Current Electricity

JEE Main level, numerical answer, medium

6 of 380

Question 6 of 380

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.s\alpha \times 10^{-3} \mathrm{~m}^{2} / 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)\left.=1.6 \times 10^{-19} \mathrm{C}\right)

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Answer: 1

For a cylindrical conductor with length 2 m, cross-section 0.2 mm² = 0.2 × 10⁻⁶ m², carrying 1.6 A at 2 V.
Resistance: R=V/I=2/1.6=1.25R = V/I = 2/1.6 = 1.25 Ω.
Resistivity: ρ=RA/L=1.25×0.2×10−6/2\rho = RA/L = 1.25 \times 0.2 \times 10^{-6}/2
=0.125×10−6= 0.125 \times 10^{-6} Ω·m.
Using ρ=1/(neμ)\rho = 1/(ne\mu), mobility: μ=1/(neρ)\mu = 1/(ne\rho)
=1/(5×1028×1.6×10−19×0.125×10−6)= 1/(5 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.125 \times 10^{-6})
=1/(1×103)=10−3= 1/(1 \times 10^3) = 10^{-3} m²/V·s.
Thus α=1\alpha = 1.

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