$A$ charged particle is subjected to acceleration in a cyclotron as shown. The charged particle undergoes an increase in its speed:

$A$ particle of mass $m$ and charge $q$ is moving in a cyclotron with magnetic field $B$. The frequency of the circular motion of the particle is proportional to

In the cyclotron, as the radius of the circular path of the charged particle increases, ($\omega$ = angular velocity, $V$ = linear velocity)

$A$ cyclotron's oscillator frequency is $20 MHz$. The operating magnetic field for accelerating protons is (charge of proton $= 1.6 \times 10^{-19} C$,mass of proton $= 1.67 \times 10^{-27} kg$). (in $T$)

If $q$ is the electric charge,$B$ is the magnetic field,$R$ is the dee radius,and $m$ is the mass of ions,the kinetic energy of the ions in a cyclotron is given by

$A$ proton is accelerating in a cyclotron where the applied magnetic field is $2 \,T$. If the potential gap is effectively $100 \,kV$, then how many revolutions does the proton have to make between the "dees" to acquire a kinetic energy of $20 \,MeV$?