A planar loop of wire rotates in a uniform magnetic field. Initially, at $t = 0$, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10$ s about an axis in its plane, then the magnitude of the induced emf will be maximum and minimum, respectively, at:
Answer: (B) $2.5$ s and $5.0$ s
$\Phi = BA\cos\omega t$ (maximum at $t = 0$), so $|\varepsilon| = BA\omega|\sin\omega t|$ with $\omega = \dfrac{2\pi}{10}$.
$|\varepsilon|$ is maximum when $\omega t = \dfrac\pi2$: $t = 2.5$ s (also $7.5$ s).
$|\varepsilon|$ is minimum (zero) when $\omega t = \pi$: $t = 5.0$ s (also $10$ s).
So maximum at $2.5$ s and minimum at $5.0$ s.
(The official answer key lists '$5.0$ s and $10.0$ s', but at both those instants the emf is zero.)
Solution by Sreeraj P, M.Sc Physics