A current of 1.6, A is flowing through a wire | vedclass

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(A) The relationship between current $i$ and drift velocity $V_d$ is given by the formula: $i = neAV_d$,where $n$ is the electron density,$e$ is the e

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$10^{-4}\, m/s$

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(A) The relationship between current $i$ and drift velocity $V_d$ is given by the formula: $i = neAV_d$,where $n$ is the electron density,$e$ is the elementary charge $(1.6 	imes 10^{-19}\, C)$,and $A$ is the cross-sectional area.Given:$i = 1.6\, A$$n = 10^{29}\, m^{-3}$$A = 1\, mm^2 = 10^{-6}\, m^2$$e = 1.6 	imes 10^{-19}\, C$Rearranging for $V_d$:$V_d = \frac{i}{neA}$Substituting the values:$V_d = \frac{1.6}{10^{29} 	imes 1.6 	imes 10^{-19} 	imes 10^{-6}}$$V_d = \frac{1.6}{1.6 	imes 10^{29-19-6}}$$V_d = \frac{1}{10^4} = 10^{-4}\, m/s$.

$A$ current of $1.6\, A$ is flowing through a wire having a cross-sectional area of $1\, mm^2$. If the density of free electrons in the material of the wire is $10^{29}\, m^{-3}$,the drift velocity of the electrons will be:

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