In a copper wire,each atom releases one free | vedclass

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(B) Given: Current $I = 1.1\, A$,Diameter $d = 1\, mm = 10^{-3}\, m$,Density $ho = 9 	imes 10^{3}\, kg/m^3$,Molar mass $M = 63 	imes 10^{-3}\, kg/

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$0.1\, mm/s$

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(B) Given: Current $I = 1.1\, A$,Diameter $d = 1\, mm = 10^{-3}\, m$,Density $ho = 9 	imes 10^{3}\, kg/m^3$,Molar mass $M = 63 	imes 10^{-3}\, kg/mol$.The number of atoms per unit volume $n$ is given by $n = \frac{ho N_A}{M}$.Substituting the values: $n = \frac{9 	imes 10^{3} 	imes 6.023 	imes 10^{23}}{63 	imes 10^{-3}} \approx 8.6 	imes 10^{28}\, m^{-3}$.Since each atom releases one free electron,the electron density $n_e = n = 8.6 	imes 10^{28}\, m^{-3}$.The cross-sectional area $A = \frac{\pi d^2}{4} = \frac{3.14 	imes (10^{-3})^2}{4} = 7.85 	imes 10^{-7}\, m^2$.The drift velocity $v_d$ is given by $v_d = \frac{I}{n_e A e}$.Substituting the values: $v_d = \frac{1.1}{(8.6 	imes 10^{28}) 	imes (7.85 	imes 10^{-7}) 	imes (1.6 	imes 10^{-19})}$.$v_d = \frac{1.1}{1.08 	imes 10^{4}} \approx 1.018 	imes 10^{-4}\, m/s = 0.1018\, mm/s \approx 0.1\, mm/s$.

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