An electron is revolving round a proton, prod | vedclass

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Question

(a)Magnetic field due to revolution of electron
$B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi i}}{r} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi .\left( {

Vedclass · Accepted Answer

$10^{17} \,rad/sec$

Vedclass · Answer

(a)Magnetic field due to revolution of electron
$B = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi i}}{r} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi .\left( {\frac{{e\omega }}{{2\pi }}} ight)}}{r} = {10^{ - 7}} 	imes \frac{{e\omega }}{r}$
$ \Rightarrow 16 = {10^{ - 7}} 	imes \frac{{1.6 	imes {{10}^{ - 19}}\omega }}{{1 	imes {{10}^{ - 10}}}} \Rightarrow \omega = {10^{17}}\,rad/\sec .$

An electron is revolving round a proton, producing a magnetic field of $16\, weber/m^2$ in a circular orbit of radius $1\,\mathop A\limits^o $. It’s angular velocity will be

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