KCET · Physics · Current Electricity
A conductor wire having \(10^{29}\) free electrons \(/ \mathrm{m}^{3}\) carries a current of \(20 \mathrm{~A}\). If the cross-section of the wire is \(1 \mathrm{~mm}^{2}\), then the drift velocity of electrons will be
\(\left(e=1.6 \times 10^{-19} \mathrm{C}\right)\)
- A \(1.25 \times 10^{-4} \mathrm{~ms}^{-1}\)
- B \(1.25 \times 10^{-3} \mathrm{~ms}^{-1}\)
- C \(1.25 \times 10^{-5} \mathrm{~ms}^{-1}\)
- D \(6.25 \times 10^{-3} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(1.25 \times 10^{-3} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Current, \(I=n A e v_{\mathrm{d}}\)
\(\therefore\) Drift velocity \(v_{\mathrm{d}}=\frac{I}{n A e}\)
\(=\frac{20}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}}\)
\(v_{d}=1.25 \times 10^{-3} \mathrm{~ms}^{-1}\)
\(\therefore\) Drift velocity \(v_{\mathrm{d}}=\frac{I}{n A e}\)
\(=\frac{20}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}}\)
\(v_{d}=1.25 \times 10^{-3} \mathrm{~ms}^{-1}\)
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