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 × 10²⁸ m^–3, the resistivity of the material is close to:

1.6 x 10^-8Ωm

1.6 x 10^-7Ωm

1.6 x 10^-6Ωm

1.6 x 10^-5Ωm

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Solution

1.6 x 10^-5Ωm

vd = 25 x 10-4 m/s ⇒ n = 8 x 10²⁸ /m³

we know that
J= nevd or I = nevd A
where, symbols have their usual meaning

rightwards double arrow space straight V over straight R space equals nev subscript straight d straight A or space fraction numerator straight V over denominator begin display style ρL over straight A end style end fraction equals space nev subscript straight d straight A or space straight V over ρL equals nev subscript straight d straight rho space equals space fraction numerator straight V over denominator nev subscript straight d straight L end fraction space equals space fraction numerator 5 over denominator 8 space straight x space 10 to the power of 28 space straight x space 16 space straight x space 10 to the power of negative 19 end exponent space straight x space 25 space straight x space 10 to the power of negative 4 end exponent space straight x space 0.1 end fraction space straight rho space equals space 16 space straight x space 10 to the power of negative 5 end exponent space Ωm

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