A \( 2.0\ \mathrm{m} \) long copper wire of area of cross-section \( 2\times10^{-6}\ \mathrm{m^{2}} \) is carrying a current of \( 2.0\ \mathrm{A} \). Assuming that the number density of free electrons in the wire is \( 10^{29}\ \mathrm{m^{-3}} \), the drift velocity of electrons is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 6.25\times10^{-2}\ \mathrm{mm/s} \)
The current is carried by electrons drifting with velocity \( v_{d} \):
\( I = neAv_{d} \)
\( \displaystyle v_{d} = \frac{I}{neA} \)
\( I = 2.0\,\mathrm{A},\ n = 10^{29}\,\mathrm{m^{-3}},\ A = 2\times10^{-6}\,\mathrm{m^{2}},\ e = 1.6\times10^{-19}\,\mathrm{C} \)
\( \displaystyle v_{d} = \frac{2.0}{10^{29}\times1.6\times10^{-19}\times2\times10^{-6}}\ \mathrm{m\,s^{-1}} = \frac{2.0}{3.2\times10^{4}}\ \mathrm{m\,s^{-1}} \)
\( v_{d} = 6.25\times10^{-5}\,\mathrm{m\,s^{-1}} \)
Converting to mm/s (1 m = \( 10^{3} \) mm):
\( v_{d} = 6.25\times10^{-5}\times10^{3}\,\mathrm{mm/s} = 6.25\times10^{-2}\,\mathrm{mm/s} \)
The length of the wire (2.0 m) is not needed, because the drift velocity depends only on the current density \( I/A \) and on \( n \).