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(D) Given: Length $l = 1 \,m$,Area $A = 5 	imes 10^{-7} \,m^{2}$,Current $I = 1 \,A$,Electron density $n = 8 	imes 10^{28} \,m^{-3}$,Charge of elect

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$6.4 	imes 10^{3} \,s$

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(D) Given: Length $l = 1 \,m$,Area $A = 5 	imes 10^{-7} \,m^{2}$,Current $I = 1 \,A$,Electron density $n = 8 	imes 10^{28} \,m^{-3}$,Charge of electron $e = 1.6 	imes 10^{-19} \,C$. The drift velocity $v_{d}$ is given by the formula $v_{d} = \frac{I}{neA}$. Substituting the values: $v_{d} = \frac{1}{8 	imes 10^{28} 	imes 1.6 	imes 10^{-19} 	imes 5 	imes 10^{-7}} = \frac{1}{64 	imes 10^{2}} = \frac{1}{6.4 	imes 10^{3}} \,m/s$. The time $T$ taken to drift across the length $l$ is $T = \frac{l}{v_{d}}$. $T = \frac{1}{1 / (6.4 	imes 10^{3})} = 6.4 	imes 10^{3} \,s$.

$A$ copper wire of length $1 \,m$ and uniform cross-sectional area $5 \times 10^{-7} \,m^{2}$ carries a current of $1 \,A$. Assuming that there are $8 \times 10^{28}$ free electrons per $m^{3}$ in copper,how long will an electron take to drift from one end of the wire to the other?

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