Two cylindrical rods of uniform crosssection area $A$ and $2A$, having free electrons per unit volume $2n$ and $n$ respectively are joined in series. A current $I$ flows through them in steady state. Then the ratio of drift velocity of free electron in left rod to drift velocity of electron in the right rod is $\left( {\frac{{{v_L}}}{{{v_R}}}} \right)$
Medium
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Since current ${\rm{I}} = neA{v_d}$ through both rods
is same $2(\mathrm{n})$ e $\mathrm{A} \mathrm{v}_{\mathrm{L}}=\mathrm{n}$ e $(2 \mathrm{A}) \mathrm{v}_{\mathrm{R}} \Rightarrow \frac{\mathrm{v}_{\mathrm{L}}}{\mathrm{v}_{\mathrm{R}}}=1$
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