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 \geq 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:

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For rotating loops in spatially varying magnetic fields, induced emf shows sinusoidal variation interrupted when the loop moves out of the field region.

Updated On: May 19, 2025

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The Correct Option is A

Solution and Explanation

To solve this problem, we need to determine the induced electromotive force (e.m.f., $V$) in a square loop rotating in a time-varying magnetic field. Let’s break it down step by step.
Step 1: Understand the Setup

The square loop lies in the $XZ$-plane at $t = 0$, with its lower edge hinged along the $X$-axis.
The loop rotates with a constant angular speed $\omega$ about the $X$-axis in the clockwise direction (as viewed from the positive $X$-axis). Clockwise rotation from this perspective means the loop rotates from the $Z$-axis toward the negative $Y$-axis.
The magnetic field exists only in the region $y \geq 0$ and is given by $\vec{B}(t) = B_0 (\cos \omega t) \hat{k}$, where $B_0$ is a constant, and $\hat{k}$ is the unit vector along the $Z$-axis. The field is zero for $y<0$.
We need to find the induced e.m.f. $V$ as a function of time and match it with the given plots.

Step 2: Define the Loop’s Position and Area Vector Assume the square loop has a side length $a$, so its area is $A = a^2$. At $t = 0$, the loop lies in the $XZ$-plane, meaning its area vector (normal to the plane) points along the $Y$-axis (positive $Y$-direction, $\hat{j}$). The loop rotates about the $X$-axis with angular speed $\omega$. The angle of rotation $\theta$ at time $t$ is: \[ \theta = \omega t \] Since the rotation is clockwise as viewed from the positive $X$-axis, the loop’s plane tilts from the $XZ$-plane (where the normal is along $+\hat{j}$) toward the negative $Y$-axis. Let’s define the position:

At $t = 0$, the loop is in the $XZ$-plane, and the area vector $\vec{A} = a^2 \hat{j}$.
After time $t$, the loop has rotated by an angle $\omega t$. The normal vector (area vector direction) rotates in the $YZ$-plane.

The normal vector’s direction after rotation: Initially (at $\theta = 0$), the normal is along $+\hat{j}$. After rotating by $\theta = \omega t$ clockwise (from $Z$ to $-Y$), we can determine the new normal using the rotation matrix about the $X$-axis. For a clockwise rotation by $\theta$ about the $X$-axis: The initial normal is $(0, 1, 0)$.

After rotation:
\[ \begin{pmatrix} 0 \\ \cos (\omega t) \\-\sin (\omega t) \end{pmatrix} \] So, the area vector becomes: \[ \vec{A}(t) = a^2 \left( \cos (\omega t) \hat{j} - \sin (\omega t)\hat{k} \right) \]

Step 3: Calculate the Magnetic Flux The magnetic flux $\Phi$ through the loop is: \[ \Phi = \vec{B} \cdot \vec{A} \] The magnetic field is $\vec{B}(t) = B_0 (\cos \omega t) \hat{k}$, and it exists only in the region $y \geq 0$. This complicates things because part of the loop may be in $y<0$, where $\vec{B} = 0$. Let’s first compute the flux assuming the field were uniform across the loop, then adjust for the $y \geq 0$ condition: \[ \vec{B}(t) \cdot \vec{A}(t) = \left( B_0 \cos \omega t \hat{k} \right) \cdot \left( a^2 \cos (\omega t) \hat{j} - a^2 \sin (\omega t) \hat{k} \right) \] \[ = (B_0 \cos \omega t) \cdot (- a^2 \sin (\omega t)) = - B_0 a^2 \cos (\omega t) \sin (\omega t) \] So, if the field were uniform: \[ \Phi_{\text{uniform}} = - B_0 a^2 \cos (\omega t) \sin (\omega t) \] However, the field is only present for $y \geq 0$. As the loop rotates, its position in the $Y$-coordinate changes:

At $t = 0$, the loop is in the $XZ$-plane ($y = 0$), so the entire loop is at the boundary of the field region.
As $t$ increases, the loop tilts, and the $Y$-coordinate of points on the loop ranges from 0 (at the hinged edge along the $X$-axis) to $y = -a \sin (\omega t)$ at the opposite edge (since it moves toward the negative $Y$-axis).

When $\omega t = 0$, the loop is entirely at $y = 0$, so it’s in the field region. When $\omega t = \pi/2$, the loop is in the $XY$-plane, with $y$ ranging from 0 to $-a$, so half the loop (on average) is in $y<0$, where $\vec{B} = 0$.

Step 4: Adjust for the Field Region The flux depends on the area of the loop in the region $y \geq 0$. Let’s parameterize the loop:

The loop’s hinged edge is along the $X$-axis (from $x = 0$ to $x = a$, $y = 0$, $z = 0$).
The opposite edge, initially at $z = a$, rotates. Its $y$-coordinate becomes $y = -a \sin (\omega t)$, and $z = a \cos (\omega t)$.

As the loop rotates, the $Y$-coordinate of a point at height $z$ (before rotation) becomes: \[ y = -z \sin (\omega t), \quad z = z \cos (\omega t) \] The original $z$-coordinate ranges from 0 to $a$. So, $y$ ranges from 0 to $-a \sin (\omega t)$. The condition $y \geq 0$ means: \[ -z \sin (\omega t) \geq 0 \implies z \leq 0 \quad \text{(if $\sin (\omega t)>0$)} \] Since $z \geq 0$, when $\sin (\omega t)>0$, no part of the loop satisfies $y \geq 0$ except at the hinge ($z = 0$). This suggests we need to compute the effective area in $y \geq 0$. Instead, let’s reconsider the flux by integrating over the loop’s surface. The fraction of the loop in $y \geq 0$ decreases as $\sin (\omega t)$ increases. At $\omega t = \pi/2$, $y$ ranges from 0 to $-a$, so half the loop is in $y<0$. The effective area in the field region depends on the angle.
Step 5: Induced e.m.f. The induced e.m.f. is given by Faraday’s Law: \[ V = -\frac{d\Phi}{dt} \] Let’s approximate the flux by considering the effective area. Notice the field and rotation have the same frequency $\omega$, suggesting resonance effects. Let’s compute $V$ using the uniform flux and adjust: \[ \Phi_{\text{uniform}} = - B_0 a^2 \cos (\omega t) \sin (\omega t) \] \[ \frac{d\Phi}{dt} = - B_0 a^2 \left[ (-\sin (\omega t)) \sin (\omega t) \omega + \cos (\omega t) (\cos (\omega t)) \omega \right] \] \[ = - B_0 a^2 \omega \left[ -\sin^2 (\omega t) + \cos^2 (\omega t) \right] = - B_0 a^2 \omega (\cos^2 (\omega t) - \sin^2 (\omega t)) \] \[ = - B_0 a^2 \omega \cos (2 \omega t) \] \[ V = -\frac{d\Phi}{dt} = B_0 a^2 \omega \cos (2 \omega t) \] This e.m.f. oscillates with frequency $2\omega$, which matches the period in the plots ($\pi/\omega$).
Step 6: Account for $y \geq 0$ Condition When $\omega t = \pi/2$, the loop is half in $y<0$, so the flux is halved, but the derivative (e.m.f.) depends on the rate of change. The $\cos (2 \omega t)$ form arises from the product of $\cos (\omega t)$ (from the field) and $\sin (\omega t)$ (from the area vector’s $Z$-component), and the $y \geq 0$ condition modulates the amplitude, not the frequency.
Step 7: Match with Plots