A metal wire has \(5\times10^{28}\) free electrons per cubic metre. The electrons drift along the wire with a speed of \(0.25\ \mathrm{mm\,s^{-1}}\). The magnitude of the current density in the wire is (Take \(e=1.6\times10^{-19}\ \mathrm{C}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2.0\times10^{6}\ \mathrm{A\,m^{-2}}\)
Current density is the charge that flows per second through unit area normal to the flow. For free electrons of number density \( n \) moving with drift speed \( v_d \), its magnitude is \( j=nev_d \).
\( \begin{aligned} n &= 5\times10^{28}\ \mathrm{m^{-3}} \\ v_d &= 0.25\ \mathrm{mm\,s^{-1}}=2.5\times10^{-4}\ \mathrm{m\,s^{-1}} \end{aligned} \)
\( \begin{aligned} j &= nev_d \\ &= 5\times10^{28}\times1.6\times10^{-19}\times2.5\times10^{-4} \\ &= 2.0\times10^{6}\ \mathrm{A\,m^{-2}} \end{aligned} \)
NCERT (2026-27):
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.”