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Physics · Electromagnetic Induction

JEE Main 2024 — 9 April, Shift 2 — Question 35

A square loop of side 15 cm being moved towards right at a constant speed of 2 cm/s2 \mathrm{~cm} / \mathrm{s} as shown in figure. The front edge enters the 50 cm wide magnetic field at t=0t=0. The value of induced emf in the loop at t=10 s\mathrm{t}=10 \mathrm{~s} will be :

Question figure
  1. Option A:

    0.3 mV

  2. Option B:

    4.5 mV

  3. Option C:

    zero

    Correct
  4. Option D:

    3 mV

Answer: C

Step-by-step solution

At t=10sec\mathrm{t}=10 \mathrm{sec} complete loop is in magnetic field therefore no change in flux

e=dϕdt=0\mathrm{e}=\frac{\mathrm{d} \phi}{\mathrm{dt}}=0 e=0\mathrm{e}=0 for complete loop

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Magnetic Flux, Faraday's Law and Lenz's Law
A square loop of side 15 cm being moved towards right at a constant… | JEE Main 2024 PYQ with Solution · DhiX AI