A coil of area 0.2 m² having 20 turns rotates about its vertical diameter with angular speed 50 rad s⁻¹ in a uniform horizontal magnetic field of 0.003 Wb m⁻². The maximum induced emf is x/10 V. Find x.
Physics · Mixed Chapter
JEE Mains

Question

A coil of area 0.2 m² having 20 turns rotates about its vertical diameter with angular speed 50 rad s⁻¹ in a uniform horizontal magnetic field of 0.003 Wb m⁻². The maximum induced emf is x/10 V. Find x.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Magnetic flux: For N turns rotating in a uniform magnetic field, Φ = NBA cos(ωt). Step 2 — Faraday’s law: e = −dΦ/dt = NBAω sin(ωt). Step 3 — The peak emf is e₀ = NBAω. Substitute N = 20, B = 0.003 T, A = 0.2 m² and ω = 50 rad/s. e₀ = 20 × 0.003 × 0.2 × 50 = 0.6 V. Step 4 — Given e₀ = x/10 V; therefore x/10 = 0.6 and x = 6. Answer: 6.