When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10⁻⁴ m s⁻¹. If the electron density in the wire is 8 × 10²⁸ m⁻³, the resistivity of the material is close to: (a) 1.6 × 10⁻⁶ Ω m (b) 1.6 × 10⁻⁵ Ω m (c) 1.6 × 10⁻⁸ Ω m (d) 1.6 × 10⁻⁷ Ω m

Answer: (b). E = V/l = 5/0.1 = 50 V/m and J = nevd ρ = E/J = (E)/(nevd) = 508 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2.5 × 10⁻⁴ ρ = 503.2 × 10⁶ ≈ 1.6 × 10⁻⁵ Ω m So the answer is (b).

When 5 V potential difference is applied across a wire of length 0.1… | JEE Main 2015

Q32JEE Main 2015Medium

Electric Current, Drift Velocity & Current Density

When $5\text{ V}$ potential difference is applied across a wire of length $0.1\text{ m},$ the drift speed of electrons is $2.5 \times 10^{-4}\text{ m s}^{-1}.$ If the electron density in the wire is $8 \times 10^{28}\text{ m}^{-3},$ the resistivity of the material is close to:

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Answer: (b)

$E = \dfrac{V}{l} = \dfrac{5}{0.1} = 50\text{ V/m}$ and $J = nev_d$
$\rho = \dfrac{E}{J} = \dfrac{E}{nev_d} = \dfrac{50}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.5 \times 10^{-4}}$
$\rho = \dfrac{50}{3.2 \times 10^{6}} \approx 1.6 \times 10^{-5}\ \Omega\text{ m}$
So the answer is (b).

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