Q 12-04-137JEE MainJEE Main 2023 (1 Feb, Shift 1)Medium
A charged particle of $2\ \mu\text{C}$ accelerated by a potential difference of $100$ V enters a region of uniform magnetic field of magnitude $4$ mT at right angle to the direction of field. The charged particle completes a semicircle of radius $3$ cm inside the magnetic field. The mass of the charged particle is ______ $\times10^{-18}$ kg.
Numerical value type. Enter your answer.
Answer: 144
$r=\dfrac{\sqrt{2mqV}}{qB}\Rightarrow m=\dfrac{qB^2r^2}{2V}=\dfrac{2\times10^{-6}\times16\times10^{-6}\times9\times10^{-4}}{200}=1.44\times10^{-16}$ kg $=144\times10^{-18}$ kg.
Solution by Sreeraj P, M.Sc Physics