A wire of \( 2.0 \) mm diameter carries a current of \( 1.0 \) A. The drift velocity of electrons is \( 1 \) mm s\( ^{-1} \). The electron density in the wire is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.99\times10^{27}\,\mathrm{m^{-3}} \)
The current in a conductor is carried by electrons drifting with speed \( v_d \): \( I = neAv_d \), where n is the number density of free electrons.
\( d = 2.0\,\mathrm{mm} \Rightarrow r = 1.0\times10^{-3}\,\mathrm{m} \)
\( A = \pi r^{2} = 3.14\times10^{-6}\,\mathrm{m^{2}} \)
\( v_d = 1\,\mathrm{mm\,s^{-1}} = 10^{-3}\,\mathrm{m\,s^{-1}} \)
\( \displaystyle n = \frac{I}{eAv_d} = \frac{1.0}{(1.6\times10^{-19})(3.14\times10^{-6})(10^{-3})} \)
\( \displaystyle n = \frac{1}{5.03\times10^{-28}} \approx 1.99\times10^{27}\,\mathrm{m^{-3}} \)