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(C) Given, cross-sectional area, $A = 1 \,cm^2 = 10^{-4} \,m^2$. Charge density, $n = 3 	imes 10^{23} \,m^{-3}$. Current through wire, $I = 24 \,mA =

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$5 	imes 10^{-3} \,m/s$

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(C) Given, cross-sectional area, $A = 1 \,cm^2 = 10^{-4} \,m^2$. Charge density, $n = 3 	imes 10^{23} \,m^{-3}$. Current through wire, $I = 24 \,mA = 24 	imes 10^{-3} \,A$. We know that for conductors, the relation between current and drift velocity is $I = n A e v_d$, where $e = 1.6 	imes 10^{-19} \,C$. Rearranging for drift velocity, $v_d = \frac{I}{n A e}$. Substituting the values: $v_d = \frac{24 	imes 10^{-3}}{3 	imes 10^{23} 	imes 10^{-4} 	imes 1.6 	imes 10^{-19}}$. $v_d = \frac{24 	imes 10^{-3}}{4.8 	imes 10^{0}} = \frac{24}{4.8} 	imes 10^{-3} = 5 	imes 10^{-3} \,m/s$.

$A$ conducting wire of cross-sectional area $1 \,cm^2$ has $3 \times 10^{23}$ charge carriers per $m^3$. If the wire carries a current of $24 \,mA$, then the drift velocity of the carriers is:

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