A conducting square loop initially lies in the XZ plane with its lower edge hinged along the X-axis.

Q: Why is the angle between the area vector and the magnetic field $(90^\circ - \omega t)$ and not just $\omega t$?

At $t=0$, the loop lies in the XZ plane, meaning its surface normal (area vector) points along the Y-axis. The magnetic field points along the Z-axis. So initially, the angle between them is $90^\circ$. As the loop rotates towards the Y-axis by an angle $\omega t$, the area vector rotates towards the Z-axis, reducing the angle between them to $(90^\circ - \omega t)$.

Q: Why does the EMF graph start at a negative value?

The EMF is given by Faraday's Law: $\varepsilon = -\frac{d\phi}{dt}$. When we differentiate our flux equation $\phi = \frac{B_0 A}{2} \sin(2\omega t)$, we get $\varepsilon = -B_0 A \omega \cos(2\omega t)$. At $t=0$, $\cos(0) = 1$, leaving us with a negative coefficient. This negative sign is a direct consequence of Lenz's Law.

Q: Why is the frequency of the EMF $2\omega$ when the loop is only rotating at $\omega$?

This is a beautiful mathematical consequence of both the field and the area projection changing simultaneously. The field changes as $\cos(\omega t)$ and the area projection changes as $\sin(\omega t)$. When you multiply them, the trigonometric identity $\sin(\theta)\cos(\theta) = \frac{1}{2}\sin(2\theta)$ naturally doubles the frequency.

Q: What exactly happens when $t = \frac{\pi}{\omega}$?

At $t = \frac{\pi}{\omega}$, the loop has rotated exactly $180^\circ$ (or $\pi$ radians). It has just crossed the XZ plane again and is about to enter the region where $y < 0$. Since the problem states the magnetic field is zero in this region, the flux instantly becomes zero and stays zero until the loop returns to the positive Y region.

Q: Could we have chosen the area vector to point in the opposite direction initially?

Yes, the choice of the area vector's initial direction (either $+Y$ or $-Y$) is arbitrary. If you chose the opposite direction, your flux would have a negative sign, and your resulting EMF would start positive. However, looking at the given options, none of the graphs start at a positive maximum and follow the correct piecewise shape, which tells us the author's intended sign convention.

Question

A conducting square loop initially lies in the $XZ$ plane with its lower edge hinged along the $X$-axis. Only in the region $y \ge 0$, there is a time dependent magnetic field pointing along the $Z$-direction, $\vec{B}(t) = B_0(\cos \omega t)\hat{K}$, where $B_0$ is a constant. The magnetic field is zero everywhere else. At time $t = 0$, the loop starts rotating with constant angular speed $\omega$ about the $X$ axis in the clockwise direction as viewed from the $+X$ axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. ($V$) in the loop as a function of time:

Sigma Percentile · Accepted Answer

### The Dual Nature of Changing Flux

When dealing with electromagnetic induction, we often encounter problems where either the magnetic field changes with time, or the area of the loop exposed to the field changes. But what happens when **both** change simultaneously? This problem is a beautiful demonstration of such a scenario.

Imagine a conducting square loop hinged along the $X$-axis, swinging like a trapdoor into the $Y-Z$ plane. The magnetic field is not only restricted to the positive $Y$ region ($y \ge 0$), but it is also pulsating according to the function $\vec{B}(t) = B_0 \cos(\omega t) \hat{K}$.

### Setting Up the Geometry

To find the induced EMF, we must rely on Faraday's Law of Induction, which states that the induced EMF is the negative rate of change of magnetic flux:

$$\varepsilon = -\frac{d\phi}{dt}$$

The magnetic flux $\phi$ is the dot product of the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$. At $t = 0$, the loop lies flat in the $X-Z$ plane. As it rotates clockwise (viewed from the $+X$ axis) with an angular speed $\omega$, the angle it makes with the $Z$-axis is $\omega t$.

Because the area vector $\vec{A}$ is always perpendicular to the surface of the loop, the angle between $\vec{A}$ and the vertical magnetic field $\vec{B}$ becomes $(90^\circ - \omega t)$.

### The Master Equation

Now, let's substitute these geometric realities into our flux equation. We must account for both the time-varying magnitude of the magnetic field and the time-varying projection of the area:

$$\phi = [B_0 \cos(\omega t)] \cdot [A \cos(90^\circ - \omega t)]$$

Since $\cos(90^\circ - \omega t) = \sin(\omega t)$, the equation simplifies to:

$$\phi = B_0 A \cos(\omega t) \sin(\omega t)$$

Using the double-angle trigonometric identity, $2 \sin	heta \cos	heta = \sin(2	heta)$, we can elegantly collapse this expression:

$$\phi = \frac{B_0 A}{2} \sin(2\omega t)$$

### Differentiating for EMF

With our flux equation streamlined, we apply calculus to find the induced EMF. Differentiating with respect to time requires the chain rule:

$$\varepsilon = -\frac{d}{dt} \left( \frac{B_0 A}{2} \sin(2\omega t) ight)$$

$$\varepsilon = -\frac{B_0 A}{2} \cdot 2\omega \cos(2\omega t)$$

$$\varepsilon = -B_0 A \omega \cos(2\omega t)$$

This result tells us two crucial things. First, the frequency of the induced EMF is **twice** the mechanical rotational frequency of the loop. Second, at $t = 0$, the EMF starts at a **negative maximum** value of $-B_0 A \omega$.

### The Spatial Constraint

If the magnetic field existed everywhere, the EMF would simply be a continuous cosine wave. However, the problem introduces a critical spatial constraint: the magnetic field only exists in the region $y \ge 0$.

The loop enters this region at $t = 0$ and exits it when it completes half a rotation, which corresponds to an angle of $\pi$ radians. The time taken for this half-rotation is $t = \frac{\pi}{\omega}$.

During the second half of its rotation, from $t = \frac{\pi}{\omega}$ to $t = \frac{2\pi}{\omega}$, the loop swings through the $y < 0$ region. Here, the magnetic field is strictly zero. Consequently, the magnetic flux is zero, and the induced EMF flatlines to zero.

### Final Conclusion

Piecing it all together, we are looking for a graph that:
1. Starts at a negative maximum.
2. Completes exactly one full cycle of a cosine wave between $t = 0$ and $t = \frac{\pi}{\omega}$.
3. Remains perfectly flat at zero from $t = \frac{\pi}{\omega}$ to $t = \frac{2\pi}{\omega}$.

Looking at the given options, **Graph (A)** flawlessly represents this piecewise behavior.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Faraday’s Law of Induction$

$Setting up the Flux Equation$

$Simplifying the Flux$

$Calculating the EMF$

$Applying Spatial Constraints$

$Matching the Graph$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

The Dual Nature of Changing Flux

Setting Up the Geometry

The Master Equation

Differentiating for EMF

The Spatial Constraint

Final Conclusion

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