$A$ cyclotron's 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? $(e = 1.60 \times 10^{-19} \; C, m_p = 1.67 \times 10^{-27} \; kg, 1 \; MeV = 1.6 \times 10^{-13} \; J)$

(A) The oscillator frequency $f$ must be equal to the cyclotron frequency of the proton.
$f = \frac{qB}{2\pi m_p} \implies B = \frac{2\pi m_p f}{q}$
Substituting the values: $B = \frac{2 \times 3.14 \times 1.67 \times 10^{-27} \times 10^7}{1.60 \times 10^{-19}} \approx 0.656 \; T \approx 0.66 \; T$.
The maximum velocity $v$ of the proton at the exit radius $r = 0.6 \; m$ is given by $v = r\omega = r(2\pi f)$.
$v = 0.6 \times 2 \times 3.14 \times 10^7 = 3.77 \times 10^7 \; m/s$.
The kinetic energy $K$ is given by $K = \frac{1}{2} m_p v^2$.
$K = \frac{1}{2} \times 1.67 \times 10^{-27} \times (3.77 \times 10^7)^2 = 1.186 \times 10^{-12} \; J$.
Converting to $MeV$: $K = \frac{1.186 \times 10^{-12}}{1.6 \times 10^{-13}} \approx 7.41 \; MeV$.