Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (Density of copper $= 9 \times 10^{3} \, \mathrm{kg} \, \mathrm{m}^{-3}$ and atomic weight = 63)
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Density of $cu = 9 \times 10^{3} \mathrm{kg/m^{3}}$ (mass of 1 m 3 of Cu)
Q 6.0 × × 10 23 atoms has a mass = 63 × × 10 –3 kg
∴ ∴ Number of electrons per m 3 are
$= \frac{6.0 \times 10^{23}}{63 \times 10^{-3}} \times 9 \times 10^{3} = 8.5 \times 10^{28}$
Now drift velocity $= V_{d} = \frac{i}{neA}$
$= \frac{1.1}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times \pi \times (0.5 \times 10^{-3})^{2}}$
$= 0.1 \times 10^{-3} \, m/sec$
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