JEE Main 2024 Physics question
A coil of 200 turns and area 0.20 m is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is volt. The value of β is .
Answer: 5
Solution
ε₀ = NBAω = 200 × 0.01 × 0.20 × π = 0.4π = 2π/5 V.