A square loop of side 2.0 cm is placed inside a long solenoid that has 50 turns per centimetre and carries a sinusoidally varying current of amplitude 2.5 A and angular frequency 700 rad s⁻¹. The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is x × 10⁻⁴ V. The value of x is __________.
(Take, π=22/7 )

Solution

Physics sir · 4 steps

The loop sits in the solenoid's uniform field, so flux is B times its area. Differentiate once.

1. Field inside. Inside a long solenoid the field is uniform: B = μ₀ n I, with n = 50 turns per cm = 5000 m⁻¹. The current is I = I₀ sin ω t, so B also varies sinusoidally.
2. Flux and emf. The axes coincide, so the field is perpendicular to the loop and Φ = μ₀ n A I₀ sin ω t. Differentiating gives ε = -μ₀ n A I₀ ω cos ω t, and the amplitude is the factor in front.

   ε₀ = μ₀ n A I₀ ω

3. Put in numbers. A = (2 × 10⁻²)² = 4 × 10⁻⁴ m². Then ε₀ = 4π × 10⁻⁷ × 5000 × 4 × 10⁻⁴ × 2.5 × 700 = 1.4π × 10⁻³ V. With π = 22/7, this is 4.4 × 10⁻³ V.
4. So. ε₀ = 44 × 10⁻⁴ V, so x = 44.

The common slip: 0.44. Using n = 50 per metre instead of converting 50 per cm to 5000 m⁻¹ makes the emf 100 times too small, giving 0.44.

The idea to keep. For a coil inside a solenoid, use the solenoid's field times the inner coil's own area. A sinusoidal current gives an emf amplitude that is the flux amplitude times ω.

Answer from the NTA final answer key. Solution by PrepWiser.