There are \( 10^{29} \) free electrons per cubic metre of copper. For a current of 16 A in a copper wire of cross-section \( 1\,\mathrm{mm^{2}} \), the drift speed of the free electrons along the wire is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 10^{-3}\,\mathrm{m\,s^{-1}} \)
Free electrons carry the current in a wire, moving with the drift speed \( v_d \). The current is \( I=neAv_d \). Here \( n \) is the number of free electrons per unit volume and \( A \) is the area of cross-section. Take \( e=1.6\times10^{-19}\,\mathrm{C} \).
\( n=10^{29}\,\mathrm{m^{-3}} \)
\( A=1\,\mathrm{mm^{2}}=10^{-6}\,\mathrm{m^{2}} \)
\( \begin{aligned} v_d &= \frac{I}{neA} \\ &= \frac{16\,\mathrm{A}}{10^{29}\,\mathrm{m^{-3}}\times1.6\times10^{-19}\,\mathrm{C}\times10^{-6}\,\mathrm{m^{2}}} \\ &= \frac{16}{1.6\times10^{4}}\,\mathrm{m\,s^{-1}} \\ &= 10^{-3}\,\mathrm{m\,s^{-1}} \end{aligned} \)