A cyclotron is used to accelerate protons. If the operating...

When the proton is rotated in a cyclotron its maximum velocity will be corresponding to the maximum radius of the circular path. We know that the radius of the circle in a magnetic field is given by,

$r = \dfrac{mv}{Bq} ⇒ v = \dfrac{Bqr}{m}$.

The kinetic energy will be,

$K = \dfrac{1}{2}m{v}^{2} = \dfrac{{B}^{2}{q}^{2}{r}^{2}}{2m} ⇒ K = \dfrac{{1}^{2} \times {\left(1 . 6 \times {10}^{- 19}\right)}^{2} \times {\left(0 . 6\right)}^{2}}{2 \times 1 . 6 \times {10}^{- 27} \times 1 . 6 \times {10}^{- 13}}MeV ⇒ K = 18MeV$