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A current of $5 \mathrm{~A}$ is passing through a metallic wire of cross-sectional area $4 \times 10^{-6} \mathrm{~m}^{2}$. If the density of the charge carriers in the wire is $5 \times$ $10^{26} \mathrm{~m}^{-3}$, the drift speed of the electrons will be $\left[\mathrm{e}=1.602 \times 10^{-19} \mathrm{C}\right]$
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Verified Answer
The correct answer is:
$1.56 \times 10^{-2} \mathrm{~ms}^{-1}$
In a metal, conduction current is due to electrons given by
$\begin{array}{l}
I=n A e v \\
\Rightarrow \text { drift velocity, } v=\frac{\mathrm{I}}{\mathrm{nAe}} \\
\Rightarrow \mathrm{v}=\frac{5}{5 \times 10^{26} \times 4 \times 10^{-6} \times 1.602 \times 10^{-19}} \\
\quad=\frac{1}{4 \times 1.602 \times 10^{1}} \\
\quad=\frac{10^{-1}}{6.408}=1.56 \times 10^{-2} \mathrm{~m} / \mathrm{s}
\end{array}$
$\begin{array}{l}
I=n A e v \\
\Rightarrow \text { drift velocity, } v=\frac{\mathrm{I}}{\mathrm{nAe}} \\
\Rightarrow \mathrm{v}=\frac{5}{5 \times 10^{26} \times 4 \times 10^{-6} \times 1.602 \times 10^{-19}} \\
\quad=\frac{1}{4 \times 1.602 \times 10^{1}} \\
\quad=\frac{10^{-1}}{6.408}=1.56 \times 10^{-2} \mathrm{~m} / \mathrm{s}
\end{array}$
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