Question

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 electrons, find the average time taken by an electron to cross the length of the wire.

Answer 1

Given:
Current through the wire, i = 2 A
Length of the wire, l = 4 m
Area of cross section, A = 1 mm² = 1 × 10^–6 m²
Number of electrons per unit volume, n = 10²⁹
We know:
$$\begin{gathered} i=nA{V}_{d}e \\ \Rightarrow V_d=\frac{i}{nAe} \\ =\frac{2}{{10}^{29}\times {10}^{-6}\times 1.6\times {10}^{-19}} \\ \Rightarrow V_{\mathrm{d}}=\frac{1}{8000}\mathrm{m}/\mathrm{s}, \end{gathered}$$
where V_d is the drift speed.
Let t be the time taken by an electron to cross the length of the wire.
$$\begin{gathered} \Rightarrow t=\frac{\text{L}\mathrm{ength}\mathrm{of}\text{the}\mathrm{wire}}{\text{D}\mathrm{rift}\mathrm{speed}} \\ =\frac{l}{V_{\mathrm{d}}} \\ \therefore t=\frac{4}{{\displaystyle \frac{1}{8000}}} \\ =32\times {10}^3\mathrm{s} \end{gathered}$$