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

Step 1: Understanding the induced emf in a loop inside a solenoid.
The induced emf in the loop is given by the equation: \[ \text{emf} = B A W N \sin(\omega t) \] Where:
- \( B \) is the magnetic field in the solenoid,
- \( A \) is the area of the loop,
- \( W \) is the angular frequency of the solenoid's current,
- \( N \) is the number of turns per unit length.
Step 2: Calculating the magnetic field \( B \) in the solenoid.
The magnetic field \( B \) inside a solenoid is given by: \[ B = \mu_0 n I \] Where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m}/\text{A} \),
- \( n = 5000 \, \text{turns/m} \),
- \( I = 2.5 \, \text{A} \).
So, the magnetic field \( B \) is: \[ B = 4\pi \times 10^{-7} \times 5000 \times 2.5 = 5\pi \times 10^{-3} \, \text{T} \] Step 3: Area of the loop.
The area of the loop is: \[ A = 2 \, \text{cm} \times 2 \, \text{cm} = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \] Step 4: Calculating the induced emf.
Substitute the values into the equation for emf: \[ \text{emf} = (5 \times 10^{-3}) \times (4 \times 10^{-4}) \times 700 \times 1 \] \[ \text{emf} = 5 \times \frac{22}{7} \times 4 \times 700 \times 10^{-7} = 44 \times 10^{-4} \, \text{V} \]