GATE 2024 EC Q.39: solution and answer

Official answer: ASource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Read the phase constant off $e^{-j\beta x}$: $\beta =\frac{2\pi }{3}\text{rad/m}$, so $\lambda =\frac{2\pi }{\beta }=3\text{m}$. The air fills $x<0$ and the conductor face is at $x=0$.

2. At a perfect conductor the reflection coefficient for $\mathbf{E}$ is $\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=-1$, because ${\eta }_2=0$ when $\sigma =\infty$. The reflected field is therefore ${\mathbf{E}}_r=-A_y{e}^{+j\beta x}{\hat{a}}_y$.

3. Total electric field in the air: $\mathbf{E}=A_y\left(e^{-j\beta x}-e^{j\beta x}\right){\hat{a}}_y=-2j{A}_y\sin (\beta x){\hat{a}}_y$. It is zero at $x=0$, as the boundary condition (tangential $\mathbf{E}$ vanishes on a perfect conductor) requires.

4. Magnetic fields from $\mathbf{H}=\frac{1}{{\eta }_0}{\hat{a}}_k\times \mathbf{E}$, where ${\hat{a}}_k$ is the direction of travel and ${\eta }_0=\sqrt{{\mu }_0/{ϵ}_0}\approx 377\Omega$ (with ${\mu }_0=4\pi \times 10^{-7}\text{H/m}$ and ${ϵ}_0\approx 8.854\times 10^{-12}\text{F/m}$). Incident wave (${\hat{a}}_k={\hat{a}}_x$): ${\mathbf{H}}_i=\frac{A_y}{{\eta }_0}{e}^{-j\beta x}{\hat{a}}_z$. Reflected wave (${\hat{a}}_k=-{\hat{a}}_x$): ${\mathbf{H}}_r=\frac{1}{{\eta }_0}(-{\hat{a}}_x)\times \left(-A_y{e}^{j\beta x}{\hat{a}}_y\right)=\frac{A_y}{{\eta }_0}{e}^{j\beta x}{\hat{a}}_z$.

5. Total magnetic field: $$\mathbf{H}=\frac{A_y}{{\eta }_0}\left(e^{-j\beta x}+e^{j\beta x}\right){\hat{a}}_z=\frac{2A_y}{{\eta }_0}\cos (\beta x){\hat{a}}_z$$ So $\mathbf{H}$ is at its maximum at the wall, where the surface current flows, while $\mathbf{E}$ is zero there.

6. Nulls of $\mathbf{H}$ are where $\cos (\beta x)=0$, that is $\beta x=-\frac{\pi }{2},-\frac{3\pi }{2},\dots$ in the air. The first one (nearest the wall) is at $\beta x=-\frac{\pi }{2}$: $$x=-\frac{\pi }{2\beta }=-\frac{\pi }{2}\cdot \frac{3}{2\pi }=-\frac{3}{4}\text{m}$$ This is $\lambda /4$ from the conductor, so the answer is option (A).