Physics · Electromagnetic Induction

JEE Main 2026 — 22 January, Evening Shift — Question 42

A conducting circular loop is rotated about its diameter at a constant angular speed of $100 \mathrm{rad} / \mathrm{s}$ in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by $30^{\circ}$ from the horizontal position, the induced EMF is 15.4 mV . The radius of the loop is $\_\_\_\_$ mm . (Take $\left.\pi=\frac{22}{7}\right)$

Answer: 14

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{E}=\mathrm{B} \omega \mathrm{Asin} \omega \mathrm{t}$ $15.4 \times 10^{-3}=\frac{1}{2} \times 100 \times \frac{22}{7} \mathrm{r}^{2} \times \frac{1}{2}$ $\mathrm{r}=\sqrt{\frac{15.4 \times 28 \times 10^{-5}}{22}}$ $\mathrm{R}=14 \mathrm{~mm}$

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Exam: JEE Main 2026
Paper: 22 January, Evening Shift
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Magnetic Flux, Faraday's Law and Lenz's Law