JEE Main 2026 — Jan 22, Shift 2 — Physics solution

JEE Main2026Jan 22, Shift 2Physics

A cylindrical conductor of length 2 m and area of cross-section 0.2 mm $^{2}$ carries an electric current of 1.6 A when its ends are connected to a 2V battery. Mobility of electrons in the conductor is $α \times 1 0^{- 3}$ m $^{2}$ /V $\cdot$ s. The value of $α$ is :

(electron concentration $= 5 \times 1 0^{28}$ /m $^{3}$ and electron charge $= 1.6 \times 1 0^{- 19}$ C)

Your answer

Solution

Current $I = n e A v_{d}$ , so $v_{d} = \frac{I}{n e A}$ .

Mobility $μ = \frac{v _{d}}{E}$ with $E = \frac{V}{ℓ}$ , so $μ = \frac{I ℓ}{n e A V}$ .

$μ = \frac{1.6 \times 2}{5 \times 1 0 ^{28} \times 1.6 \times 1 0 ^{- 19} \times 0.2 \times 1 0 ^{- 6} \times 2}$ .

Denominator: $5 \times 1.6 \times 0.2 \times 2 \times 1 0^{28 - 19 - 6} = 3.2 \times 1 0^{3}$ .

$μ = \frac{3.2}{3.2 \times 1 0 ^{3}} = 1 0^{- 3}$ m $^{2}$ /V $\cdot$ s.

So $α = 1$ .

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