Question

A) If you can build a cyclotron with twice the radius, by what factor would the allowed maximum particle energy increase? Assume that the magnetic field remains the same.

B)If you can build a cyclotron with twice the radius and with the magnetic field twice as strong, by what factor would the allowed maximum particle energy increase?

both are multiple choice with as options.

2

4

8

16

Answer 1

Concepts and reason

The required concepts to solve the problem are centripetal force, Lorentz force and kinetic energy.

Firstly, find the relation between the radius of the cyclotron and the velocity of the particle. From this velocity, find the increase in the energy of the particle. Since, the magnetic field changes along with the radius so finally find the change in velocity and also the increase in the energy of the particle.

Fundamentals

A cyclotron is a particle accelerator which is used to accelerate a charged particle by an alternating current in a magnetic field. The cyclotron frequency can be found using the equation

$f = \frac{{qB}}{{2\pi m}}$

Here, $q$ is the charge of the particle, $B$ is the magnetic field strength and $m$ is the mass of the charged particle.

The motion of the charged particle in the cyclotron is usually circular. So, the charged particle experiences a centripetal force which is equated as below,

${F_C} = \frac{{m{v^2}}}{r}$

Here, $m$is the mass of the particle, $v$is the velocity with which the particle is travelling, and $r$ is the radius of the circular path followed by the particle.

This centripetal force required to keep the particle in circular motion provided by the Lorentz force ${F_B}$ given by,

${F_B} = qvB$

The kinetic energy of the particle is,

$KE = \frac{1}{2}m{v^2}$

Here, $m$is the mass of the particle and$v$ is the velocity of the particle.

(A)

As the force for the rotation is provided by the Lorentz force, we get

$\begin{array}{c}\\{F_{\rm{C}}} = {F_{\rm{B}}}\\\\\frac{{m{v^2}}}{r} = qvB\\\\v = \frac{{qBr}}{m}\\\end{array}$

As the radius of the cyclotron becomes twice and the other quantities remain constant, the velocity doubles.

$\begin{array}{c}\\v' = \frac{{qB\left( {2r} \right)}}{m}\\\\ = \frac{{2qBr}}{m}\\\\ = 2v\\\end{array}$

The initial kinetic energy of the particle is,

$KE = \frac{1}{2}m{v^2}$

As the radius doubles, the velocity doubles. So, the maximum particle energy increase is,

$\begin{array}{c}\\K{E_{\max }} = \frac{1}{2}m{\left( {v'} \right)^2}\\\\ = \frac{1}{2}m{\left( {2v} \right)^2}\\\\ = 4\left( {\frac{1}{2}m{v^2}} \right)\\\\ = 4KE\\\end{array}$

(B)

The velocity of a charged particle in a cyclotron is given by,

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

As the radius and the magnetic field becomes twice the initial value, the velocity becomes 4 times.

$\begin{array}{c}\\v' = \frac{{q\left( {2B} \right)\left( {2r} \right)}}{m}\\\\ = \frac{{4qBr}}{m}\\\\ = 4v\\\end{array}$

The initial kinetic energy of the particle is,

$KE = \frac{1}{2}m{v^2}$

As the radius and magnetic field strength becomes twice, the velocity becomes twice. So, the maximum particle energy increase is,

$\begin{array}{c}\\K{E_{\max }} = \frac{1}{2}m{\left( {v'} \right)^2}\\\\ = \frac{1}{2}m{\left( {4v} \right)^2}\\\\ = 16\left( {\frac{1}{2}m{v^2}} \right)\\\\ = 16KE\\\end{array}$

Ans: Part A

The allowed maximum particle energy is increased by a factor of 4.

Part B

The allowed maximum particle energy is increased by a factor of 16.