Practice question
Question
A copper wire of length \( 2 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2
\) carries a current of \( 2 \, \text{A} \). If the number density of free electrons in copper is \( 8.5
\times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \), what is the drift
speed of the electrons?
Explanation
**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Drift speed is given by v_d = (I/n e A) . Given: I = 2 A , n = 8.5 × 10²⁸ m⁻³ , e = 1.6 × 10⁻¹⁹ C , A = 2 × 10⁻⁶ m² . Substitute: v_d = (2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2 × 10⁻⁶) . Calculate: v_d = (2/2.72 × 10⁴) = 7.35 × 10⁻⁵ m/s . Applying I = n e
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