The figure shows an arbitrarily shaped planar conducting loop A in the XY plane. Two nonintersecting regions with areas $a_1$  and  $a_2$  within the loop are subjected to magnetic fields ${\mathbf{B}}_1=\frac{m}{\sqrt{2}}\sin \left(\omega t\right)(\hat{x}+\hat{z})$ , and ${\mathbf{B}}_2=-\frac{n}{\sqrt{2}}\cos \left(2\omega t+\frac{\pi }{4}\right)(\hat{y}+\hat{z})$ , respectively. What is the expression for the induced rms voltage in loop A?

The correct answer is:

Step-by-Step Explanation:

1. Analyze the Geometry from the Figure:
As shown in the coordinate system of the provided image, the axes are labeled X, Y, and Z. The conducting loop A lies entirely in the XY plane. Within this loop, there are two distinct, non-intersecting shaded regions labeled with areas a₁ and a₂.
Since the loop A lies in the XY plane, its area vector is normal to the plane, pointing along the Z-axis:
$$\mathbf{dA}=dA\hat{z}$$

2. Determine the Magnetic Flux:
The total magnetic flux $\Phi$ passing through the loop A is the sum of the fluxes through the two regions of area a₁ and a₂. Because the area vector points in the Z direction, only the z-components of the magnetic fields B₁ and B₂ contribute to the flux:
$$\Phi ={\Phi }_1+{\Phi }_2={\mathbf{B}}_1·\left(a_1\hat{z}\right)+{\mathbf{B}}_2·\left(a_2\hat{z}\right)$$
Given the magnetic fields:
$${\mathbf{B}}_1=\frac{m}{\sqrt{2}}\sin \left(\omega t\right)(\hat{x}+\hat{z})\Rightarrow {\mathbf{B}}_1·\hat{z}=\frac{m}{\sqrt{2}}\sin \left(\omega t\right)$$
$${\mathbf{B}}_2=-\frac{n}{\sqrt{2}}\cos (2\omega t+\frac{\pi }{4})(\hat{y}+\hat{z})\Rightarrow {\mathbf{B}}_2·\hat{z}=-\frac{n}{\sqrt{2}}\cos (2\omega t+\frac{\pi }{4})$$
Substituting these back into the flux equation:
$$\Phi =\frac{a_{1}m}{\sqrt{2}}\sin \left(\omega t\right)-\frac{a_{2}n}{\sqrt{2}}\cos (2\omega t+\frac{\pi }{4})$$

3. Calculate the Induced EMF (Voltage):
By Faraday's Law of Induction, the induced EMF $V\left(t\right)$ is:
$$V\left(t\right)=-\frac{d\Phi }{dt}$$
Differentiating each term with respect to time:
$$V\left(t\right)=-\frac{a_{1}m\omega }{\sqrt{2}}\cos \left(\omega t\right)-\sqrt{2}{a}_{2}n\omega \sin (2\omega t+\frac{\pi }{4})$$

4. Find the Root-Mean-Square (RMS) Voltage:
The voltage expression consists of two components at different harmonic frequencies: one at frequency $\omega$ and the other at frequency $2\omega$.
Since sine/cosine functions with different frequencies are orthogonal, the average of their cross-product over a period is zero. Thus, the total mean-square value is the sum of the mean-square values of each independent component:
$$⟨V^2⟩=⟨V_1^2⟩+⟨V_2^2⟩$$
For a sinusoidal term of amplitude $V_0$, the average squared value is $\frac{1}{2}{V}_0^2$:
$$⟨V_1^2⟩=\frac{1}{2}{(\frac{}{}}^{}$$
a 1 m ω 2 ) 2 = a 1 2 m 2 ω 2 4
$$⟨V_2^2⟩=\frac{1}{2}{\left(\sqrt{2}{a}_{2}n\omega \right)}^2=a_2^2{n}^2{\omega }^2$$
Combining the terms under a common denominator of 2 to match the correct option structure:
$$V_{\mathrm{rms}}=\sqrt{\frac{a_1^2{\omega }^2{m}^2+4a_2^2{\omega }^2{n}^2}{2}}$$