Question

Your goal is to accelerate the particles to kinetic energy K. What minimum radius R of the cyclotron is required? Express your answer in terms of m, q, B, and K.

Answer 1

Concepts and reason

The concept of the cyclotron is used here.

For solving this question, first, substitute the expression for angular velocity and linear velocity in the kinetic energy formula and then substitute the formula for the angular velocity of a particle in a cyclotron in the expression for kinetic energy.

Fundamentals

The cyclotron is used to accelerate a particle. It is made up of two semicircular plates through which the magnetic field is passed in the perpendicular direction. The charged particle is made to enter the cyclotron in the direction perpendicular to both plane of plates and the magnetic field. An electric field is also applied to the cyclotron which changes its direction after a fixed interval of time. When the particle enters the cyclotron, the magnetic field bends the path of particle continuously and makes it follow a circular path. The electric field accelerates the particle and because of this, the charged particle is forced to move in a bigger circle.

The angular velocity of the particle in a cyclotron is given as:

$\omega = \frac{{qB}}{m}$

Here, $\omega$ is the angular velocity, q is the charge of the particle, B is the strength of magnetic field and m is the mass of the particle.

The relation between linear velocity and angular velocity is given as:

$v = \omega r$ …… (1)

Here, v is the linear velocity and r is the radius of the path in which particle is rotating.

The kinetic energy of the particle is given as:

$K = \frac{1}{2}m{v^2}$ …… (2)

Here, K is the kinetic energy of the particle.

The radius in terms of charge, magnetic field strength, mass and kinetic energy is determined as follows:

Substitute $\frac{{qB}}{m}$ for $\omega$ in equation (1),

$v = \frac{{qBr}}{m}$

Substitute $\frac{{qBr}}{m}$ for v in equation (2),

$\begin{array}{l}\\K = \frac{1}{2}m{\left( {\frac{{qBr}}{m}} \right)^2}\\\\K = \frac{{{q^2}{B^2}{r^2}}}{{2m}}\\\end{array}$

Rearranging the above equation.

$\begin{array}{l}\\{r^2} = \frac{{2mK}}{{{q^2}{B^2}}}\\\\r = \sqrt {\frac{{2mK}}{{{q^2}{B^2}}}} \\\\r = \frac{{\sqrt {2mK} }}{{qB}}\\\end{array}$

Ans:

The radius in terms of charge, magnetic field strength, mass and kinetic energy is $\frac{{\sqrt {2mK} }}{{qB}}$