A potential difference of 0.4 V is applied across a copper wire of length 2 m and cross-sectional area 1 mm2. The wire has \(8\times10^{28}\) free electrons per cubic metre. Their mobility is \(5\times10^{-3}\ \mathrm{m^{2}\,V^{-1}\,s^{-1}}\). The current in the wire is (Take \(e=1.6\times10^{-19}\ \mathrm{C}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 12.8 A
Mobility is the drift speed per unit electric field, so \( v_d=\mu E \). The drifting electrons give a current \( I=neAv_d \). The field in the wire is the potential difference per unit length.
\( \begin{aligned} E &= \frac{V}{l} \\ &= \frac{0.4}{2} \\ &= 0.2\ \mathrm{V\,m^{-1}} \end{aligned} \)
\( \begin{aligned} v_d &= \mu E \\ &= 5\times10^{-3}\times0.2 \\ &= 1\times10^{-3}\ \mathrm{m\,s^{-1}} \end{aligned} \)
With \( A=1\ \mathrm{mm^{2}}=10^{-6}\ \mathrm{m^{2}} \),
\( \begin{aligned} I &= neAv_d \\ &= 8\times10^{28}\times1.6\times10^{-19}\times10^{-6}\times10^{-3} \\ &= 12.8\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Mobility – “An important quantity is the mobility μ defined as the magnitude of the drift velocity per unit electric field:”
\( \mu=\frac{|v_d|}{E} \)\( \qquad (3.24) \)
Summary – “If j is normal to a cross-sectional area A and is constant over the area, the magnitude of the current I through the area is nevd A.”