Question

An electron having energy $10\text{eV}$ is circulating in a plane at right angle to a uniform field of magnetic induction ${10}^{-4}{\text{Wb/m}}^2(=1.0\text{gauss})$, the orbital radius of electron is

• A. $11\text{cm}$
• B. $18\text{cm}$
• C. $12\text{cm}$
• D. $16\text{cm}$

Answer 1

The correct option is A $11\text{cm}$

Given:
$\text{KE}$ of electron, $K=10\text{eV};B={10}^{-4}{\text{Wb/m}}^2$

Radius of circular path of a charged particle having $K$ kinetic energy in a uniform magnetic field is

$R=\frac{\sqrt{2mK}}{qB}$

Here, $m=9.1\times {10}^{-31}\text{kg}=$ mass of electron
$q=1.6\times {10}^{-19}\text{C}$

$R=\frac{\sqrt{2\times 9.1\times {10}^{-31}\times (10\times 1.6\times {10}^{-19})}}{1.6\times {10}^{-19}\times {10}^{-4}}$

$R=\left[\sqrt{\frac{2\times 9.1\times {10}^{-31}\times 10}{1.6\times {10}^{-19}}}\right]\times \frac{1}{{10}^{-4}}$

$\therefore R=10.66\times {10}^{-2}\text{m}=11\text{cm}$

Therefore, option $\left(a\right)$ is correct.