Q 12-06-146JEE MainJEE Main 2018 (16 Apr, Shift 1)Easy
A coil of cross-sectional area $A$ having $n$ turns is placed in a uniform magnetic field $B$. When it is rotated with an angular velocity $\omega$, the maximum e.m.f. induced in the coil will be:
Answer: (C) $nBA\omega$
Flux linkage: $\Phi = nBA\cos\omega t$.
$$\varepsilon = -\frac{d\Phi}{dt} = nBA\omega\sin\omega t \quad\Rightarrow\quad \varepsilon_\text{max} = nBA\omega$$
Solution by Sreeraj P, M.Sc Physics