Solution:
The induced emf is given by \(e = BA\omega \sin\theta\). Given \(e = \frac{\sqrt{3}}{2}BA\omega\), we have \(\sin\theta = \frac{\sqrt{3}}{2}\), which yields \(\theta = \frac{\pi}{3}\).
Induced emf produced in a coil of area \(A\) rotating with angular velocity \(\omega\) in a magnetic field \(B\) is equal to \[\frac{\sqrt{3}BA\omega}{2}\], when the angle between the area vector of the coil and direction of magnetic field is
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