A small piece of metal wire is dragged in the region of magnetic flux $8 × 10^{-4} Wb$ in 0.5 s.The length of the wire is perpendicular to magnetic field. The emf induced in the wire would be

1.6 mV

The correct answer is Option (3) → 1.6 mV

Given:

Φ = 8×10⁻⁴ Wb, Δt = 0.5 s

Faraday's law of electromagnetic induction gives:

$\mathcal{E} = \frac{\Delta \Phi}{\Delta t}$

Substitute the values:

$\mathcal{E} = \frac{8 \times 10^{-4}}{0.5} = 1.6 \times 10^{-3}\ \text{V}$

Answer: 1.6×10⁻³ V