Q 12-04-155JEE MainJEE Main 2022 (26 Jul, Shift 2)Medium
A velocity selector consists of electric field $\vec E = E\hat k$ and magnetic field $\vec B = B\hat j$ with $B = 12\ \text{mT}$. The value of $E$ required for an electron of energy $728\ \text{eV}$ moving along the positive $x$-axis to pass undeflected is (Given, mass of electron $= 9.1\times10^{-31}\ \text{kg}$)
Answer: (A) $192\ \text{kV m}^{-1}$
$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2\times728\times1.6\times10^{-19}}{9.1\times10^{-31}}} = \sqrt{2.56\times10^{14}} = 1.6\times10^7\ \text{m s}^{-1}$.
Undeflected: $E = vB = 1.6\times10^7\times12\times10^{-3} = 1.92\times10^5\ \text{V m}^{-1} = 192\ \text{kV m}^{-1}$.
Solution by Sreeraj P, M.Sc Physics