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Q.3 A conducting square loop initially lies in the

$X Z$ 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, $\vec{B} (t) = B_{0} (\cos ω t) \hat{k}$ , 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 $ω$ 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?

Options:
(No plots provided, but the solution will derive the correct functional form.)

Solution:

1. Understanding the Problem

A conducting square loop lies initially in the $X Z$ plane, hinged along the $X$ -axis.
A magnetic field $\vec{B} (t) = B_{0} \cos (ω t) \hat{k}$ exists only in the region $y \geq 0$ .
At $t = 0$ , the loop starts rotating about the $X$ -axis with angular speed $ω$ (clockwise as viewed from $+ X$ ).
We need to find the induced e.m.f. $V (t)$ in the loop.

2. Key Physics Concepts

Faraday's Law of Induction: The induced e.m.f. is given by the rate of change of magnetic flux through the loop:
$V = - \frac{d Φ_{B}}{d t}$
Magnetic Flux: $Φ_{B} = \int \vec{B} \cdot d \vec{A}$ , where $d \vec{A}$ is the area vector of the loop.
Rotation of the Loop: The loop rotates with angular velocity $ω$ , changing the angle between $\vec{B}$ and $d \vec{A}$ .

3. Step-by-Step Analysis

Step 1: Define the Geometry

Let the side length of the square loop be $L$ .
The loop rotates about the $X$ -axis, so at time $t$ , its plane makes an angle $θ = ω t$ with the $X Z$ -plane (since it starts at $θ = 0$ at $t = 0$ ).

Step 2: Magnetic Flux Calculation

The magnetic field $\vec{B} (t) = B_{0} \cos (ω t) \hat{k}$ exists only for $y \geq 0$ .
The loop is in the $X Z$ -plane at $t = 0$ . As it rotates, the area vector $d \vec{A}$ changes:
- At $t = 0$ , $d \vec{A} = L^{2} \hat{k}$ .
- At time $t$ , $d \vec{A} = L^{2} \cos (ω t) \hat{k}$ (since the loop rotates in the $X Z$ -plane).
However, the magnetic field is only present in $y \geq 0$ , so the effective area exposed to $\vec{B}$ is:
$A_{eff} = \frac{L^{2}}{2} (1 + \cos (2 ω t))$
(This is because only half the loop is in $y \geq 0$ at any time, and the projection varies with rotation.)
The magnetic flux is:
$Φ_{B} = \vec{B} \cdot {\vec{A}}_{eff} = B_{0} \cos (ω t) \cdot \frac{L^{2}}{2} (1 + \cos (2 ω t))$ $Φ_{B} = \frac{B_{0} L^{2}}{2} \cos (ω t) (1 + \cos (2 ω t))$

Step 3: Simplify the Flux Expression

Using the trigonometric identity $\cos (2 x) = 2 \cos^{2} x - 1$ :

Φ_{B} = \frac{B_{0} L^{2}}{2} \cos (ω t) (1 + 2 \cos^{2} (ω t) - 1) = \frac{B_{0} L^{2}}{2} \cos (ω t) \cdot 2 \cos^{2} (ω t) = B_{0} L^{2} \cos^{3} (ω t)

Step 4: Compute the Induced e.m.f.

Using Faraday's law:

V = - \frac{d Φ_{B}}{d t} = - B_{0} L^{2} \frac{d}{d t} [\cos^{3} (ω t)]

V = - B_{0} L^{2} \cdot 3 \cos^{2} (ω t) \cdot (- ω \sin (ω t)) = 3 B_{0} L^{2} ω \cos^{2} (ω t) \sin (ω t)

Simplify using $\cos^{2} x \sin x = \frac{1}{4} (\sin (3 x) + \sin x)$ :

V = \frac{3}{4} B_{0} L^{2} ω (\sin (3 ω t) + \sin (ω t))

Step 5: Interpretation

The induced e.m.f. $V (t)$ is a superposition of two sine waves with frequencies $ω$ and $3 ω$ . The dominant term is $\sin (ω t)$ , but the exact plot would show a combination of these frequencies.

6. Final Answer

The induced e.m.f. $V (t)$ varies as:

V (t) \propto \sin (ω t) + higher harmonics

Given the options (not provided here), the correct plot would reflect this time-dependent behavior, typically showing a periodic oscillation with contributions from multiple frequencies.

Note: The exact plot would depend on the specific options, but the derived form is:

V (t) = \frac{3}{4} B_{0} L^{2} ω (\sin (3 ω t) + \sin (ω t))

Thus, the correct plot would show a combination of these sinusoidal components.