The number of free electrons per 100 mm of ordinary copper wire is \( 2 \times 10^{21} \). The average drift speed of electrons is 0.25 mm/s. The amount of current flowing would be: \( (e = 1.6 \times 10^{-19}\,\mathrm{C}) \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 0.8\,\mathrm{A} \)
The current is the charge crossing a section per second. The free electrons move along the wire with the drift speed \( v_d \), so the electrons in a length \( v_d \) pass a section every second. If \( N \) is the number of free electrons per unit length,
\( \displaystyle I = N\,e\,v_{d} \)
\( \displaystyle N = \frac{2\times10^{21}}{100\,\mathrm{mm}} = \frac{2\times10^{21}}{0.1\,\mathrm{m}} = 2\times10^{22}\,\mathrm{m^{-1}} \)
\( \displaystyle v_{d} = 0.25\,\mathrm{mm\,s^{-1}} = 2.5\times10^{-4}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle I = 2\times10^{22} \times 1.6\times10^{-19} \times 2.5\times10^{-4} = 0.8\,\mathrm{A} \)