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:-

\[i = nA V_d e\]

\[ \Rightarrow  V_d  = \frac{i}{nAe}\]

\[ = \frac{2}{{10}^{29} \times {10}^{- 6} \times 1 . 6 \times {10}^{- 19}}\]

\[ \Rightarrow  V_d  = \frac{1}{8000}  m/s,\]

where V_d is the drift speed.

Let t be the time taken by an electron to cross the length of the wire.

\[\Rightarrow t = \frac{\text{Length of the wire}}{\text{Drift speed}}\] 

\[ = \frac{l}{V_d}\] 

\[ \therefore   t = \frac{4}{\frac{1}{8000}}\] 

\[ = 32 \times  {10}^3   s\]