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(D) The relationship between current $I$ and drift velocity $v_d$ is given by the formula: $I = n A e v_d$,where $n$ is the number density of electron

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$1.25$

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(D) The relationship between current $I$ and drift velocity $v_d$ is given by the formula: $I = n A e v_d$,where $n$ is the number density of electrons,$A$ is the cross-sectional area,and $e$ is the elementary charge $(1.6 	imes 10^{-19} \,C)$.Given:$I = 6.4 \,A$$A = 4 	imes 10^{-7} \,m^2$$n = 8 	imes 10^{28} \,m^{-3}$$e = 1.6 	imes 10^{-19} \,C$Rearranging the formula to solve for $v_d$:$v_d = \frac{I}{n A e}$Substituting the values:$v_d = \frac{6.4}{(8 	imes 10^{28}) 	imes (4 	imes 10^{-7}) 	imes (1.6 	imes 10^{-19})}$$v_d = \frac{6.4}{32 	imes 10^{21} 	imes 1.6 	imes 10^{-19}}$$v_d = \frac{6.4}{51.2 	imes 10^2} = \frac{6.4}{5120} = 0.00125 \,m/s$Expressing this in terms of $10^{-3} \,m/s$:$v_d = 1.25 	imes 10^{-3} \,m/s$.Thus,the drift velocity is $1.25$.

The area of cross-section of a copper wire is $4 \times 10^{-7} \,m^2$ and the number of free electrons per cubic metre in copper is $8 \times 10^{28}$. If the wire carries a current of $6.4 \,A$,then the drift velocity of the electrons (in $10^{-3} \,m \,s^{-1}$) is:

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