A long copper wire has a resistance of \( 10\,\Omega \) and an area of cross-section \( 1\,\mathrm{mm^{2}} \). A potential difference of \( 10\,\mathrm{V} \) exists across the wire. If the number of electrons per cubic meter in copper is \( 8\times10^{28} \), the drift velocity will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 0.78\times10^{-4}\,\mathrm{m/s} \)
The current in a conductor is related to the drift velocity by \( I = neAv_d \). The current itself comes from Ohm’s law.
\( I = V/R = 10\,\mathrm{V}/10\,\Omega = 1\,\mathrm{A} \)
\( A = 1\,\mathrm{mm^{2}} = 1\times10^{-6}\,\mathrm{m^{2}} \)
\( \displaystyle v_d = \frac{I}{neA} = \frac{1\,\mathrm{A}}{(8\times10^{28}\,\mathrm{m^{-3}})(1.6\times10^{-19}\,\mathrm{C})(1\times10^{-6}\,\mathrm{m^{2}})} \)
\( \displaystyle v_d = \frac{1}{1.28\times10^{4}}\,\mathrm{m/s} \approx 0.78\times10^{-4}\,\mathrm{m/s} \)