Q 12-04-159JEE MainJEE Main 2022 (27 Jul, Shift 2)Medium
A cyclotron is used to accelerate protons. If the operating magnetic field is $1.0$ T and the radius of the cyclotron 'dees' is $60$ cm, the kinetic energy of the accelerated protons in MeV will be :
[use $m_p = 1.6\times10^{-27}$ kg, $e = 1.6\times10^{-19}$ C]
Answer: (B) $18$
At the edge of the dees $r = \dfrac{mv}{eB}$, so
$$K = \frac{e^2B^2r^2}{2m} = \frac{(1.6\times10^{-19})^2(1)^2(0.6)^2}{2\times1.6\times10^{-27}} = 2.88\times10^{-12}\ \text{J}$$
$$K = \frac{2.88\times10^{-12}}{1.6\times10^{-13}}\ \text{MeV} = 18\ \text{MeV}$$
Solution by Sreeraj P, M.Sc Physics