Question

A current of 1.0 A exists in a copper wire of cross - section 1.0 $m{m}^2$. Assuming one free electron per atom calculate the drift speed of the free electrons in the wire. The density of copper is 9000 $kg{m}^{-3}$.

Answer 1

$i=1A,$

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

$f_{cu}=9000{\text{kg/m}}^3$

$\text{Molecular mass has}{N}_0\text{atoms}$

$=mkg\text{has}\left(\frac{N_0}{m}\right)\text{atoms}$

$=\frac{\frac{N_{0}A}{9000}}{63.5{\times }^{-3}}$

$\text{No. of atoms= No. of electrons.}$

$n=\frac{\text{Number of electrons}}{\text{Unit Volume}}$

$=\frac{N.p.}{m.l}=\frac{N_{0}p}{M}$

$=\frac{6\times {10}^{23}\times 9000}{63.5\times {10}^{-3}}$

$i=V_{d}nAe$

$V_d=\frac{1}{\frac{6\times {10}^{28}\times 9000}{63.5\times {10}^{-3}}}\times {10}^{-6}\times 1.6\times {10}^{-19}$

$=\frac{63.5\times {10}^{-3}}{6\times {10}^{23}\times 9000\times {10}^{-6}\times 1.6\times {10}^{-19}}$

$=\frac{63.5\times {10}^{-3}}{6\times {10}^{26}9\times {10}^{-6}\times 1.6\times {10}^{-19}}$

$=\frac{63.5\times {10}^{-3}}{6\times 9\times 16}$

$=0.073\times {10}^{-3}m/s=0.073mm/s$