Q: Consider a wire of length 4 m and cross-sectional area 1 mm2 carrying a current of 2 A. If each cubic metre of the material contains 1029 free electrics, find the average time taken by an electron to cross the length of the wire.
Sol: $\large v_d = \frac{I}{n A e}$
$\large v_d = \frac{2}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}}$
= 1.25 × 10-4 ms-1
Average time taken by an electron to cross the length of wire
$\large t = \frac{l}{v_d} = \frac{4}{1.25 \times 10^{-4}}$
= 3.2 × 104 sec