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(D) The relationship between current $I$,cross-sectional area $A$,and drift velocity $v_d$ is given by $I = n e A v_d$.Since the area $A = \pi (d/2)^2

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$4V$

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(D) The relationship between current $I$,cross-sectional area $A$,and drift velocity $v_d$ is given by $I = n e A v_d$.Since the area $A = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we can write $I = n e \left( \frac{\pi d^2}{4} \right) v_d$.For the first wire: $I = n e \left( \frac{\pi d^2}{4} \right) V$.For the second wire with diameter $d' = d/2$: $I = n e \left( \frac{\pi (d/2)^2}{4} \right) v' = n e \left( \frac{\pi d^2}{16} \right) v'$.Equating the two expressions for $I$: $n e \left( \frac{\pi d^2}{4} \right) V = n e \left( \frac{\pi d^2}{16} \right) v'$.Simplifying,we get $\frac{V}{4} = \frac{v'}{16}$,which implies $v' = 4V$.

$A$ current $I$ flows through a uniform wire of diameter $d$ when the mean electron drift velocity is $V$. The same current will flow through a wire of diameter $d/2$ made of the same material if the mean drift velocity of the electron is:

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