KCET · Physics · Current Electricity
A current of \(2 \mathrm{~A}\) is passing through a metal wire of cross-sectional area \(2 \times 10^{-6} \mathrm{~m}^{2}\). If the number density of free electrons in the wire is \(5 \times 10^{26} \mathrm{~m}^{-3}\), the drift speed of electrons is (Given, \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\) )
- A \(\frac{1}{32} \mathrm{~ms}^{-1}\)
- B \(\frac{1}{16} \mathrm{~ms}^{-1}\)
- C \(\frac{1}{40} \mathrm{~ms}^{-1}\)
- D \(\frac{1}{80} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{80} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Drift velocity of electrons
\(\begin{aligned}v_{\mathrm{d}} &=\frac{\mathrm{I}}{n e A} \\&=\frac{2}{5 \times 10^{26} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}} \end{aligned}\)
or \(\quad \mathrm{v}_{\mathrm{d}}=\frac{1}{80} \mathrm{~ms}^{-1}\)
\(\begin{aligned}v_{\mathrm{d}} &=\frac{\mathrm{I}}{n e A} \\&=\frac{2}{5 \times 10^{26} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}} \end{aligned}\)
or \(\quad \mathrm{v}_{\mathrm{d}}=\frac{1}{80} \mathrm{~ms}^{-1}\)
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