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An aluminium (Al) rod with area of cross-section \(4 \times 10^{-6} \mathrm{~m}^2\) has a current of 5 ampere, flowing through it. Find the drift velocity of electron in the rod. Density of \(\mathrm{Al}=2.7 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\) and Atomic wt. \(=27\). Assume that each \(\mathrm{Al}\) atom provides one electron
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Verified Answer
The correct answer is:
\(1.29 \times 10^{-4} \mathrm{~m} / \mathrm{s}\)
Electron density, \(n=\frac{d \times N}{M}\) where \(\mathrm{d}\) = density, \(\mathrm{N}=\) Avogadro number, \(\mathrm{M}=\) atomic weight
\(\begin{aligned}\text {so, }
n & =\frac{2.7 \times 10^3 \times 6.02 \times 10^{26}}{27}(\text { atom } / \mathrm{kg}) \\
& =6.02 \times 10^{28} \text { electrons } / \mathrm{m}^3
\end{aligned}\)
\(\therefore \quad\) Drift velocity, \(v_d=\frac{I}{n e A}\)
\(\begin{aligned} \text { or }
v_d & =\frac{5}{6.02 \times 10^{28} \times 1.6 \times 10^{-19} \times 4 \times 10^{-6}} \\
& =1.29 \times 10^{-4} \mathrm{~m} / \mathrm{s}
\end{aligned}\)
\(\begin{aligned}\text {so, }
n & =\frac{2.7 \times 10^3 \times 6.02 \times 10^{26}}{27}(\text { atom } / \mathrm{kg}) \\
& =6.02 \times 10^{28} \text { electrons } / \mathrm{m}^3
\end{aligned}\)
\(\therefore \quad\) Drift velocity, \(v_d=\frac{I}{n e A}\)
\(\begin{aligned} \text { or }
v_d & =\frac{5}{6.02 \times 10^{28} \times 1.6 \times 10^{-19} \times 4 \times 10^{-6}} \\
& =1.29 \times 10^{-4} \mathrm{~m} / \mathrm{s}
\end{aligned}\)
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