Question

A particle of mass $m$ and charge $Q$ is accelerating in a cyclotron. If magnetic field is $B$ and radius of cyclotron is $r$ then the kinetic energy of particle-
(A). $\dfrac{QBr}{2m}$
(B). $\dfrac{{{Q}^{2}}{{B}^{2}}r}{2m}$
(C). $\dfrac{{{Q}^{2}}{{B}^{2}}{{r}^{2}}}{2m}$
(D). $\dfrac{{{Q}^{2}}{{B}^{2}}{{r}^{2}}}{m}$

Answer 1

The cyclotron accelerates charged particles at very high speeds by the application of Lorentz force acting on the particles due to magnetic field. The particle follows a spiral path. The kinetic energy of the particle is due to its spiral motion; therefore we can calculate kinetic energy by substituting velocity in terms of angular velocity or frequency in the relation between kinetic energy mass and velocity.

Formulas Used:
$f=\dfrac{qB}{2\pi m}$
$K=\dfrac{1}{2}m{{v}^{2}}$
$v=\omega r$

Complete answer:
A cyclotron is a device that accelerates charged particles at very high speeds. It works on the principle that a charged particle experiences a force due to the perpendicular magnetic field which makes it move in a spiral motion.
The frequency of the cyclotron is given by-
$f=\dfrac{qB}{2\pi m}$ - (1)
Here,$f$ is the frequency of cyclotron
$q$ is the magnitude of charge on the charged particle, given $q=Q$
$B$ is the magnitude of magnetic field
$m$ is the mass of the charged particle
The kinetic energy of the charged particle will be-
$K=\dfrac{1}{2}m{{v}^{2}}$ - (2)
Here,$K$ is the kinetic energy
$v$ is the velocity of the charged particle

We know that,
$v=\omega r$
Here, $\omega$ is the angular velocity
$\omega =2\pi f$ , therefore from eq (1), the value of $\omega$is-

& \omega =2\pi \times \dfrac{QB}{2\pi m} \\\ & \Rightarrow \omega =\dfrac{QB}{m} \\\ \end{aligned}$$ Substituting the value of $$\omega $$in$$v$$, we get, $$v=\dfrac{QBr}{m}$$ Substituting in eq (2), we get, $$\begin{aligned} & K=\dfrac{1}{2}m{{\left( \dfrac{QBr}{m} \right)}^{2}} \\\ & K=\dfrac{{{Q}^{2}}{{B}^{2}}{{r}^{2}}}{2m} \\\ \end{aligned}$$ Therefore, the kinetic energy of the charged particle in a cyclotron is $$\dfrac{{{Q}^{2}}{{B}^{2}}{{r}^{2}}}{2m}$$. **So, the correct option is (D).** **Note:** In a cyclotron, the frequency is independent of the time period. A cyclotron can only accelerate charged particles. Here, the centripetal force is provided by the Lorentz force acting on the charged particle due to the magnetic field. It is used in many nuclear reactions and is also used in particle therapy to cure cancer.