Question

The number density of free electrons in a copper conductor estimated in Example $3.1$ is $8.5\times {10}^{28}{m}^{-3}.$ How long does an electron take to drift from one end of a wire $3.0m$ lomg to its other end? The area of cross-section of the wire is $2.0\times {10}^{-6}{m}^2$ and it is carrying a current of $3.0A.$

Answer 1

Given,

Number density of free electrons,$n=8.5\times {10}^{28}{m}^{-3}$

Length of the copper wire,$l=3.0m$

Area of cross section of the wire$,A=2.0\times {10}^{-6}{m}^2$

current carried by the wire$,i=3.0A$

And, the current is given by$,l=neA{v}_d$

Where,$e=Electriccharge=1.6\times {10}^{-19}C$

$v_d=$ drift velocity

Time taken by the electron to drift from one end to the other of the
wire of length $l$ will be,

$t=\frac{l}{v_d}\Rightarrow t=\frac{t\times nAe}{l}$

Substituting the values we get,

$\Rightarrow t=\frac{3.0\times 8.5\times {10}^{28}\times 2.0\times {10}^{-6}\times 1.6\times {10}^{-19}}{3.0}$

$t=2.72\times {10}^{4}s$