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

Step-by-step solution

It is given that the current is varying sinusoidally. The current can be written as I = I 0 sin ω t . The magnetic field through the solenoid B = μ 0 n I The flux through the square is ϕ = μ 0 n I A The emf is ε = μ 0 n A × d ( I 0 sin ω t ) d t ⇒ ε = μ 0 n A I 0 ω cos ω t The amplitude of the emf is, ε = μ 0 n A I 0 ω = 4 π × 10 - 7 × 50 10 - 2 × 4 × 10 - 4 × 2 . 5 × 700 ⇒ ε = 44 × 10