Question

Every atom makes one free electron in copper. If $1.1 \, A$ Current is flowing in the wire of copper having $1 \, mm$ 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. $0.1 \, mm \, s^{- 1}$
• B. $0.2 \, mm \, s^{- 1}$
• C. $0.3 \, mm \, s^{- 1}$
• D. $0.2 \, cm \, s^{- 1}$

Answer 1

Answer: (A)

$v _{d ⁡}=\frac{i ⁡}{n ⁡ Ae}$ ;
where $n =\frac{N_{A} \rho }{M ⁡}$
$=\frac{\text{6.023} \times 10^{26} \times 9 \times 10^{3}}{63}=\text{0.860}\times 10^{2 9}$
$=8.6\times 10^{28}$
and $A=\frac{\pi D^2}{4}=\frac{22}{7} \times \frac{\left(10^{-3}\right)^2}{4} \mathrm{~m}^2$;
$=\frac{11}{14}\times 10^{- 6 \, }m^{2}$
$v_{d}=\frac{1.1}{8.6 \times 10^{28} \times \frac{11}{14} \times 10^{- 6} \times 1.6 \, \times 10^{- 19}}$
$=\frac{1}{\text{9.6} \times 1 0^{+ 3}}=\frac{1 0 0 \times 1 0^{- 4}}{9 6}=\text{1.0}\times 10^{- 4} \, m \, s^{- 1}$
$=0.1 \, mm \, s^{- 1}$