A square loop of side 2 . 0   cm is placed inside a long solenoid that has 50 turns per centimetre and…

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=227

Solution

It is given that the current is varying sinusoidally. The current can be written as I=I0sinωt.

The magnetic field through the solenoid B=μ0nI

The flux through the square is ϕ=μ0nIA

The emf is

 ε=μ0nA×d(I0sinωt)dtε=μ0nAI0ωcosωt

The amplitude of the emf is,

ε=μ0nAI0ω=4π×10-7×5010-2×4×10-4×2.5×700ε=44×10-4 V

Asked in: JEE Main 2023 (10 Apr Shift 2)

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