Electromagnetic Induction

JEE Main level, single correct, medium

40 of 153

Question 40 of 153

At the centre of a fixed large circular coil of radius R\mathrm{R}, a much smaller circular coil of radius rr is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity ω\omega about an axis along their common diameter. Calculate the emf induced in the smaller coil after a time tt of its start of rotation.

Show answer and solution

Answer: (C) μ0I2Rωπr2sin⁡ωt\frac{\mu_0 I}{2 R} \omega \pi r^2 \sin \omega t

According to Faraday's law of electromagnetic induction,
e=−dϕdt and ϕ=BAcos⁡ωt=Bπr2cos⁡ωt⇒e=−ddt(πr2Bcos⁡ωt)=πr2Bsin⁡ωt(ω)∴e=μ0I2Rπωr2sin⁡ωt(∵B=μ0I2R)\begin{array}{r} e=-\frac{d \phi}{d t} \text { and } \phi=B A \cos \omega t=B \pi r^2 \cos \omega t \\ \Rightarrow \quad e=-\frac{d}{d t}\left(\pi r^2 B \cos \omega t\right)=\pi r^2 B \sin \omega t(\omega) \\ \therefore e=\frac{\mu_0 I}{2 R} \pi \omega r^2 \sin \omega t\left(\because B=\frac{\mu_0 I}{2 R}\right) \end{array}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Electromagnetic induction simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Electromagnetic Induction (Magnetic Flux, Faraday's and Lenz law), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.