The number density of free electrons in a cop | vedclass

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(C) Given:Number density of free electrons,$n = 8.5 	imes 10^{28} \; m^{-3}$Length of the wire,$l = 3.0 \; m$Area of cross-section,$A = 2.0 	imes 10

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$2.7 	imes 10^{4} \; s$

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(C) Given:Number density of free electrons,$n = 8.5 	imes 10^{28} \; m^{-3}$Length of the wire,$l = 3.0 \; m$Area of cross-section,$A = 2.0 	imes 10^{-6} \; m^2$Current,$I = 3.0 \; A$Charge of an electron,$e = 1.6 	imes 10^{-19} \; C$The relation for current is $I = nAe v_d$,where $v_d$ is the drift velocity.Since $v_d = \frac{l}{t}$,we have $I = nAe \frac{l}{t}$.Rearranging for time $t$,we get $t = \frac{nAe l}{I}$.Substituting the values:$t = \frac{(8.5 	imes 10^{28}) 	imes (2.0 	imes 10^{-6}) 	imes (1.6 	imes 10^{-19}) 	imes 3.0}{3.0}$$t = 8.5 	imes 2.0 	imes 1.6 	imes 10^{(28 - 6 - 19)}$$t = 27.2 	imes 10^3 = 2.72 	imes 10^4 \; s$.Rounding to two significant figures,$t = 2.7 	imes 10^4 \; s$.

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