A cyclotron's oscillator frequency is 10 MHz. What should be the opertaing 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=1.60xx10^(-19)C, m_(p)=1.67xx10^(-27)kg, 1 MeV = 1.6xx10^(-13) J)`.

Explanation

The oscillator frequency should be same as proton.s cyclotron frequency.
Using Eqs. (4.5) and (4.6 (a)) we have
`B=2pimv//q=1.67xx10^(-27)xx10^(7)//(1.6xx10^(-19))=0.66T`
Final velocity of protons is
`v=rxx2piv=0.6mxx6.3xx10^(7)=3.78xx10^(7)m//s`