The oscillating frequency of a cyclotron is $10 \, MHz$. If the radius of its dee is $0.5 \, m$,the kinetic energy of a proton,which is accelerated by the cyclotron,is ...... $MeV$.

$A$ cyclotron's oscillator frequency is $10 \text{ MHz}$ and the operating magnetic field is $0.66 \text{ T}$. If the radius of its dees is $60 \text{ cm}$, then the kinetic energy of the proton beam produced by the accelerator is: (in $\text{ MeV}$)

$A$ charged particle moves along a circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charged particle increases to $4$ times its initial value. What will be the ratio of the new radius to the original radius of the circular path of the charged particle?

$A$ cyclotron is used to accelerate protons. If the operating magnetic field is $1.0\,T$ and the radius of the cyclotron 'dees' is $60\,cm$,the kinetic energy of the accelerated protons in $MeV$ will be.
[use $m_{p} = 1.6 \times 10^{-27}\,kg, e = 1.6 \times 10^{-19}\,C$]

If the maximum value of accelerating potential provided by a radio frequency oscillator is $12 \, kV$,the number of revolutions made by a proton in a cyclotron to achieve one-sixth of the speed of light is ....... .
$[m_p = 1.67 \times 10^{-27} \, kg, e = 1.6 \times 10^{-19} \, C, c = 3 \times 10^8 \, m/s]$

Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency $(rf)$ field were doubled.