Physics · 2022
JEE Main · 25 July 2022, Shift 2 · Q41
An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 × 10^-4 Wbm^-2. The frequency of revolution of the electron…
An electron with energy $\displaystyle 0.1$ keV moves at right angle to the earth's magnetic field of $\displaystyle 1 \times 10^{-4} \mathrm{Wbm}^{-2}$. The frequency of revolution of the electron will be (Take mass of electron $\displaystyle =9.0 \times 10^{-31} \mathrm{~kg}$ )
Official answer
(3)
$\displaystyle 2.8 \times 10^6 \mathrm{~Hz}$
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JEE Main 2022 Physics question, with the answer from NTA’s final answer key. Where our answers come from.