A metal wire of cross-sectional area 0.5 mm² and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance 1 Ω. The density, atomic mass and electrical conductivity of the metal are 6.35 × 10³ kg m⁻³, 63.5 g mol⁻¹ and 2 × 10⁸ S m⁻¹, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s⁻¹) of the electrons in the wire is: [Take Avogadro’s number as 6 × 10²³ and charge of the electron as 1.6 × 10⁻¹⁹ C.]
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Question

A metal wire of cross-sectional area 0.5 mm² and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance 1 Ω. The density, atomic mass and electrical conductivity of the metal are 6.35 × 10³ kg m⁻³, 63.5 g mol⁻¹ and 2 × 10⁸ S m⁻¹, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s⁻¹) of the electrons in the wire is: [Take Avogadro’s number as 6 × 10²³ and charge of the electron as 1.6 × 10⁻¹⁹ C.]

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G1 Physics · Accepted Answer

Wire resistance R = l/(σA) = 100/[2 × 10⁸ × 0.5 × 10⁻⁶] = 1 Ω. Including the battery internal resistance, total resistance = 2 Ω, so current I = 2/2 = 1 A. Number density n = (ρ/M)N(A) = [6.35 × 10³/0.0635](6 × 10²³) = 6 × 10²⁸ m⁻³. Using I = neAv(A), drift speed v(d) = 1/[6 × 10²⁸ × 1.6 × 10⁻¹⁹ × 0.5 × 10⁻⁶] ≈ 2.08 × 10⁻⁴ m s⁻¹ = 0.208 mm s⁻¹.