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(B) In a series circuit,the current $i$ flowing through the conductor remains constant at every cross-section.Since $i = n e A v_{d}$,where $n$ is the

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Drift velocity of electron increases.

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(B) In a series circuit,the current $i$ flowing through the conductor remains constant at every cross-section.Since $i = n e A v_{d}$,where $n$ is the number density of electrons,$e$ is the charge,$A$ is the cross-sectional area,and $v_{d}$ is the drift velocity,we have $v_{d} = \frac{i}{n e A}$.As we move from $P$ to $Q$,the radius $r$ decreases,so the cross-sectional area $A = \pi r^{2}$ decreases. Since $i$ is constant,the drift velocity $v_{d}$ must increase.From the relation $v_{d} = \frac{e E 	au}{m}$,where $E$ is the electric field,$	au$ is the relaxation time,and $m$ is the mass of the electron,we see that $v_{d} \propto E$. Therefore,as $v_{d}$ increases,the electric field $E$ also increases.Since the current $i$ is constant throughout the conductor,the electron current does not change.Thus,as we move from $P$ to $Q$,the drift velocity of electrons increases and the electric field increases.

In the given figure,a battery of emf $E$ is connected across a conductor $PQ$ of length $L$ and different area of cross-sections having radii $r_{1}$ and $r_{2}$ $(r_{2} < r_{1})$. Choose the correct option as one moves from $P$ to $Q$:

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