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(B) Given: Length of wire $l = 3 \,m$,Cross-sectional area $A = 6.14 	imes 10^{-6} \,m^2$,Current $I = 6 \,A$,Atomic weight $M = 108 \,g/mol = 0.108

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$10^{-4} \,m/s$

Vedclass · Answer

(B) Given: Length of wire $l = 3 \,m$,Cross-sectional area $A = 6.14 	imes 10^{-6} \,m^2$,Current $I = 6 \,A$,Atomic weight $M = 108 \,g/mol = 0.108 \,kg/mol$,Density $ho = 10500 \,kg/m^3$,Avogadro number $N_A = 6.023 	imes 10^{23} /mol$,Charge of electron $e = 1.6 	imes 10^{-19} \,C$.The number of atoms per unit volume $n$ is given by $n = \frac{ho N_A}{M}$.Since each atom contributes one electron,the number of free electrons per unit volume is $n = \frac{10500 	imes 6.023 	imes 10^{23}}{0.108} \approx 5.86 	imes 10^{28} \,m^{-3}$.The drift velocity $v_d$ is given by the formula $I = n e A v_d$.Rearranging for $v_d$: $v_d = \frac{I}{n e A}$.Substituting the values: $v_d = \frac{6}{(5.86 	imes 10^{28}) 	imes (1.6 	imes 10^{-19}) 	imes (6.14 	imes 10^{-6})}$.$v_d = \frac{6}{5.86 	imes 1.6 	imes 6.14 	imes 10^{3}} \approx \frac{6}{57.56 	imes 10^3} \approx 1.04 	imes 10^{-4} \,m/s$.Thus,the drift velocity is close to $10^{-4} \,m/s$.

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