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 $3.0 m$ long 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 $3.0 A.$
Exercise
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Number density of free electrons in a copper conductor, $n = 8.5 \times 10^{28}m^{-3 }$ Length of the copper wire, $l = 3.0 m$
Area of cross$-$section of the wire, $A = 2.0 \times 10^{-6} m^2$
Current carried by the wire, $I = 3.0 A,$ which is given by the relation,
$I = nAeV_d$
where,
$e =$ Electric charge $= 1.6 \times 10^{-19} C$
$\text{V}_\text{d}=\text{Drift velocity}=\frac{\text{Length of the wire}}{\text{Times taken to cover}}=\frac{\text{l}}{\text{t}}$
$\text{I}=\text{n Ae}\frac{\text{l}}{\text{t}}$
$\text{I}=\frac{\text{n.A.e.l}}{\text{t}}$
$\text{t}=\frac{\text{n.A.e.l}}{\text{I}}$
$=\frac{3\times8.5\times10^{28}\times2\times10^{-6}\times1.6\times10^{-19}}{3.0}$
$=2.7\times10^4\text{s}$
Therefore, the time taken by an electron to drift from one end of the wire to the other is $2.7\times10^4\text{s}.$
Area of cross$-$section of the wire, $A = 2.0 \times 10^{-6} m^2$
Current carried by the wire, $I = 3.0 A,$ which is given by the relation,
$I = nAeV_d$
where,
$e =$ Electric charge $= 1.6 \times 10^{-19} C$
$\text{V}_\text{d}=\text{Drift velocity}=\frac{\text{Length of the wire}}{\text{Times taken to cover}}=\frac{\text{l}}{\text{t}}$
$\text{I}=\text{n Ae}\frac{\text{l}}{\text{t}}$
$\text{I}=\frac{\text{n.A.e.l}}{\text{t}}$
$\text{t}=\frac{\text{n.A.e.l}}{\text{I}}$
$=\frac{3\times8.5\times10^{28}\times2\times10^{-6}\times1.6\times10^{-19}}{3.0}$
$=2.7\times10^4\text{s}$
Therefore, the time taken by an electron to drift from one end of the wire to the other is $2.7\times10^4\text{s}.$
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