JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Current Electricity

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 2 V battery. Mobility of electrons in the conductor is

α \times 1 0^{- 3} m^{2} / V

.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

)

Answer: 1

Solution

The drift speed is

v_{d} = \frac{I}{n e A}

, and the mobility is

μ = \frac{v _{d}}{E} = \frac{v _{d} ℓ}{V} = \frac{I ℓ}{n e A V}

. With

I = 1.6 A

ℓ = 2 m

n = 5 \times 1 0^{28} m^{- 3}

e = 1.6 \times 1 0^{- 19} C

A = 0.2 m m^{2} = 0.2 \times 1 0^{- 6} m^{2}

and

V = 2 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} = \frac{3.2}{3.2 \times 1 0 ^{3}} = 1 \times 1 0^{- 3} m^{2} / (V s) .

Hence

α = 1

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Integer answer