Question

A cyclotron's oscillator frequency is $10MHz$. What should be the operating magnetic field for accelerating protons? If the radius $r$ its 'dees' is $60cm$, calculate the kinetic energy (in $MeV$) of the proton beem produces by the accelerator.

Answer 1

The oscillator frequency should be same as proton cyclotron frequency, then magnetic field,
$B=\frac{2\pi mv}{q}$
$=\frac{3\times 3.14\times 1.67\times {10}^{-27}\times {10}^7}{1.6\times {10}^{-19}}=0.66T$
$v=r\omega =r\times 2\pi v$
$=0.6\times 2\times 3.14\times {10}^7=3.78\times {10}^{7}m/s$
So, Kinetic energy, $KE=\frac{1}{2}m{v}^2$
$=\frac{1}{2}\times 1.67\times {10}^{-27}\times (3.78\times {10}^7{)}^{2}J$
$=\frac{1}{2}\times \frac{1.67\times {10}^{-27}\times 14.3\times {10}^{14}}{1.3\times {10}^{-19}\times {10}^6}MeV=7.4MeV$