Question

The electron with charge $(q=1.6\times {10}^{-19}\text{C})$ moves in an orbit of radius $5\times {10}^{-11}\text{m}$ with a speed of $2.2\times {10}^6{\text{ms}}^{-1}$, around an atom. The equivalent current is:

• A. $1.12\times {10}^{-6}\text{A}$
• B. $1.12\times {10}^{-3}\text{A}$
• C. $1.12\times {10}^{-9}\text{A}$
• D. $1.12\text{A}$

Answer 1

The correct option is B $1.12\times {10}^{-3}\text{A}$

Given:
$q=1.6\times {10}^{-19}\text{C};r=5\times {10}^{-11}\text{m};v=2.2\times {10}^6{\text{ms}}^{-1}$

As we know that

$I=qf=q(\frac{v}{2\pi r})$

putting the given value

$\Rightarrow I=\frac{1.6\times {10}^{-19}(2.2\times {10}^6)}{\left(2\pi \right)(5\times {10}^{-11})}$

$\Rightarrow I=1.12\times {10}^{-3}\text{A}$

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

Why this Question ? The current at time $t$ is $$I=\underset{\Delta t\to 0}{lim}\frac{\Delta Q}{\Delta t}=\frac{dQ}{dt}$$ Thus, electric current through an area is the rate of transfer of charge from one side of the area to the another.