Q 12-04-185JEE MainJEE Main 2020 (6 Sep, Shift 1)Medium
An electron is moving along the $+x$ direction with a velocity of $6\times10^6\ \text{m s}^{-1}$. It enters a region of uniform electric field of $300$ V/cm pointing along the $+y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$ direction will be:
Answer: (C) $5\times10^{-3}$ T, along $+z$ direction
The electric force on the electron is along $-y$, so the magnetic force $-e(\vec v\times\vec B)$ must be along $+y$, i.e. $\vec v\times\vec B$ must be along $-y$. Since $\hat i\times\hat k = -\hat j$, $\vec B$ is along $+z$.
Magnitude: $eE = evB \Rightarrow B = \dfrac Ev = \dfrac{3\times10^4}{6\times10^6} = 5\times10^{-3}$ T.
Solution by Sreeraj P, M.Sc Physics