Question

A copper wire of uniform cross-sectional area carries a current of 3.4 A. The drift velocity of conduction electrons is 0.2 mm/s. If the number density of electrons in copper is \(8.5 \times 10^{28} \, \text{m}^{-3}\), find the area of cross-section of the wire.

Hint:

Drift velocity is a small average velocity of electrons due to the electric field, and it is directly related to the current. The cross-sectional area can be determined using the formula \( I = n \times A \times e \times v_d \), where the other quantities are known.

Answer 1

The relationship between current \(I\), drift velocity \(v_d\), electron number density \(n\), electron charge \(e\), and cross-sectional area \(A\) is given by: \[ I = n \times A \times e \times v_d \] The provided values are: \[ I = 3.4 \, \text{A}, \quad n = 8.5 \times 10^{28} \, \text{m}^{-3}, \quad v_d = 0.2 \, \text{mm/s} = 0.2 \times 10^{-3} \, \text{m/s}, \quad e = 1.6 \times 10^{-19} \, \text{C} \] To find the area \(A\), the formula is rearranged to: \[ A = \frac{I}{n \times e \times v_d} \] Substituting the values yields: \[ A = \frac{3.4}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (0.2 \times 10^{-3})} \] Calculating \(A\): \[ A = \frac{3.4}{2.72 \times 10^{7}} = 1.25 \times 10^{-7} \, \text{m}^2 \] Therefore, the cross-sectional area of the wire is: \[ A = 1.25 \times 10^{-7} \, \text{m}^2 \]