Q 12-04-184JEE MainJEE Main 2020 (6 Sep, Shift 1)Medium
A particle of charge $q$ and mass $m$ is moving with a velocity $-v\hat i\ (v \neq 0)$ towards a large screen placed in the $Y$-$Z$ plane at a distance $d$. If there is a magnetic field $\vec B = B_0\hat k$, the maximum value of $v$ for which the particle will not hit the screen is:
Answer: (C) $\dfrac{qdB_0}{m}$
The velocity is perpendicular to $\vec B$, so the particle moves on a circle of radius $r = \dfrac{mv}{qB_0}$ in the $x$-$y$ plane.
It starts moving straight towards the screen, so the farthest it can travel towards the screen is $r$. It just fails to reach the screen if $r \le d$:
$$\frac{mv}{qB_0} \le d \Rightarrow v \le \frac{qdB_0}{m}$$
Solution by Sreeraj P, M.Sc Physics