A potential difference of 0.064 V is applied across a metal wire of length 2 m. The conductivity of the metal is \(5\times10^{7}\ \mathrm{S\,m^{-1}}\). It has \(10^{29}\) free electrons per cubic metre. The drift speed of the electrons is (Take \(e=1.6\times10^{-19}\ \mathrm{C}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1\times10^{-4}\ \mathrm{m\,s^{-1}}\)
In terms of current density, Ohm’s law reads \( j=\sigma E \). The same current density is carried by drifting electrons, \( j=nev_d \). The field in the wire is the potential difference per unit length.
\( \begin{aligned} E &= \frac{V}{l} \\ &= \frac{0.064}{2} \\ &= 0.032\ \mathrm{V\,m^{-1}} \end{aligned} \)
\( \begin{aligned} j &= \sigma E \\ &= 5\times10^{7}\times0.032 \\ &= 1.6\times10^{6}\ \mathrm{A\,m^{-2}} \end{aligned} \)
\( \begin{aligned} v_d &= \frac{j}{ne} \\ &= \frac{1.6\times10^{6}}{10^{29}\times1.6\times10^{-19}} \\ &= 1\times10^{-4}\ \mathrm{m\,s^{-1}} \end{aligned} \)
NCERT (2026-27):
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.”
Summary – “Current density j gives the amount of charge flowing per second per unit area normal to the flow, j = nq vd where n is the number density (number per unit volume) of charge carriers each of charge q, and vd is the drift velocity of the charge carriers. For electrons q = – e.”