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. We have
`B = 2pi mv//q = 6.3 xx 1.67 xx 10^(-27) xx 10^7 // (1.6 xx 10^(-19) ) = 0.66 T`
Final velocity of protons is
`v = r xx 2pi v = 0.6 m xx 6.3 xx 10^7 = 3.78 xx 10^7 m//s`
`E = 1//2 mv^2 = 1.67 xx 10^(-27) xx 14.3 xx 10^14 // (2 xx 1.6 xx 10^(-13)) = 7MeV` .