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

In a cyclotron,the time taken by an ion to describe a semicircular path in a dee is

$A$ 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 $T$)

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

In the cyclotron,as the radius of the circular path of the charged particle increases,what happens to the angular velocity $(\omega)$ and linear velocity $(v)$?

An alternating electric field of frequency $f$ is applied across the radius $R$ of a cyclotron to accelerate protons (mass $m$). The operating magnetic field $B$ used and kinetic energy $(K.E.)$ of the proton beam produced by it are respectively ($e=$ charge on proton).