A copper wire of cross-sectional area 1.0 mm | vedclass

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(C) The mass of $1 \ m^3$ of copper is $8990 \ kg = 8990 	imes 10^3 \ g$.The number of moles in $1 \ m^3$ is $n_{moles} = \frac{8990 	imes 10^3}{63.

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(C) The mass of $1 \ m^3$ of copper is $8990 \ kg = 8990 	imes 10^3 \ g$.The number of moles in $1 \ m^3$ is $n_{moles} = \frac{8990 	imes 10^3}{63.5} \approx 1.415 	imes 10^5 \ mol$.Since each mole contains $N_A = 6.022 	imes 10^{23}$ atoms and each atom contributes one free electron,the number density of free electrons $n$ is:$n = (1.415 	imes 10^5) 	imes (6.022 	imes 10^{23}) \approx 8.52 	imes 10^{28} \ m^{-3}$.Using the formula for current $I = neAv_d$,the drift velocity $v_d$ is:$v_d = \frac{I}{neA}$.Given $I = 1.34 \ A$,$A = 1.0 \ mm^2 = 10^{-6} \ m^2$,and $e = 1.6 	imes 10^{-19} \ C$:$v_d = \frac{1.34}{(8.52 	imes 10^{28}) 	imes (1.6 	imes 10^{-19}) 	imes (10^{-6})}$$v_d \approx \frac{1.34}{1.3632 	imes 10^4} \approx 0.98 	imes 10^{-4} \ m/s$.Converting to $mm/s$:$v_d \approx 0.1 \ mm/s$.

$A$ copper wire of cross-sectional area $1.0 \ mm^2$ carries a current of $1.34 \ A$. Assuming that each copper atom contributes one free electron,calculate the drift velocity of the free electrons in the wire in $mm/s$. (Given: density of copper = $8990 \ kg/m^3$,atomic mass = $63.50 \ g/mol$)

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