Q 12-04-199JEE MainJEE Main 2019 (9 Jan, Shift 2)Easy
A particle having the same charge as of electron moves in a circular path of radius $0.5\ \text{cm}$ under the influence of a magnetic field of $0.5\ \text{T}$. If an electric field of $100\ \text{V/m}$ makes it to move in a straight path, then the mass of the particle is (Given charge of electron $= 1.6\times10^{-19}\ \text{C}$)
Answer: (C) $2.0\times10^{-24}\ \text{kg}$
Undeflected motion needs $qE = qvB$, so $v = \dfrac EB = \dfrac{100}{0.5} = 200\ \text{m/s}$.
In the magnetic field alone, $r = \dfrac{mv}{qB}$:
$$m = \frac{qBr}{v} = \frac{1.6\times10^{-19}\times0.5\times0.005}{200} = 2.0\times10^{-24}\ \text{kg}$$
Solution by Sreeraj P, M.Sc Physics