An electron is moving in a circular path unde | vedclass

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Question

$	ext { Here, } B=3.57 	imes 10^{-2}\, \mathrm{T}$

$\frac{e}{m}=1.76 	imes 10^{11}\, \mathrm{C\,kg}^{-1}$

Frequency of revolution of the elec

Vedclass · Accepted Answer

$1 \,GHz$

Vedclass · Answer

$	ext { Here, } B=3.57 	imes 10^{-2}\, \mathrm{T}$

$\frac{e}{m}=1.76 	imes 10^{11}\, \mathrm{C\,kg}^{-1}$

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 $(ii)$,

$v=\frac{1}{2 \pi} 	imes \frac{e B}{m}=\frac{1}{2 	imes 3.14} 	imes 1.76 	imes 10^{11} 	imes 3.57 	imes 10^{-2}$

$=10^{9} \,\mathrm{Hz}=1 \,\mathrm{GHz}$

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

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