When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10^–4 ms^–1. If the electron density in the wire is 8 × 10^28 m^–3, the resistivity of the…

Question

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10^–4 ms^–1. If the electron density in the wire is 8 × 10^28 m^–3, the resistivity of the material is close to:

Accepted Answer

Correct answer: D. (d) We know I= neAv_d ,I= VR and R= lA So VR= neAv_d V nev_dl= RAl ⇒ = V nev_dl = 58 10^28 1.6 10^-19 2.5 10^-4 0.1 Ωm =1.56 10^-5 m ≈1.6 10^-5 Ωm

$When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 \times 1 0^{-4} m s^{-1} . If the electron density in the wire is 8 \times 1 0^{28} m^{-3}, the resistivity of the material is close to:$

Solution

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