Question

Every atom makes one free electron in copper. If $1.1A$ Current is flowing in the wire of copper having $1mm$ diameter, then the drift velocity (approx.) will be (density of copper $=9\times 10^{3}kg{m}^{-3}$ and atomic weight of copper = $63$ )

• A. $0.1mm{s}^{-1}$
• B. $0.2mm{s}^{-1}$
• C. $0.3mm{s}^{-1}$
• D. $0.2cm{s}^{-1}$

Answer 1

Answer: (A)

$v_d=\frac{i}{nAe}$ ;
where $n=\frac{N_A\rho }{M}$
$=\frac{\text{6.023}\times 10^{26}\times 9\times 10^3}{63}=\text{0.860}\times 10^{29}$
$=8.6\times 10^{28}$
and $A=\frac{\pi D^2}{4}=\frac{22}{7}\times \frac{{\left(10^{-3}\right)}^2}{4}{\mathrm{m}}^2$;
$=\frac{11}{14}\times 10^{-6}{m}^2$
$v_d=\frac{1.1}{8.6\times 10^{28}\times \frac{11}{14}\times 10^{-6}\times 1.6\times 10^{-19}}$
$=\frac{1}{\text{9.6}\times 10^{+3}}=\frac{100\times 10^{-4}}{96}=\text{1.0}\times 10^{-4}m{s}^{-1}$
$=0.1mm{s}^{-1}$