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 freqency.
Using `Eqs. R=mv//qb and omega =2pi v=(qB)/(M)` we have
`B=2pi mv//q=6.3xx1.67xx10^(-27)xx10^(7)//(1.6xx10^(-19))=0.66T`
Final velocity of protons is
`v=rxx2pi v=0.6 m xx6.3xx10^(7)=3.78xx10^(7) m//s`.
`E=1//1 mv^(2)=1.67xx10^(-27)xx14.3xx10^(14)//(2xx1.6xx10^(-13))=7 MeV`.