A cylindrical conductor of length 2m and area of cross-section 0.2 mm² carries an electric current of 1.6 A when its ends are connected to a 2V battery. Mobility of electrons in the conductor is α × 10⁻³ m²/V.s. The value of α is :
(electron concentration = 5 × 10²⁸/m³ and electron charge = 1.6 × 10⁻¹⁹ C)

Question

A cylindrical conductor of length 2m and area of cross-section 0.2 mm² carries an electric current of 1.6 A when its ends are connected to a 2V battery. Mobility of electrons in the conductor is α × 10⁻³ m²/V.s. The value of α is :

(electron concentration = 5 × 10²⁸/m³ and electron charge = 1.6 × 10⁻¹⁹ C)

G1 Physics · Accepted Answer

Electric field inside the conductor is E = V/l. Electron drift speed is v(d) = μE = μV/l. Current is I = neA v(d). Therefore, I = neA(μV/l). Hence, μ = Il/(neAV). Substitute: I = 1.6 A, l = 2 m, n = 5 × 10²⁸ m⁻³, e = 1.6 × 10⁻¹⁹ C, A = 0.2 mm² = 2 × 10⁻⁷ m², V = 2 V. Thus, μ = (1.6×2)/ [(5×10²⁸)(1.6×10⁻¹⁹)(2×10⁻⁷)(2)] = 1 × 10⁻³ m²/(V·s). Therefore, α = 1.