A conducting square loop initially lies in the XZ-plane with its lower edge hinged along the X-axis.Only in the region y ≥ 0, there is a time dependent magnetic field along Z-direction: B(t) = B₀ cos(ωt) k̂  where B₀ is a constant. The magnetic field is zero elsewhere. At t = 0, the loop starts rotating with constant angular speed ω about the X-axis in clockwise direction (as seen from +X-axis).Ignoring self-inductance and gravity, which plot correctly represents the induced emf (V) vs time?

To determine the correct plot of the induced electromotive force ($V$) versus time ($t$), we analyze the motion of the loop and the magnetic field region step-by-step:

1. Identifying the Region and Magnetic Field:
The magnetic field is given as:
$\overrightarrow{B}\left(t\right)=B_{0}cos\left(\omega t\right)\hat{k}$ for $y\ge 0$
$\overrightarrow{B}\left(t\right)=0$ for $y<0$

2. Motion of the Loop:
Initially at $t=0$, the loop lies in the XZ-plane ($y=0$). It starts rotating about the X-axis clockwise (as seen from the positive X-axis) with a constant angular velocity $\omega$.
- During the time interval $t\in \left[0,\frac{\pi }{\omega }\right]$, the loop rotates through the region $y\ge 0$.
- During the time interval $t\in \left[\frac{\pi }{\omega },\frac{2\pi }{\mathrm{��}}\right]$, the loop is in the region $y<0$ where the magnetic field is zero.

3. Magnetic Flux through the Loop:
Let $A$ be the area of the square loop.
At time $t$, when the loop is in the region $y\ge 0$, the angle between the normal vector of the loop $\overrightarrow{A}$ and the Z-axis (direction of the magnetic field $\overrightarrow{B}$) is $\left(\frac{\pi }{2}-\omega t\right)$.
The magnetic flux $\Phi \left(t\right)$ is:
$\Phi \left(t\right)=\overrightarrow{B}·\overrightarrow{A}=B\left(t\right)Acos\left(\frac{\pi }{2}-\omega t\right)$
$\Phi \left(t\right)=\left[B_{0}cos\left(\omega t\right)\right]Asin\left(\omega t\right)=\frac{1}{2}{B}_{0}Asin\left(2\omega t\right)$

4. Induced EMF ($V$):
By Faraday's law of induction, the induced electromotive force is:
$V=-\frac{d\Phi }{dt}$
$V=-\frac{d}{dt}\left[\frac{1}{2}{B}_{0}Asin\left(2\omega t\right)\right]=-B_{0}A\omega cos\left(2\omega t\right)$

Evaluating this expression over key times:
- At $t=0$, $V=-B_{0}A\omega$ (negative non-zero value).
- At $t=\frac{\pi }{4\omega }$, $V=0$.
- At $t=\frac{\pi }{2\omega }$, $V=B_{0}A\omega$ (maximum positive value).
- At $t=\frac{3\pi }{4\omega }$, $V=0$.
- At $t=\frac{\pi }{\omega }$, $V=-B_{0}A\omega$.

For the interval $t\in \left[\frac{\pi }{\omega },\frac{2\pi }{\omega }\right]$:
The loop is in the region where the magnetic field is zero, meaning $\Phi =0$ and thus $V=0$.

This periodic behavior of $V$ matches the plot shown in Image 1.