Q 12-04-189JEE MainJEE Main 2020 (8 Jan, Shift 1)Medium
A proton with kinetic energy of $1$ MeV moves from south to north. It gets an acceleration of $10^{12}\ \text{m/s}^2$ by an applied magnetic field (west to east). The value of the magnetic field is: (Rest mass of proton is $1.6\times10^{-27}$ kg)
Answer: (A) $0.71$ mT
Speed: $\tfrac12mv^2 = 1\ \text{MeV} = 1.6\times10^{-13}$ J
$$v = \sqrt{\frac{2\times1.6\times10^{-13}}{1.6\times10^{-27}}} = \sqrt2\times10^7\ \text{m/s}$$
The field is perpendicular to the velocity, so $ma = qvB$:
$$B = \frac{ma}{qv} = \frac{1.6\times10^{-27}\times10^{12}}{1.6\times10^{-19}\times1.41\times10^7} \approx 7.1\times10^{-4}\ \text{T} = 0.71\ \text{mT}$$
Solution by Sreeraj P, M.Sc Physics