Question

The diameter of a copper wire is 2 mm, a steady current of 6.25 A is generated by $8.5\times {10}^{28}/m^3$ electrons flowing through it. Calculate drift velocity of conduction electrons.

Answer 1

$\begin{array}{c}i=neAV\\ \Rightarrow 6.25=8.5\times {10}^{28}\times 1.6\times {10}^{-19}.\left(\pi \times {\left(\frac{2\times {10}^{-3}}{2}\right)}^2\right).V\\ \Rightarrow 6.25=8.5\times 1.6\times {10}^9\times \pi \times {10}^{-6}.V\\ \Rightarrow 6.25=8.5\times 1.6\times {10}^3\times \pi \times V\\ \Rightarrow V=\frac{6.25}{8.5\times 1.6\times {10}^3\pi }\\ \Rightarrow V=\frac{6.25}{8.5\times 1.6\pi }\times {10}^{-3}\\ \Rightarrow 0.14\times {10}^{-3}\\ \Rightarrow V=0.14\times {10}^{-3}m/s\end{array}$