Q 12-04-150JEE MainJEE Main 2022 (25 Jul, Shift 2)Medium
An electron with energy $0.1\ \text{keV}$ moves at right angle to the earth's magnetic field of $1\times10^{-4}\ \text{Wb m}^{-2}$. The frequency of revolution of the electron will be (Take mass of electron $= 9.0\times10^{-31}\ \text{kg}$)
Answer: (C) $2.8\times10^6\ \text{Hz}$
The cyclotron frequency does not depend on the speed:
$$f = \frac{eB}{2\pi m} = \frac{1.6\times10^{-19}\times10^{-4}}{2\pi\times9\times10^{-31}}\approx2.8\times10^6\ \text{Hz}$$
Solution by Sreeraj P, M.Sc Physics