A metal has \(10^{29}\) free electrons per cubic metre. The relaxation time of these electrons is \(2.25\times10^{-14}\ \mathrm{s}\). An electric field of \(0.05\ \mathrm{V\,m^{-1}}\) is set up in a wire of this metal. The current density in the wire is (Take \(e=1.6\times10^{-19}\ \mathrm{C}\) and \(m=9\times10^{-31}\ \mathrm{kg}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3.2\times10^{6}\ \mathrm{A\,m^{-2}}\)
The drift picture of electrons gives the conductivity of a metal as \( \sigma=ne^{2}\tau/m \). Ohm’s law written for current density is \( j=\sigma E \).
\( \begin{aligned} \sigma &= \frac{ne^{2}\tau}{m} \\ &= \frac{10^{29}\times(1.6\times10^{-19})^{2}\times2.25\times10^{-14}}{9\times10^{-31}} \\ &= 6.4\times10^{7}\ \mathrm{S\,m^{-1}} \end{aligned} \)
\( \begin{aligned} j &= \sigma E \\ &= 6.4\times10^{7}\times0.05 \\ &= 3.2\times10^{6}\ \mathrm{A\,m^{-2}} \end{aligned} \)
NCERT (2026-27):
Drift of Electrons and the Origin of Resistivity – “… if we identify the conductivity σ as”
\( \sigma=\frac{ne^{2}}{m}\tau \)\( \qquad (3.23) \)
Ohm’s Law – “Thus, the last equation can be written as, E = jρ … or, j = σ E … where σ ≡1/ρ is called the conductivity.”