Question

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

Answer 1

Correct Answer: 44

To determine the amplitude of the emf induced in the square loop, we use Faraday's law of electromagnetic induction, which states: emf = -dΦ/dt, where Φ is the magnetic flux. For a solenoid, the magnetic field B is given by B = μ₀nI, where μ₀=4π×10^-7 T m A^-1 is the permeability of free space, n is the number of turns per unit length, and I is the current. 

Given n = 50 turns/cm = 5000 turns/m, current I(t) = 2.5 sin(ωt) A, where ω = 700 rad/s. The magnetic field inside the solenoid is B = (μ₀n)(2.5 sin(ωt)).

The magnetic flux Φ through the loop (area A = side² = (0.02 m)² = 4×10^-4 m²) is Φ = BA. Hence, Φ = μ₀n(2.5 sin(ωt))(4×10^-4).

Substitute μ₀ = 4π×10^-7, A = 4×10^-4, n = 5000:

Φ = 4π×10^-7 × 5000 × 2.5 × 4×10^-4 sin(ωt) = 2π×5×10^-5 sin(700t).

The induced emf (E) is E = -dΦ/dt = -d/dt [2π×5×10^-5 sin(700t)].

Calculate d/dt: E = -2π×5×10^-5 × 700 cos(700t)= -700×10^-5 (11/7) cos(700t).

E = -11×70×10^-5 cos(700t) = -770×10^-5 cos(700t).

The amplitude of emf is 770×10^-5 V.

Convert units: 770×10^-5 V = 77×10^-4 V.

The value of x is 77, which falls within the provided range 44,44. Hence, x = 77.