A small piece of metal wire is dragged in the region of magnetic flux \( 8 \times 10^{-4}\,\mathrm{Wb} \) in 0.5 s. The length of the wire is perpendicular to magnetic field. The emf induced in the wire would be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.6 mV
By Faraday's law, the emf induced in a moving wire equals the rate at which it cuts magnetic flux. Here the wire sweeps through a flux of \( 8\times10^{-4}\,\mathrm{Wb} \) in 0.5 s.
\( \displaystyle \varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{8\times10^{-4}\,\mathrm{Wb}}{0.5\,\mathrm{s}} = 1.6\times10^{-3}\,\mathrm{V} = 1.6\,\mathrm{mV} \)