A circular loop of radius 20 cm and resistance 2Ω is placed in a time varying magnetic field B = (2t 2+2t+3) T . At t=0 , for the plane of the loop being perpendicular to the magnetic field, the induced current in the loop at t=3 s is α/50 A . The value of α is (Take π = 22/7 )

Physics · Electromagnetic Induction

JEE Main 2026 — 5 April, Evening Shift — Question 24

A circular loop of radius $20\mathrm{cm}$ and resistance $2\Omega$ is placed in a time varying magnetic field $B = (2t^2+2t+3)\mathrm{T}$ . At $t=0$ , for the plane of the loop being perpendicular to the magnetic field, the induced current in the loop at $t=3\mathrm{s}$ is $\frac{\alpha}{50}\mathrm{A}$ . The value of $\alpha$ is (Take $\pi = 22/7$ )

Answer: 44

Numerical answer — enter this value.

Step-by-step solution

$\phi = B \pi r^2$ , $\epsilon = d\phi/dt = \pi r^2 dB/dt$ , $dB/dt = 4t+2$ . At t=3, $dB/dt = 14$ . $\epsilon = \pi (0.2)^2 \times 14 = \pi \times 0.04 \times 14 = 0.56\pi$ . $i = \epsilon/R = 0.56\pi/2 = 0.28\pi = 0.28\times 22/7 = 0.88$ A = $44/50$ A, so $\alpha = 44$ .

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Magnetic Flux, Faraday's Law and Lenz's Law

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Answer: 44

Step-by-step solution

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