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)

A cylindrical conductor of length 2 m and area of cross-section 0.2 mm² carries an electric current of 1.6 A

Q. A cylindrical conductor of length $2\ \text{m}$ and area of cross-section $0.2\ \text{mm}^2$ carries an electric current of $1.6\ \text{A}$ when its ends are connected to a $2\ \text{V}$ battery. Mobility of electrons in the conductor is $\alpha\times10^{-3}\ \text{m}^2/\text{V·s}$. The value of $\alpha$ is :

(electron concentration $= 5\times10^{28}\ /\text{m}^3$ and electron charge $= 1.6\times10^{-19}\ \text{C}$)

Correct Answer: 1

Explanation

This problem is based on current electricity and mobility of charge carriers, a very common topic in JEE Main, JEE Advanced and IIT JEE.

Electric field in the conductor is given by

$$ E = \frac{V}{L} $$

$$ E = \frac{2}{2} = 1\ \text{V/m} $$

Current flowing through a conductor is related to mobility by

$$ I = n q A \mu E $$

Given:

$$ I = 1.6\ \text{A},\quad n = 5\times10^{28}\ /\text{m}^3,\quad q = 1.6\times10^{-19}\ \text{C} $$

Area of cross-section:

$$ A = 0.2\ \text{mm}^2 = 0.2\times10^{-6}\ \text{m}^2 $$

Substituting values:

$$ 1.6 = (5\times10^{28})(1.6\times10^{-19})(0.2\times10^{-6})\mu(1) $$

$$ 1.6 = 1600\,\mu $$

$$ \mu = 10^{-3}\ \text{m}^2/\text{V·s} $$

Comparing with $\alpha\times10^{-3}$,

$$ \alpha = 1 $$

Explanation

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