Question

A proton of energy $100\text{eV}$ is moving perpendicular to a magnetic field of ${10}^{-4}\text{T}$. The cyclotron frequency of the proton in $\text{rad /sec}$ is:

• A. $2.80\times {10}^6$
• B. $9.6\times {10}^3$
• C. $5.6\times {10}^6$
• D. $1.76\times {10}^6$

Answer 1

The correct option is B $9.6\times {10}^3$

Given
Energy of proton, $E=100\text{eV}=100\times 1.6\times {10}^{-19}\text{J}$
$B={10}^{-4}\text{T}$

The frequency of cyclotron is given by,
$$f=\frac{qB}{2\pi m}$$As,
$q=1.6\times {10}^{-19}\text{C}$

$m=1.67\times {10}^{-27}\text{kg}$

$\Rightarrow f=\frac{1.6\times {10}^{-19}\times {10}^{-4}}{2\times 3.14\times 1.67\times {10}^{-27}}=1.525\times {10}^3\text{Hz}$

The relation between frequency and angular frequency is,
$$\omega =2\pi f$$$\Rightarrow \omega =2\times \pi \times 1.525\times {10}^3=9.6\times {10}^3\text{rad/s}$

Hence, option $\left(b\right)$ is the correct answer.