Question

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?
($q=1.6\times {10}^{-19}C$, $m_e=9.1\times {10}^{-31}kg$)

• A. $1.4\times {10}^{5}Hz$
• B. $1.4\times {10}^{7}Hz$
• C. $1.4\times {10}^{6}Hz$
• D. $1.4\times {10}^{9}Hz$

Answer 1

Answer: (B)

The correct option is B $1.4\times {10}^{7}Hz$
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.