GATE 2026 EE Q.51: solution and answer

Official answer: ASource: Official answer key, GATE 2026 EE (IIT Guwahati)

1. The loop lies in the XY plane, so its area vector is along $\hat{z}$ (the sign depends on the direction of traversal). Only the $z$-component of each field links flux: $\Phi =\int \mathbf{B}\cdot d\mathbf{S}=\int B_{z}dS$. The $\hat{x}$ and $\hat{y}$ components lie in the plane of the loop, so they link no flux.

2. Take each field as uniform over its patch. The $z$-components are $B_{1z}=\frac{m}{\sqrt{2}}\sin (\omega t)$ and $B_{2z}=-\frac{n}{\sqrt{2}}\cos (2\omega t+\pi /4)$. The fluxes are therefore ${\Phi }_1=a_1\frac{m}{\sqrt{2}}\sin (\omega t)$ and ${\Phi }_2=-a_2\frac{n}{\sqrt{2}}\cos (2\omega t+\pi /4)$.

3. By Faraday's law, $e=-\frac{d\Phi }{dt}$ with $\Phi ={\Phi }_1+{\Phi }_2$. Since $\frac{d{\Phi }_2}{dt}=\frac{2a_{2}n\omega }{\sqrt{2}}\sin (2\omega t+\pi /4)$,
   $$e(t)=-\frac{a_{1}m\omega }{\sqrt{2}}\cos (\omega t)-\frac{2a_{2}n\omega }{\sqrt{2}}\sin (2\omega t+\pi /4)$$

4. The two terms are sinusoids at different frequencies, $\omega$ and $2\omega$. Their peak values are $E_1=\frac{a_{1}m\omega }{\sqrt{2}}$ and $E_2=\frac{2a_{2}n\omega }{\sqrt{2}}$.

5. Over the common period $2\pi /\omega$, the average of the cross product $\cos (\omega t)\sin (2\omega t+\pi /4)$ is zero, because sinusoids at different frequencies are orthogonal. The mean-square values therefore add: $E_{rms}^2=\frac{E_1^2}{2}+\frac{E_2^2}{2}$.

6. Substituting the peaks:
   $$E_{rms}^2=\frac{1}{2}\cdot \frac{a_1^2{m}^2{\omega }^2}{2}+\frac{1}{2}\cdot \frac{4a_2^2{n}^2{\omega }^2}{2}=\frac{a_1^2{\omega }^2{m}^2+4a_2^2{\omega }^2{n}^2}{4}$$

7. Therefore $E_{rms}=\sqrt{\frac{a_1^2{\omega }^2{m}^2+4a_2^2{\omega }^2{n}^2}{4}}$, which is option (A). The direction of traversal only fixes the overall sign of $e(t)$, so it does not affect the rms value.