Practice question
Question
A copper wire of cross-sectional area \( 6 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1.8
\, \text{A} \). If \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \) and \( e = 1.6 \times 10^{-19} \,
\text{C} \), what is the drift speed?
Explanation
**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. Drift speed: v_d = (I/n e A) . Substitute: v_d = (1.8/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 6 × 10⁻⁷) . Calculate: v_d = (1.8/8.16 × 10³) ≈ 2.21 × 10⁻⁴ m/s . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields 2.21 × 10⁻⁴
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.