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(C) The current $I$ flowing through a series circuit remains constant at all cross-sections of the wire,so $I$ is constant.From the relation $I = neAv

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Drift speed of electrons at point $A$ is lesser than that at point $B$.

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(C) The current $I$ flowing through a series circuit remains constant at all cross-sections of the wire,so $I$ is constant.From the relation $I = neAv_d$,where $n$ is the number density of electrons,$e$ is the charge of an electron,$A$ is the cross-sectional area,and $v_d$ is the drift speed.Since $I$,$n$,and $e$ are constant,we have $v_d \propto \frac{1}{A}$.As the cross-sectional area $A$ decreases along the wire from $A$ to $B$,the drift speed $v_d$ must increase.Therefore,the drift speed at point $A$ is less than the drift speed at point $B$,i.e.,$(v_d)_A < (v_d)_B$.

Suppose a current-carrying wire has a cross-sectional area that gradually becomes smaller along the wire,having the shape of a very long cone as shown in the figure. Choose the correct statement.

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