Question 3

The magnetic field is $\vec{B}(t) = B_0 \cos(\omega t) \hat{k}$ only for $y \ge 0$. The loop rotates about the $X$-axis with angular speed $\omega$. At time $t$, the angle of rotation is $\theta = \omega t$.

1. For $0 \le t \le \pi/\omega$, the loop is in the $y \ge 0$ region. The area vector is $\vec{A} = L^2(\sin \omega t \hat{j} + \cos \omega t \hat{k})$.
2. The magnetic flux is $\Phi = \vec{B} \cdot \vec{A} = (B_0 \cos \omega t \hat{k}) \cdot (L^2 \cos \omega t \hat{k}) = B_0 L^2 \cos^2 \omega t$.
3. Induced e.m.f. $V = -\frac{d\Phi}{dt} = -B_0 L^2 \frac{d}{dt}(\cos^2 \omega t) = -B_0 L^2 (2 \cos \omega t)(-\omega \sin \omega t) = B_0 L^2 \omega \sin(2\omega t)$. This corresponds to one full sine cycle from $t=0$ to $t=\pi/\omega$.
4. For $\pi/\omega < t < 2\pi/\omega$, the loop is in the $y < 0$ region where $\vec{B} = 0$, thus $\Phi = 0$ and $V = 0$. This matches the graph in option (A).