Q 12-04-018JEE MainHard
A cyclotron has a magnetic field of $1$ T and dees of radius $0.5$ m. Find the maximum kinetic energy, in MeV, of the protons it produces. ($m_p = 1.67 \times 10^{-27}$ kg, $e = 1.6 \times 10^{-19}$ C)
Numerical value type. Enter your answer.
Answer: 12
The protons leave when their orbit radius reaches $R$: $v = \dfrac{eBR}{m}$.
$$K = \frac{e^2B^2R^2}{2m} = \frac{(1.6 \times 10^{-19})^2 \times 1 \times 0.25}{2 \times 1.67 \times 10^{-27}} \approx 1.92 \times 10^{-12}\ \text{J}$$
In MeV: $\dfrac{1.92 \times 10^{-12}}{1.6 \times 10^{-13}} \approx 12$ MeV.
Solution by Sreeraj P, M.Sc Physics