JEE Main 2024 Physics question
A square loop of side 10 cm and resistance 0.7 Ω is placed vertically in east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then magnitude of induced emf is V. The value of x is .
Answer: 2
Solution
The field makes 45° with the normal of the loop, so ε = ΔΦ/Δt = (BA cos45° − 0)/1 = 0.2 × 10⁻² × (1/√2) = √2 × 10⁻³ V, hence x = 2.