A square loop of side 2.0 cm is placed inside a long solenoid having 50 turns per centimetre. The solenoid current has amplitude 2.5 A and angular frequency 700 rad s⁻¹. The loop and solenoid axes coincide. The induced-emf amplitude is x × 10⁻⁴ V. Take π = 22/7. Find x.
Physics · Circular Motion
JEE Mains 10 April 2023 Shift 2 Evening Shift

Question

A square loop of side 2.0 cm is placed inside a long solenoid having 50 turns per centimetre. The solenoid current has amplitude 2.5 A and angular frequency 700 rad s⁻¹. The loop and solenoid axes coincide. The induced-emf amplitude is x × 10⁻⁴ V. Take π = 22/7. Find x.

Physics · Circular Motion

JEE Mains 10 April 2023 Shift 2 Evening Shift

G1 Physics · Accepted Answer

Area of square loop: A = (0.02)² = 4×10⁻⁴ m² Turns per metre of solenoid: n = 50 per cm = 5000 m⁻¹ Magnetic-field amplitude: B₀ = μ₀nI₀ = (4π×10⁻⁷)(5000)(2.5) = 5π×10⁻³ T Flux amplitude: Φ₀ = B₀A Induced-emf amplitude: ε₀ = ωΦ₀ = 700(5π×10⁻³)(4×10⁻⁴) Using π = 22/7: ε₀ = 44×10⁻⁴ V Therefore x = 44. Final value: 44