Electrostatics

JEE Main level, numerical answer, medium

90 of 311

Question 90 of 311

A 1m1m long copper wire carries a current of 1A1A. If the cross section of the wire is 2.0mm22.0{mm}^{2} and the resistivity of copper is 1.7×10−8Ωm1.7\times {10}^{-8}\Omega m. The force experienced by moving electron in the wire is _____×10−23N\times {10}^{-23}N
(Charge of electron=1.6×10−19C=1.6\times {10}^{-19}C )

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Answer: 136

Given here, length l=1ml=1m, current i=1Ai=1A, area A=2×10−6mA=2\times {10}^{-6}m and resistivity ρ=1.7×10−8Ωm\rho =1.7\times {10}^{-8}\Omega m.
Resistance of wire is R=ρlA=1.7×10−8×12×10−5=1.72×10−2ΩR=\dfrac{\rho l}{A}=\dfrac{1.7\times {10}^{-8}\times 1}{2\times {10}^{-5}}=\dfrac{1.7}{2}\times {10}^{-2}\OmegaNow, electric field in the wire is E=ρJE=\rho J, where, J=iAJ=\dfrac{i}{A} is current density.
Thus, E=iρA=iR=1×1.72×10−2Vm−1E=\dfrac{i\rho }{A}=iR=1\times \dfrac{1.7}{2}\times {10}^{-2}V{m}^{-1}
Electric force in terms of electric field is given by F=qEF=qE
Therefore, force experienced by moving electron is F=1.6×10−19×1.72×10−2F=1.6\times {10}^{-19}\times \dfrac{1.7}{2}\times {10}^{-2}
=1.36×10−21=1.36\times {10}^{-21}=136×10−23N=136\times {10}^{-23}N

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