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(C) The current $I$ in a wire is given by the relation $I = neAv_d$,where $n$ is the number density of free electrons,$e$ is the charge of an electron

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$I/2$

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(C) The current $I$ in a wire is given by the relation $I = neAv_d$,where $n$ is the number density of free electrons,$e$ is the charge of an electron,$A$ is the cross-sectional area,and $v_d$ is the drift velocity.Since the wire has a circular cross-section,$A = \pi r^2$.Therefore,$I = ne(\pi r^2)v_d$.This implies $I \propto r^2 v_d$.For the first wire,$I_1 = I$,$r_1 = r$,and $v_{d1} = V$.For the second wire,$r_2 = r/2$ and $v_{d2} = 2V$.Taking the ratio: $\frac{I_2}{I_1} = \frac{r_2^2 v_{d2}}{r_1^2 v_{d1}} = \frac{(r/2)^2 (2V)}{r^2 V} = \frac{(r^2/4) (2V)}{r^2 V} = \frac{2}{4} = \frac{1}{2}$.Thus,$I_2 = I_1 / 2 = I/2$.

In a wire of circular cross-section with radius $r$,free electrons travel with a drift velocity $V$ when a current $I$ flows through the wire. What is the current in another wire of half the radius and of the same material when the drift velocity is $2V$?

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