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4.Moving Charges and Magnetism
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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
A
$62.8 \,MHz$
B
$6.28 \,MHz$
C
$1 \,GHz$
D
$100 \,MHz$
(NEET-2016)
Solution
$\text { Here, } B=3.57 \times 10^{-2}\, \mathrm{T}$
$\frac{e}{m}=1.76 \times 10^{11}\, \mathrm{C\,kg}^{-1}$
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 $(ii)$,
$v=\frac{1}{2 \pi} \times \frac{e B}{m}=\frac{1}{2 \times 3.14} \times 1.76 \times 10^{11} \times 3.57 \times 10^{-2}$
$=10^{9} \,\mathrm{Hz}=1 \,\mathrm{GHz}$
Standard 12
Physics
