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(C) The formula for current is $I = n e A v_d$,where $I$ is the current,$n$ is the number density,$e$ is the elementary charge,$A$ is the cross-sectio

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$0.15$

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(C) The formula for current is $I = n e A v_d$,where $I$ is the current,$n$ is the number density,$e$ is the elementary charge,$A$ is the cross-sectional area,and $v_d$ is the drift velocity.Given: $I = 1.7\, A$,$d = 1.02\, mm \implies r = 0.51 	imes 10^{-3}\, m$,$n = 8.5 	imes 10^{28}\, m^{-3}$,$e = 1.6 	imes 10^{-19}\, C$.The area $A = \pi r^2 = \pi 	imes (0.51 	imes 10^{-3})^2 \approx 8.17 	imes 10^{-7}\, m^2$.Rearranging for $v_d$: $v_d = \frac{I}{n e A}$.Substituting the values: $v_d = \frac{1.7}{(8.5 	imes 10^{28}) 	imes (1.6 	imes 10^{-19}) 	imes (\pi 	imes (0.51 	imes 10^{-3})^2)}$.$v_d = \frac{1.7}{8.5 	imes 1.6 	imes \pi 	imes 0.2601 	imes 10^{-7}} \approx \frac{1.7}{11.12} 	imes 10^{-3} \approx 0.15 	imes 10^{-3}\, m/s$.Thus,$v_d = 0.15\, mm/s$.

$A$ copper wire of diameter $1.02\, mm$ carries a current of $1.7\, A$. Find the drift velocity $(v_d)$ of electrons in the wire. Given $n$,the number density of electrons in copper $= 8.5 \times 10^{28} /m^3$. (Express the answer in $mm/s$)

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