Physics · Electromagnetic Induction

JEE Main 2024 — 29 January, Shift 1 — Question 51

A square loop of side 10 cm and resistance $0.7 \Omega$ is placed vertically in east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then, magnitude of induced emf is $\sqrt{\mathrm{x}} \times 10^{-3} \mathrm{~V}$ . The value of $x$ is $\qquad$ .

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\vec{A}=(0.1)^{2} \hat{j}$

$\overrightarrow{\mathrm{B}}=\frac{0.2}{\sqrt{2}} \hat{\mathrm{i}}+\frac{0.2}{\sqrt{2}} \hat{\mathrm{j}}$

Magnitude of induced emf $\mathrm{e}=\frac{\Delta \phi}{\Delta \mathrm{t}}=\frac{\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}-0}{1}=\sqrt{2} \times 10^{-3} \mathrm{~V}$

Answer key and solution verified before publishing.

Practise Electromagnetic Induction

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Magnetic Flux, Faraday's Law and Lenz's Law