Q: Consider a wire of length 4 m and cross-sectional area 1 mm^{2} carrying a current of 2 A. If each cubic metre of the material contains 10^{29} 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 × 10^{4} sec