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Physics · Electromagnetic Induction

JEE Main 2026 — 6 April, Evening Shift — Question 11

A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as B=0.4sin⁡(300t)B = 0.4 \sin(300t) Tesla. The normal to the plane of loop makes an angle of 60° with the field. The maximum induced emf produced in the loop is ______ mV.

  1. Option A:

    12

  2. Option B:

    18

  3. Option C:

    21

  4. Option D:

    24

    Correct

Answer: D

Step-by-step solution

Area A=(0.02)2=4×10−4A = (0.02)^2 = 4\times10^{-4} m². Flux ϕ=BAcos⁡θ=0.4sin⁡(300t)×4×10−4×0.5=8×10−5sin⁡(300t)\phi = BA\cos\theta = 0.4\sin(300t)\times4\times10^{-4}\times0.5 = 8\times10^{-5}\sin(300t). Induced emf E=∣dϕ/dt∣=8×10−5×300cos⁡(300t)\mathcal{E} = |d\phi/dt| = 8\times10^{-5}\times300\cos(300t), maximum = 2.4×10−22.4\times10^{-2} V = 24 mV.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Evening Shift
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Magnetic Flux, Faraday's Law and Lenz's Law
A square loop of side 2 cm is placed in a time varying magnetic field… | JEE Main 2026 PYQ with Solution · DhiX AI