Question

The number density of free electrons in a copper conductor estimated in Example 3.1 is $8.5 × 10^{28} m^{−3}$. 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 $2.0 × 10^{−6} m^2$ and it is carrying a current of 3.0 A.

Answer 1

Number density of free electrons in a copper conductor, $n = 8.5 × 10^{28} m^{−3}$ Length of the copper wire, $l = 3.0 m$
Area of cross-section of the wire,$A = 2.0 × 10^{−6} m^2$
Current carried by the wire,$I = 3.0 A$, which is given by the relation,
$I = nAeV_d$
Where,
e = Electric charge = $1.6 × 10^{−19} C$
$V_d = Drift\space velocity =\frac{ Length \space of \space the \space wire (I)}{Time \space taken\space to\space cover l (t)}$
$I = nAe\frac{l}{t}$
$t = \frac{nAel}{I}$
$t = \frac{3 \times 8.5 \times 10^{28} \times 2 \times10^{-6} \times 1.6 \times 10^{-19}}{3.0}$
$t = 2.7 \times 10^{4} s$
Therefore, the time taken by an electron to drift from one end of the wire to the other is $2.7 \times 10^{4} s.$