Consider a wire carrying a current of 10,A wi | vedclass

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Question

(A) The relation between current $I$ and drift velocity $v_d$ is given by the formula:$I = n e A v_d$Where:$I = 10\,A$ (current)$n = 9 	imes 10^{28}\

Vedclass · Accepted Answer

$6.94 	imes 10^{-6}\,m/s$

Vedclass · Answer

(A) The relation between current $I$ and drift velocity $v_d$ is given by the formula:$I = n e A v_d$Where:$I = 10\,A$ (current)$n = 9 	imes 10^{28}\,m^{-3}$ (number density of electrons)$e = 1.6 	imes 10^{-19}\,C$ (charge of an electron)$A = 1\,cm^2 = 1 	imes 10^{-4}\,m^2$ (cross-sectional area)Rearranging for drift velocity $v_d$:$v_d = \frac{I}{n e A}$Substituting the values:$v_d = \frac{10}{(9 	imes 10^{28}) 	imes (1.6 	imes 10^{-19}) 	imes (1 	imes 10^{-4})}$$v_d = \frac{10}{9 	imes 1.6 	imes 10^{28-19-4}}$$v_d = \frac{10}{14.4 	imes 10^5}$$v_d = \frac{10}{1.44 	imes 10^6} \approx 6.94 	imes 10^{-6}\,m/s$

Consider a wire carrying a current of $10\,A$ with a cross-sectional area of $1\,cm^2$. If the number of electrons per unit volume is $9 \times 10^{28}\,m^{-3}$,find the drift velocity of the electrons.

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