Question

A cyclotron is used to accelerate protons to a kinetic energy of $5\;{\text{MeV}}$. If the strength of the magnetic field in the cyclotron is $2\;{\text{T}}$ , find the radius and the frequency needed for the applied alternating voltage of the cyclotron.
(Given: velocity of proton $ = 3 \times {10^7}\;{\text{m/s}}$ )

Answer 1

Hint: For the cyclotron the centripetal force of the proton and the Lorentz force must be equal. The Lorentz force causes the cyclotron to bend. The expression for kinetic energy can be rearranged to get an expression for the radius of the cyclotron.


Complete step by step answer:
Given the kinetic energy of the accelerating protons is $K.E = 5\;{\text{MeV}}$ and strength of magnetic field in the cyclotron is $B = 2\;{\text{T}}$.
The expression for kinetic energy is given as,
$K.E = \dfrac{1}{2}m{v^2}$
Where, $m$ is the mass and $v$ is the velocity.
Let’s convert ${\text{MeV}}$ to Joules.
We know that $1\;{\text{MeV}} = 1.602 \times {10^{ - 13}}\;{\text{J}}$
Therefore,
$\
  \dfrac{1}{2}{m_p}{v^2} = 5\;{\text{MeV}} \\
   = {\text{5}} \times {\text{1}}{\text{.602}} \times {\text{1}}{{\text{0}}^{ - 13}}\;{\text{J}} \\
  \dfrac{1}{2}{m_p}{v^2}{\text{ = 8}}{\text{.01}} \times {\text{1}}{{\text{0}}^{ - 13}}\;{\text{J}} \\
\ $
Where, ${m_p}$ is the mass of the proton.
The centripetal force of the accelerating protons is given as,
${F_C} = \dfrac{{{m_p}{v^2}}}{r}$
Where, ${m_p}$ is the mass of the proton, $v$ is the velocity of the proton and $r$ is the radius of the cyclotron.
And the expression for the Lorentz force when the magnetic field is applied is given as,
${F_l} = {q_P}vB$
Where, ${q_P}$ is the charge of the proton, $v$ is the velocity of the proton and $B$ is the magnetic field.
For a cyclotron motion the centripetal force and the Lorentz force must be equal.
Therefore,
$\dfrac{{{m_p}{v^2}}}{r} = {q_P}vB............\left( 1 \right)$
Let’s rearrange this equation by dividing both sides by $2$ .
$\dfrac{1}{2}{m_p}{v^2}\dfrac{1}{r} = \dfrac{{{q_P}vB}}{2}$
Replace $\dfrac{1}{2}{m_p}{v^2} = K.E$.
$\
  K.E \times \dfrac{1}{r} = \dfrac{{{q_P}vB}}{2} \\
  r = \dfrac{{2 \times K.E}}{{{q_P}vB}} \\
\ $
Substituting the values, we get
$\
  r = \dfrac{{2 \times {\text{8}}{\text{.01}} \times {\text{1}}{{\text{0}}^{ - 13}}\;{\text{J}}}}{{1.6 \times {{10}^{ - 19}}\;{\text{C}} \times 3 \times {{10}^7}\;{\text{m/s}} \times 2\;{\text{T}}}} \\
   = 0.167\;{\text{m}} \\
\ $
Thus the radius of the cyclotron is $0.167\;{\text{m}}$.
From equation $\left( 1 \right)$,
$v = \dfrac{{{q_p}rB}}{{{m_p}}}$
The expression for angular velocity is given as,
$\omega = 2\pi f$
Where $f$ is the frequency of a cyclotron.
The relation connecting angular velocity $\omega $ and linear velocity $v$ is given as,
$v = r\omega $
Substitute the above results,
$\
  \dfrac{{{q_p}rB}}{{{m_p}}} = r \times 2\pi f \\
  f = \dfrac{{{q_p}B}}{{2\pi {m_p}}} \\
\ $
Substitute the values in above expression.
$\
  f = \dfrac{{1.6 \times {{10}^{ - 19}}\;C \times 2\;T}}{{2\pi \times 1.67 \times {{10}^{ - 27}}\;{\text{kg}}}} \\
   = 3.05 \times {10^7}\;{\text{Hz}} \\
\ $
Thus the frequency of cyclotron is $3.05 \times {10^7}\;{\text{Hz}}$.

Note: We have to note that the frequency of the cyclotron and the velocity of the particle does not depend on the radius of the cyclotron. The magnetic field is the major factor for the cyclotron motion hence the Lorentz force is important.