JEE Main 2024PhysicsElectromagnetic InductionHardNumerical

JEE Main 2024 — Electromagnetic Induction Question with Solution

JEE Main 2024 (27 Jan Shift 1)

Question

Two coils have mutual inductance 0.002 H. The current changes in the first coil according to the relation i=i0sinωt, where i0=5 A and ω=50π rad s-1. The maximum value of emf in the second coil is πα V. The value of α is

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Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

The amount of flux induced in the second coil due to the first one is given by

ϕ2=Mi=Mi0sinωt   ...1

where, M is the mutual inductance of the coils.

The formula to calculate the emf induced can be written as

ε=-dϕ2dt   ...2

From equation (2), it follows that,

εmax=Mi0ω=0.002×5×50π V=π2 V

Hence, α=2.

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About this question

This is a previous-year question from JEE Main 2024, covering the Electromagnetic Induction chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.