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Physics · Current Electricity

JEE Main 2026 — 22 January, Evening Shift — Question 45

A cylindrical conductor of length 2 m and area of cross-section 0.2 mm20.2 \mathrm{~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 α×10−3 m2/V\alpha \times 10^{-3} \mathrm{~m}^{2} / \mathrm{V}.s. The value of α\alpha is : (electron concentration =5×1028/m3=5 \times 10^{28} / \mathrm{m}^{3} and electron charge =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C} )

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

Vℓ=μE=μ×Vℓ\mathrm{V}_{\ell}=\mu \mathrm{E}=\mu \times \frac{\mathrm{V}}{\ell}

I=neAVd\mathrm{I}=\mathrm{neAV}_{\mathrm{d}} Vd=IneA\mathrm{V}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}} μ=Iℓ NneA \mu=\frac{\mathrm{I} \ell}{\text { NneA }} μ=1.6×32×5×1026×1.6×10−19×2×10−7\mu=\frac{1.6 \times 3}{2 \times 5 \times 10^{26} \times 1.6 \times 10^{-19} \times 2 \times 10^{-7}} μ=1×10−3 m2/v⋅ s\mu=1 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{v} \cdot \mathrm{~s} α=1\alpha=1

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Evening Shift
Subject
Physics
Chapter
Current Electricity
Topic
Current in Conductors, Drift Speed, Resistivity and Relaxation Time

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