An electron is moving in a circular path under influence of a transverse magnetic field of 3.57 × 1

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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

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