Practice free →
HomeJEE MainPhysicsElectromagnetic Induction › A 100-turn coil of area 200 cm² is rotated at 50…

A 100-turn coil of area 200 cm² is rotated at 50 rev/s in a field of 0.05 T. Peak EMF:

A100 V
B31.4 V
C6.28 V
D314 V
Answer & Solution
Correct answer: B. 31.4 V
ε₀ = N B A ω = 100 × 0.05 × 200 × 10⁻⁴ × (2π × 50) = 100 × 0.05 × 0.02 × 100π ≈ 31.4 V.
Solve this in the app — JEE Main practice & 24k+ MCQs →
Related questions