GATE 2026 EC Q.24: solution and answer

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

1. Direction of travel. Write ${\hat{a}}_x\equiv \hat{x}$ and ${\hat{a}}_y\equiv \hat{y}$. Both components carry the phase $(\omega t-kz)$, so the wave travels along $+{\hat{a}}_z$.

2. Amplitudes and phase. Write $\sin (\omega t-kz)=\cos (\omega t-kz-90^{\circ })$. The two components have equal amplitude $E_0$ and differ in phase by $90^{\circ }$. The magnitude is constant: $|\mathbf{E}{|}^2=E_0^2\left[{\cos }^2(\omega t-kz)+{\sin }^2(\omega t-kz)\right]=E_0^2$, so the tip of $\mathbf{E}$ traces a circle. The wave is circularly polarized, not linearly polarized, which rules out (C) and (D).

3. Sense of rotation. Fix the plane $z=0$, so $\mathbf{E}=E_0\left[{\hat{a}}_x\cos \omega t+{\hat{a}}_y\sin \omega t\right]$. At $\omega t=0$, $\mathbf{E}=E_0{\hat{a}}_x$. At $\omega t=90^{\circ }$, $\mathbf{E}=E_0{\hat{a}}_y$. As time goes on, the field turns from $+{\hat{a}}_x$ towards $+{\hat{a}}_y$.

4. Right-hand rule (IEEE convention). In a right-handed system, curl the fingers of the right hand from ${\hat{a}}_x$ to ${\hat{a}}_y$ and the thumb points along ${\hat{a}}_x\times {\hat{a}}_y=+{\hat{a}}_z$, which is the direction of travel. The rotation and the propagation obey the right-hand rule, so the wave is right-hand circularly polarized.

5. Phasor check. Since $\sin (\omega t-kz)=\cos (\omega t-kz-90^{\circ })$, the phasor is ${\mathbf{E}}_s=E_0({\hat{a}}_x-j{\hat{a}}_y)e^{-jkz}$. The $y$-component lags the $x$-component by $90^{\circ }$ in a wave travelling along $+z$. That is the standard form of a right-hand circularly polarized wave, so the answer is (A).