Question

A proton of mass $1.6\times {10}^{-27}\text{kg}$ and charge $1.6\times {10}^{-19}\text{C}$ is shot perpendicular to a uniform magnetic field with a velocity $5\times {10}^6\text{m/s}.$ If the magnetic field is $1\text{T}$. The frequency of revolution is

• A. $1.59\times {10}^7\text{rev}/\text{s}$
• B. $1.69\times {10}^7\text{rev}/\text{s}$
• C. $1.89\times {10}^7\text{rev}/\text{s}$
• D. $1.79\times {10}^7\text{rev}/\text{s}$

Answer 1

The correct option is A $1.59\times {10}^7\text{rev}/\text{s}$

Given,
$\text{Mass,}m=1.6\times {10}^{-27}\text{kg}\text{Charge,}q=1.6\times {10}^{-19}\text{C}\text{Velocity,}u=5\times {10}^6\text{m}/\text{s}\text{Field,}B=1\text{T}$


Charge is shot perpendicular to the magnetic field.

Let $T$ be the time period of revolution of charge$$\therefore T=\frac{2\pi m}{qB}$$Substituting the values we get,

$T=\frac{2\pi \times 1.6\times {10}^{-27}}{1.6\times {10}^{-19}\times 1}=2\pi \times {10}^{-8}\text{s}$

So, the frequency of revolution:

$f=\frac{1}{T}=\frac{{10}^8}{2\pi }=1.59\times {10}^7\text{rev}/\text{s}$

Hence, option (a) is the correct answer.