Question

Consider a wire of length $4m$ and cross-sectional area $1m{m}^3$ carrying a current of $2A$. If each cubic metre of the material contains ${10}^{29}$ free electrons, find the average time taken by an electron to cross the length of the wire.

Answer 1

$\ell =4m,A=1m{m}^2=1\times {10}^{-6}{m}^2$

$l=2A,n/V={10}^{29},t=?$

$i=nA{V}_{d}e$

$\Rightarrow i={10}^{29}\times 1\times {10}^{-6}\times V_d\times 1.6\times {10}^{-19}$

$\Rightarrow V_d=\frac{2}{{10}^{29}\times {10}^{-6}\times 1.6\times {10}^{-19}}$

$V_d=\frac{1}{0.8\times {10}^4}=\frac{1}{8000}$

$t=\frac{\ell }{V_d}=\frac{4}{1/8000}=4\times 8000$

$t=32000=3.2\times {10}^{4}sec.$