A cyclotron oscillator frequency is $10 MHz$. What should be the operating magnetic field for accelerating protons? If the radius of its dees is $60 cm$, what is the kinetic energy (in $MeV$) of the proton beam produced by the accelerator?

$\left(e=1.60\times {10}^{-19} C, m_p=1.67\times {10}^{-27} kg, 1 MeV=1.6\times {10}^{-13 }J\right)$

Magnetic field Cyclotron's oscillator frequency should be same as the proton's revolution frequency (in circular path)

$$\begin{gathered} f=\frac{Bq}{2\mathrm{\pi m}} \\ B=\frac{2\mathrm{\pi mf}}{q} \end{gathered}$$

Substituting the values in SI units, we have

$$\begin{gathered} B=\frac{\left(2\right)\left({\displaystyle \raisebox{1ex}{$22$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}\right)\left(1.67\times {10}^{-27}\right)\left(10\times {10}^6\right)}{1.6\times {10}^{-19}} \\ =0.67 T \end{gathered}$$

Kinetic energy Let final velocity of proton just after leaving the cyclotron is $v$. Then, radius of dee should be equal to 

$R=\frac{mv}{Bq}$ or $v=\frac{BqR}{m}$

Therefore, Kinetic energy of proton, $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(\frac{BqR}{m}\right)}^2=\frac{B^2{q}^2{R}^2}{2m}$

Substituting the values we get,

$$\begin{gathered} K=\frac{{\left(0.67\right)}^2\left(1.6\times {10}^{-19}\right){\left(0.60\right)}^2}{2\times 1.67\times {10}^{-27}} \\ =1.2\times {10}^{-12} J \\ =\frac{1.2\times {10}^{-12}}{\left(1.6\times {10}^{-19}\right)\left({10}^6\right)}MeV \\ =7.5 MeV \end{gathered}$$