The number density of free electrons in copper is nearly 8 × 10²⁸ m⁻³. A copper wire has its area of cross section = 2 × 10⁻⁶ m² and is carrying a current of 3.2 A. The drift speed of the electrons is ___________ × 10⁻⁶ m s⁻¹ Numerical answer type. Work it out, then open the answer.

Answer: 125. vd = I/neA vd = 3.28 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2 × 10⁻⁶ = 3.22.56 × 10⁴ vd = 1.25 × 10⁻⁴ = 125 × 10⁻⁶ m s⁻¹ Answer: 125

The number density of free electrons in copper is nearly 8 × 10²⁸… | JEE Main 2023 (8 Apr, Shift 2)

Q10JEE Main 2023 · 8 APR · SHIFT 2Easy

Electric Current, Drift Velocity & Current Density

The number density of free electrons in copper is nearly $8 \times 10^{28}\text{ m}^{-3}$ . A copper wire has its area of cross section $= 2 \times 10^{-6}\text{ m}^{2}$ and is carrying a current of $3.2\text{ A}$ . The drift speed of the electrons is ___________ $\times 10^{-6}\text{ m s}^{-1}$

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 125

$v_d = \dfrac{I}{neA}$
$v_d = \dfrac{3.2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}} = \dfrac{3.2}{2.56 \times 10^{4}}$
$v_d = 1.25 \times 10^{-4} = 125 \times 10^{-6}\text{ m s}^{-1}$
Answer: 125

Report a mistake

Electric Current, Drift Velocity & Current Density question from JEE Main 2023 (8 Apr, Shift 2)

More questions on Electric Current, Drift Velocity & Current Density