GATE EC 2024 Set 1 — Question 39

The incident electric field is given by $\vec{E}_i(x) = A_0 \hat{a}_y e^{-j\frac{2\pi x}{3}}$.
From this, the wave number is $k = \frac{2\pi}{3}$.
The wavelength $\lambda = \frac{2\pi}{k} = \frac{2\pi}{2\pi/3} = 3$ metres.The wave is incident normally on an ideal electric conductor at $x=0$. For an ideal electric conductor, the reflection coefficient for the electric field is $\Gamma_E = -1$, and for the magnetic field is $\Gamma_H = +1$.The incident magnetic field $\vec{H}_i(x)$ is related to $\vec{E}_i(x)$ by $\vec{H}_i(x) = \frac{1}{\eta_0} (\hat{k} \times \vec{E}_i(x))$, where $\eta_0$ is the intrinsic impedance of air and $\hat{k}$ is the direction of propagation ($\hat{a}_x$).
$\vec{H}_i(x) = \frac{1}{\eta_0} (\hat{a}_x \times A_0 \hat{a}_y e^{-j\frac{2\pi x}{3}}) = \frac{A_0}{\eta_0} \hat{a}_z e^{-j\frac{2\pi x}{3}}$.The reflected magnetic field $\vec{H}_r(x)$ propagates in the $-\hat{a}_x$ direction and has a reflection coefficient of $+1$.
So, $\vec{H}_r(x) = \frac{A_0}{\eta_0} \hat{a}_z e^{j\frac{2\pi x}{3}}$.The total magnetic field $\vec{H}_{total}(x)$ is the sum of the incident and reflected magnetic fields:
$\vec{H}_{total}(x) = \vec{H}_i(x) + \vec{H}_r(x) = \frac{A_0}{\eta_0} \hat{a}_z (e^{-j\frac{2\pi x}{3}} + e^{j\frac{2\pi x}{3}})$.
Using Euler's formula, $e^{-j\theta} + e^{j\theta} = 2\cos(\theta)$.
So, $\vec{H}_{total}(x) = \frac{2A_0}{\eta_0} \hat{a}_z \cos\left(\frac{2\pi x}{3}\right)$.We need to find the position of the first null of the total magnetic field, which means $\vec{H}_{total}(x) = 0$.
This occurs when $\cos\left(\frac{2\pi x}{3}\right) = 0$.
The general solution for $\cos(\theta) = 0$ is $\theta = (n + \frac{1}{2})\pi$, where $n$ is an integer.
So, $\frac{2\pi x}{3} = (n + \frac{1}{2})\pi$.
$\frac{2x}{3} = n + \frac{1}{2}$.
$x = \frac{3}{2} \left(n + \frac{1}{2}\right)$.We are looking for the first null measured from $x=0$. Since $x$ must be positive (in the air region, $x>0$), we take $n=0$.
For $n=0$:
$x = \frac{3}{2} \left(0 + \frac{1}{2}\right) = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$ metres.The final answer is $\boxed{A}$