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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2018
  • Hard
  • MCQ

At the centre of a fixed large circular coil of radius \(R\), a much smaller circular coil of radius \(r\) 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 \(\omega \) about an axis along their common diameter. Calculate the \(emf\) induced in the smaller coil after a time \(t\) of its start of rotation

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Answer & Solution

Correct Answer

(C) \(\frac{{{\mu _0}I}}{{2R}}\,\omega \pi {r^2}\,\sin \,\omega t\)

Step-by-step Solution

Detailed explanation

According to Faraday's law of electromagnetic induction,

\(e=-\frac{d \phi}{d t} \text { and } \phi=B A \cos \omega t=B \pi r^{2} \cos \omega t\)

\(\Rightarrow \quad e=-\frac{d}{d t}\left(\pi r^{2} B \cos \omega t\right)=\pi r^{2} B \sin \omega t(\omega)\)

\(\therefore e=\frac{\mu_{0} I}{2 R} \pi \omega r^{2} \sin \omega t\left(\because B=\frac{\mu_{0} I}{2 R}\right)\)

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