A cylindrical conductor of length 2 m and area of cross-section 0.2 mm² carries an electric current of 1.6 A when its ends are connected to a 2 V 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) Numerical answer type. Work it out, then open the answer.

Answer: 1. E = V/l = 2/2 = 1 V/m vd = I/neA = 1.65 × 10²⁸ × 1.6 × 10⁻¹⁹ × 0.2 × 10⁻⁶ = 10⁻³ m/s μ = vd/E = 1 × 10⁻³ m²/V s So α = 1

A cylindrical conductor of length 2 m and area of cross-section 0.2… | JEE Main 2026 (22 Jan, Shift 2)

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Electric Current, Drift Velocity & Current Density

A cylindrical conductor of length 2 m and area of cross-section $0.2\text{ mm}^{2}$ carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3}\text{ m}^{2}\text{/V s}$ . The value of $\alpha$ is:

(electron concentration $= 5 \times 10^{28}\text{ /m}^{3}$ and electron charge $= 1.6 \times 10^{-19}\text{ C}$ )

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

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Answer: 1

$E = \dfrac{V}{l} = \dfrac{2}{2} = 1\text{ V/m}$
$v_d = \dfrac{I}{neA} = \dfrac{1.6}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.2 \times 10^{-6}} = 10^{-3}\text{ m/s}$
$\mu = \dfrac{v_d}{E} = 1 \times 10^{-3}\text{ m}^{2}\text{/V s}$
So $\alpha = 1$

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