A metal wire of cross-sectional area 2 mm2 carries a current of 2.56 A. The metal has a density of \(8.0\times10^{3}\ \mathrm{kg\,m^{-3}}\) and an atomic mass of 60 u. Each atom gives one free electron. The drift speed of the free electrons is (Take Avogadro’s number \(=6\times10^{23}\ \mathrm{mol^{-1}}\) and \(e=1.6\times10^{-19}\ \mathrm{C}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(1\times10^{-4}\ \mathrm{m\,s^{-1}}\)
The current is carried by the free electrons, \( I=neAv_d \). So we first need \( n \), the number of free electrons per cubic metre. With one free electron per atom, \( n \) is the number of atoms in one cubic metre.
One cubic metre of the metal has a mass of \( 8.0\times10^{3}\ \mathrm{kg}=8.0\times10^{6}\ \mathrm{g} \). A mass of 60 g holds \( 6\times10^{23} \) atoms.
\( \begin{aligned} n &= \frac{8.0\times10^{6}}{60}\times6\times10^{23} \\ &= 8\times10^{28}\ \mathrm{m^{-3}} \end{aligned} \)
With \( A=2\ \mathrm{mm^{2}}=2\times10^{-6}\ \mathrm{m^{2}} \),
\( \begin{aligned} v_d &= \frac{I}{neA} \\ &= \frac{2.56}{8\times10^{28}\times1.6\times10^{-19}\times2\times10^{-6}} \\ &= 1\times10^{-4}\ \mathrm{m\,s^{-1}} \end{aligned} \)
NCERT (2026-27):
Example 3.1 – “… The drift speed vd is given by … vd = (I/neA) … The density of conduction electrons, n is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one).”