In a cyclotron,the angular frequency of a charged particle is independent of

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

$A$ 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 $\text{ MeV}$)

The oscillator frequency of a cyclotron is $10 \,MHz$. What should be the operating magnetic field for accelerating protons (in $,T$)? (Given: mass of proton $m = 1.67 \times 10^{-27} \,kg$,charge of proton $q = 1.6 \times 10^{-19} \,C$)

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

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