a
(a) If in time \(t\). the rod turns by an angle \(\theta\), the area generated by the rotation of rod will be

\( = \frac{1}{2}l \times l\theta \) \( = \frac{1}{2}{l^2}\theta \)

So the flux linked with the area generated by the rotation of rod

\(\phi = B\;\left( {\frac{1}{2}{l^2}\theta } \right)\cos 0 = \frac{1}{2}B{l^2}\theta = \frac{1}{2}B{l^2}\omega \,t\)

And so \(e = \frac{{d\phi }}{{dt}} = \frac{d}{{dt}}\left( {\frac{1}{2}B{l^2}\omega t} \right) = \frac{1}{2}B{l^2}\omega \)