Question

In a conductor, every atom contributes one free electron and the current flowing through it is $16A$. The area of cross section, length and density of this conductor are ${10}^{-6}{m}^2$, $1m$ and $5\times {10}^{3}kg/m^3$. The drift velocity of the electrons $($in $mm/s)$ is (Atomic weight = $60$, $N_A=6\times {10}^{23}$, $e=1.6\times {10}^{-19}C$

Answer 1

Given : $M=0.060kg,A={10}^{-6}{m}^2,\rho =5000kg/m^3$
$I=16A$
Number of atoms per unit volume $\frac{n}{V}=\frac{m{N}_A}{VM}=\frac{\rho N_A}{M}$
$\therefore$ $\frac{n}{V}=\frac{5000\times 6\times {10}^{23}}{0.060}=5\times {10}^{28}$ atoms per unit volume
As one $Cu$ atom contributes one free electron ,thus electron density $n_e=n=5\times {10}^{28}$ electrons per unit volume
Current density $J=\frac{I}{A}=\frac{16}{{10}^{-6}}=1.6\times {10}^{7}A/m^2$
Drift velocity $v_d=\frac{J}{e{n}_e}=\frac{1.6\times {10}^7}{1.6\times {10}^{-19}\times 5\times {10}^{28}}=0.002m/s=2mm/s$