An electron with energy $0.1\,ke\,V$ moves at right angle to the earth's magnetic field of $1 \times 10^{-4}\,Wbm ^{-2}$. The frequency of revolution of the electron will be. (Take mass of electron $=9.0 \times 10^{-31}\,kg$ )
JEE MAIN 2022, Easy
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$f =\frac{1}{ T }=\frac{ eB }{2 \pi m }$
$=\frac{1.6 \times 10^{-19} \times 10^{-4}}{2 \pi \times 9 \times 10^{-31}}=2.8 \times 10^{6}\,Hz$
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