Explanation

(d) For perpendicular magnetic field magnetic force is provided by the force so,
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$\frac{mv^{2}}{r} = qvB$
$ie,\ \ r = \frac{mv}{qB}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
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
$T = \frac{2\pi}{\omega} = \frac{2\pi m}{qB}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Given, $B = 3.534 \times 10^{- 5}\ T,$
$q = 1.6 \times 10^{- 19}\ C,\ m = 9.1 \times 10^{- 31}\ kg,\ T = ?$
From Eq. (ii), we get
$\therefore\ \ \ T = \frac{2 \times 3.14 \times 9.1 \times 10^{- 31}}{3.534 \times 10^{- 5} \times 1.6 \times 10^{- 19}} = 1 \times 10^{- 6}\ s = 1\mu s$