Induced EMF in Rotating Square Loop in Time Dependent Magnetic Field

To determine the induced electromotive force (e.m.f.) $V(t)$ in the loop, we analyze the magnetic flux through the loop as a function of time.

1. Orientation and Region of the Loop

• The square loop of side $L$ (and area $A = L^2$) has its bottom edge along the $X$-axis and initially lies in the $XZ$-plane.
• At $t = 0$, it begins rotating with a constant angular speed $\omega$ clockwise as viewed from the positive $X$-axis.
• The position vector of the plane of the loop at time $t$ is at an angle $\theta(t) = \omega t$ with the $+Z$-axis, tilting towards the $+Y$-axis.
• The unit normal vector $\hat{n}$ to the surface of the loop is given by: $\hat{n}(t) = \cos(\omega t)\hat{j} - \sin(\omega t)\hat{k}$
• The magnetic field $\vec{B}(t) = B_0 \cos(\omega t)\hat{k}$ is non-zero only in the region $y \ge 0$, and is zero for $y < 0$.

2. Flux Calculation for Different Time Intervals

Interval 1: $0 \le t \le \frac{\pi}{\omega}$ During this half of the rotation period, the loop is completely within the region $y \ge 0$, where the magnetic field exists. The magnetic flux $\Phi(t)$ through the loop is: $\Phi(t) = \vec{B}(t) \cdot \vec{A}(t) = \left[B_0 \cos(\omega t)\hat{k}\right] \cdot \left[A(\cos(\omega t)\hat{j} - \sin(\omega t)\hat{k})\right]$ $\Phi(t) = -B_0 A \sin(\omega t)\cos(\omega t) = -\frac{1}{2} B_0 A \sin(2\omega t)$

According to Faraday's law of electromagnetic induction, the induced e.m.f. $V(t)$ is: $V(t) = -\frac{d\Phi}{dt} = -\frac{d}{dt}\left(-\frac{1}{2} B_0 A \sin(2\omega t)\right) = B_0 A \omega \cos(2\omega t)$

Thus, for $t \in \left[0, \frac{\pi}{\omega}\right]$:

• At $t = 0$: $V(0) = B_0 A \omega > 0$
• At $t = \frac{\pi}{4\omega}$: $V = 0$
• At $t = \frac{\pi}{2\omega}$: $V = -B_0 A \omega$
• At $t = \frac{3\pi}{4\omega}$: $V = 0$
• At $t = \frac{\pi}{\omega}$: $V = B_0 A \omega$

This corresponds to one complete cosine cycle with a positive starting value.

Interval 2: $\frac{\pi}{\omega} \le t \le \frac{2\pi}{\omega}$ During this half of the rotation, the loop moves through the region $y < 0$, where the magnetic field is zero ($\vec{B} = 0$). Consequently: $\Phi(t) = 0 \implies V(t) = 0$

3. Conclusion

The induced e.m.f. $V(t)$ is a full cosine wave oscillation starting at a positive peak in the interval $\left[0, \frac{\pi}{\omega}\right]$, and remains strictly zero in the interval $\left[\frac{\pi}{\omega}, \frac{2\pi}{\omega}\right]$, repeating periodically every $T = \frac{2\pi}{\omega}$.

This behavior is correctly depicted in graph (A).

Correct Answer: (A)