\(e=-\frac{d \phi}{d t} \text { and } \phi=B A \cos \omega t=B \pi r^{2} \cos \omega t\)
\(\Rightarrow \quad e=-\frac{d}{d t}\left(\pi r^{2} B \cos \omega t\right)=\pi r^{2} B \sin \omega t(\omega)\)
\(\therefore e=\frac{\mu_{0} I}{2 R} \pi \omega r^{2} \sin \omega t\left(\because B=\frac{\mu_{0} I}{2 R}\right)\)