Question

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0\times {10}^{-7}{m}^2$ carrying a current of $1.5A$. Assume the density of conduction electrons to be $9\times {10}^{28}{m}^{-3}$.

Answer 1

Given: a copper wire of cross-sectional area $2.5\times {10}^{-7}{m}^2$carrying a current of $2.7A$. The density of conduction electrons to be $9\times {10}^{28}{m}^{-3}$

To find the average drift speed of conduction electrons

Solution:

We know that drift velocity,

$V_d=\frac{I}{nq\times A}$, where $I=2.7A$ is the current, $n=9\times {10}^{28}{m}^{-3}$ is the charge density, $q=1.6\times {10}^{-19}$ is charge electron and $A=2.5\times {10}^{-7}{m}^2$ is cross-sectional area

That is, $V_d=\frac{2.7}{9\times {10}^{28}\times 1.6\times {10}^{-19}\times 2.5\times {10}^{-7}}⟹V_d=\frac{2.7}{14.4\times {10}^9\times 2.5\times {10}^{-7}}⟹V_d=0.075\times {10}^{-2}⟹V_d=7.5\times {10}^{-4}m/s$

is the required averagedrift velocity of conduction electrons.