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JEE Main 201815 AprilEvening ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Evening Shift
Mode
Online
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2018 · 15 April · Evening Shift · Online · Physics · Electromagnetic Induction

At the centre of a fixed large circular coil of radius R , a much...

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.

Options

  1. A\frac{\mu_0\mathrm{I}}{2 \mathrm{R}} \omega \mathrm{r}^2 \sin \omega \mathrm{t}
  2. Bμ0I4Rωπr2sin⁡ωt\frac{\mu_0 I}{4 R} \omega \pi r^2 \sin \omega t
  3. Cμ0I2Rωπr2sin⁡ωt\frac{\mu_0 I}{2 R} \omega \pi r^2 \sin \omega t
  4. D\frac{\mu_0\mathrm{I}}{4 \mathrm{R}} \omega \mathrm{r}^2 \sin \omega \mathrm{t}

Correct answer

C. μ0I2Rωπr2sin⁡ωt\frac{\mu_0 I}{2 R} \omega \pi r^2 \sin \omega t

Step-by-step solution

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}
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