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InterviewSolution
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When 5V potential difference is applied across a wire of length0.1 m, the drift speed of electrons is 2.5x104ms1 If theelectron density in the wire is 8x1028m 3, the resistivity of thematerial is close toJEE (Main) 2015) |
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Answer» Given that the drift speed of the electrons is Vd = 2.5×10^−4 m/s, charge of the electron is e= 1.6×10^−19 C, electron density is n = 8.0×10^28 per m3. So, the current in the wire is,I = neAVdI = (8.0×10^28)×(1.6×10^−19)×A×(2.5×10^−4)I = A{32×105} amp Resistance of the wire is, R=VI=5/(32×10^5×A) Ω So the resistivity is,ρ = (A/l)*R= A/0.1×(5/32×10^5×A) = 5×10^−4/32⇒ρ=1.56×10^−5 Ω m answer is 1.6*10^-5 |
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