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JEE Mains · Physics · STD 12 -6. Electromagnetic induction
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
- A \(\frac{{{\mu _0}I}}{{2R}}\,\omega {r^2}\,\sin \,\omega t\)
- B \(\frac{{{\mu _0}I}}{{4R}}\,\omega \pi {r^2}\,\sin \,\omega t\)
- C \(\frac{{{\mu _0}I}}{{2R}}\,\omega \pi {r^2}\,\sin \,\omega t\)
- D \(\frac{{{\mu _0}I}}{{4R}}\,\omega {r^2}\,\sin \,\omega t\)
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, \(e=-\frac{d \phi}{d t} \text { and } \phi=B A \cos \omega t=B \pi r^{2} \cos \omega t\) \(\Rightarrow \quad e=-\frac{d}{d t}\left(\pi r^{2} B \cos \omega t\right)=\pi r^{2} B \sin \omega t(\omega)\)…
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