JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Magnetic Effects of Current

The proton is to reach a speed $v=\frac{c}{6}=5\times 10^7$ m/s, at which its kinetic energy is $\frac{1}{2}{m}_p{v}^2=\frac{1}{2}(1.67\times 10^{-27})(5\times 10^7{)}^2=2.09\times 10^{-12}$ J. In a cyclotron the particle is accelerated twice in each revolution, once at each gap crossing, so it gains $2eV=2(1.6\times 10^{-19})(12\times 10^3)=3.84\times 10^{-15}$ J per revolution. The number of revolutions needed is the ratio of these energies, $\frac{2.09\times 10^{-12}}{3.84\times 10^{-15}}\approx 543.6$, so the proton completes about $543$ revolutions.