A 1.0 m long copper wire of radius 3.5 mm carries a current of 0.77 A. Assuming the number of conduction electrons to be ~8 × 1028 m−3, the average drift velocity of conduction electrons will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.56 × 10−6 m/s
The drift velocity follows from the relation between current and carrier motion, \( I = neAv_d \).
\( \displaystyle v_d = \frac{I}{neA} \)
\( \displaystyle A = \pi r^{2} = 3.14 \times (3.5\times10^{-3})^{2} \approx 3.85\times10^{-5}\,\mathrm{m^{2}} \)
\( \displaystyle neA = 8\times10^{28} \times 1.6\times10^{-19} \times 3.85\times10^{-5} \approx 4.93\times10^{5}\,\mathrm{C\,m^{-1}} \)
\( \displaystyle v_d = \frac{0.77}{4.93\times10^{5}} \)
\( \displaystyle v_d \approx 1.56\times10^{-6}\,\mathrm{m\,s^{-1}} \)
The length of the wire is not needed.
Note: NTA dropped this question in the Final Answer Key. The original NTA paper printed Option (2) as 1.56 × 10−5 m/s, while the drift velocity from the given data is 1.56 × 10−6 m/s, so none of the printed options was correct. For the benefit of students, Option (2) has been changed to 1.56 × 10−6 m/s, so that the question has one correct answer.