A current of 2 A flows through a wire of cross-sectional area 25.0 mm². The free-electron number density is 2.0 × 10²⁸ m⁻³. The drift velocity is ______ × 10⁻⁶ m s⁻¹. [e = 1.6 × 10⁻¹⁹ C]
Physics · Circular Motion
JEE Mains 8 April 2023 Shift 1 Morning Shift

Question

A current of 2 A flows through a wire of cross-sectional area 25.0 mm². The free-electron number density is 2.0 × 10²⁸ m⁻³. The drift velocity is ______ × 10⁻⁶ m s⁻¹. [e = 1.6 × 10⁻¹⁹ C]

Physics · Circular Motion

JEE Mains 8 April 2023 Shift 1 Morning Shift

G1 Physics · Accepted Answer

Drift current relation is I = neAv(d) Therefore, v(d) = I/(neA) Given: I = 2 A n = 2.0×10²⁸ m⁻³ e = 1.6×10⁻¹⁹ C A = 25.0 mm² = 25×10⁻⁶ m² So, v(d) = 2 / [(2×10²⁸)(1.6×10⁻¹⁹)(25×10⁻⁶)] = 2.5×10⁻⁵ m s⁻¹ = 25×10⁻⁶ m s⁻¹ Final value: 25