Question

In a copper wire of diameter 2mm, 3x10²³ electrons flow in one second. Calculate the (a) charge (b) current and (c) current density ​

Answer 1

Given:-

diameter of wire, d=2 mm

flow of electrons per second=$3\times10^{23}$

To Find:-

charge, current and current density

Explanation:-

We know that there are $6.25\times10^{18}$ number of electrons in 1 C charge.

$\Rightarrow \text{charge on 1 electron}=\dfrac{1}{6.25\times10^{18}}C\\\\\text{Therefore, charge on }3\times10^{23}\text{ electrons}=\dfrac{3\times10^{23}}{6.25\times10^{18}}C=48000C$

So, 48000 C charge is flowing in a wire per second and current(I)=$\dfrac{charge}{time}$

$\Rightarrow \text{current, I}=48000A$

$\text{Now,}\\\\\text{cross-sectional area, A}=\dfrac{\pi}{4}\times d^{2}=\dfrac{\pi}{4}\times(2^{2})=3.14mm^{2}\\\\\Rightarrow\text{current density, J}=\dfrac{I}{A}=\dfrac{48000}{3.14}=15278.87A/mm^{2}$