An electron is moving in a cyclotron at a speed of $3.2\times 10^{7}m{s}^{-1}$ in a magnetic field of $5\times 10^{-4}$ T perpendicular to it. What is the frequency of this electron?
($q=1.6\times 10^{-19}C$, $m_e=9.1\times 10^{-31}kg$)

Given,

$v=3.2\times 10^{7}m/s$

$B=5\times 10^{-4}T$

$m_e=9.1\times 10^{-31}kg$

$q=1.6\times 10^{-19}C$

$f=?$

The frequency of the electron in cyclotron  is given by

$f=\frac{qB}{2\pi m_e}$

$f=\frac{1.6\times 10^{-19}\times 5\times 10^{-4}}{2\times 3.14\times 9.1\times 10^{-31}}$

$f=1.4\times 10^{7}Hz$

The correct option is B.