For a perpendicular magnetic field, magnetic force is provided by the force. So,




$\frac{m{v}^2}{r}=qvB$

$i.e. r=\frac{mv}{qB} ...\left(\mathrm{i}\right)$

As in uniform circular motion $v=r\omega$ so the angular frequency of circular motion will be given by

$\omega =\frac{v}{r}=\frac{qB}{m}$     [Using Eq. (i)]

and hence, the time period

$T=\frac{2\pi }{\omega }=\frac{2\pi m}{qB} ...\left(\mathrm{i}\mathrm{i}\right)$

Given, $B=3.534\times {10}^{-5} \mathrm{T}$

$q=1.6\times {10}^{-19} \mathrm{C},\mathrm{ }m=9.1\times {10}^{-31} \mathrm{k}\mathrm{g},\mathrm{ }T=?$

From Eq. (ii), we get

$T=\frac{2\times 3.14\times 9.1\times {10}^{-31}}{3.534\times {10}^{-5}\times 1.6\times {10}^{-19}}=1\times {10}^{-6} \mathrm{s}=1 \mathrm{\mu }\mathrm{s}$