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
Answer & explanation
Correct answer: option 3
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