An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2}\, T $. If the value of $e/m$ is $1.76 \times 10^{11}\, C/kg $, the frequency of revolution of the electron is
NEET 2016, Medium
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$\text { Here, } B=3.57 \times 10^{-2}\, \mathrm{T}$
$\frac{e}{m}=1.76 \times 10^{11}\, \mathrm{C\,kg}^{-1}$
Frequency of revolution of the electron,
$v=\frac{1}{T}=\frac{v}{2 \pi r}.........(i)$
Also, $\frac{m v^{2}}{r}=e v B \Rightarrow \frac{v}{r}=\frac{e B}{m}.........(ii)$
From eqns. $(i)$ and $(ii)$,
$v=\frac{1}{2 \pi} \times \frac{e B}{m}=\frac{1}{2 \times 3.14} \times 1.76 \times 10^{11} \times 3.57 \times 10^{-2}$
$=10^{9} \,\mathrm{Hz}=1 \,\mathrm{GHz}$
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