{Magnetic force}$
$\frac{mv^2}{r} = qvB$
Solving for the radius $r$:
$r = \frac{mv}{qB} \quad \dots (1)$
Since linear momentum $p = mv$,we can write:
$r = \frac{p}{qB}$
This indicates that as the momentum increases,the radius of the circular path also increases.
Using the relation between linear velocity and angular velocity,$v = \omega r$:
$\frac{v}{\omega} = \frac{mv}{qB}$
$\omega = \frac{qB}{m}$
where $\omega$ is the angular frequency.
The frequency of revolution $f$ is given by:
$f = \frac{\omega}{2\pi} = \frac{qB}{2\pi m}$
This is known as the cyclotron frequency,which is independent of the speed of the particle and the radius of the orbit. This principle is used in a cyclotron to accelerate charged particles.