Uniform Plane Waves – GATE ECE ElectromagneticsPractice Questions & PYQs

At perfect conductor,

As we know that $H_{\min }$ occurs at $E_{\max }$ .Hence,

So, $2 \beta x_{\max }=2 n \pi+\theta_{\Gamma}$ $\Rightarrow \quad \frac{4 \pi}{\lambda} x_{\max }=2 n \pi+\pi$ $\Rightarrow \quad x_{\max }=(2 n+1) \frac{\lambda}{4} ; n=0,1,2,3 \ldots$ So, for $1^{\text {st }}$ value of $x$ for $\mathrm{H}$ -field to be zero is $x_{\max }=\frac{\lambda}{4} \quad[\text { at } n=0]$ Now, from the equation given,

Hence, $x_{\max }=\frac{3}{4}$ If the reference is not at interference then $x=-\frac{3}{4}$ .

Given, $\quad \vec{E}=\hat{a}_{x} C_{1 x} \cos (\omega t-\beta z)+\hat{a}_{y} C_{1 y} \cos (\omega t-\beta z+\theta)$ at $z=0$ $\vec{E}=C_{1 x} \cos \omega t \hat{a}_{x}+C_{1 y} \cos (\omega t+\theta) \hat{a}_{y}$ Going by options, Option (A) $\quad \vec{E}=\cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 4) \hat{a}_{y}$ at $t=0, \omega t=0, \vec{E}=\hat{a}_{x}+\frac{1}{\sqrt{2}} \hat{a}_{y}$ at $t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}$ $\Rightarrow$ Hence, it is LHEP. Option (B) $\quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 2) \hat{a}_{y}$ at $t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}$ at $t=T / 4, \omega t=\pi / 2, \vec{E}=-1 \hat{a}_{y}$ $\Rightarrow$ Hence, it is LHEP. Option (C) $\quad \vec{E}=\cos \omega t \hat{a}_{x}+2 \cos (\omega t+3 \pi / 2) \hat{a}_{y}$ at $t=0, \quad \quad \omega t=0, \vec{E}=\hat{a}_{x}$ at $t=T / 4, \omega t=\pi / 2, \vec{E}=2 \hat{a}_{y}$ $\Rightarrow$ Hence, it is RHEP. Option (D) $\quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+3 \pi / 4) \hat{a}_{y}$ at $t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}-\frac{1}{\sqrt{2}} \hat{a}_{y}$ at $t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}=\frac{-1}{\sqrt{2}} \hat{a}_{y}$ $\Rightarrow$ Hence, it is LHEP. $\therefore$ Option (A), (B) and (D) are correct.

As we know, wave equation is given by $\frac{\partial^2 f(x,t)}{\partial x^2}=\frac{C^2d^2f(x,y)}{dt^2}$ Here, option (A) and (C) are satisfying the above standard wave equation.

For uniform plane wave $\hat{a}_{H}\cdot \hat{a}_{\rho }=0$ $\hat{a}_{H}$ is unit vector in magnetic field direction $\hat{a}_{\rho }$ is unit vector in power flow direction $\hat{a}_{H }=\frac{1\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z}}{\sqrt{1^{2}+2^{2}+b^{2}}}$ $\hat{a}_{\rho }=\frac{-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z}}{\sqrt{3^{2}+1^{2}+1^{2}}}$ $\hat{a}_{H}\cdot \hat{a}_{\rho }=0 (\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z})\cdot (-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z})=0$ $-3+2+b=0$ $b=1$

The wave has $E_{i}$ direction and propagation direction both in same plane (ZX). The wave is plane of incidence (P) polarized.

No reflection in P polarized wave under Brewster's angle of incidence.

Attenuation constant,

$\vec{E}$ field direction $\Rightarrow\left(\hat{a}_{x}+\hat{a}_{y}+j 2 \hat{a}_{z}\right)$ Propagation direction $\Rightarrow k \hat{a}_{x}-k \hat{a}_{y}$ $\vec{E}$ is perpendicular to propagation $\vec{E} \cdot \vec{P}=0$ Component in $\hat{a}_{z}$ has magnitude of 2. Component in X-Y plane has magnitude of $\sqrt{2}.$ These two components are out of phase by $90^{\circ}$ and have unequal amplitudes. So, it is elliptically polarized wave.

Critical angle,

Rate of change of magnetic field results in induced voltage, $\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$

Left circularly polarized, due to direction change after the reflection from conductor.

$\beta(\omega)$ is known so $V_{g}$ can be calculated. $v_{p}=\omega / \beta$ can be calculated. Polarization can be identified. $\mu_{r}$ and $\epsilon_{r}$ cannot be found, due to which power flux cannot be calculated as power flux $P=\frac{1}{2} \frac{|E|^{2}}{\eta}, \text{ where }\eta=120 \pi \times \sqrt{\frac{\mu_{r}}{\epsilon_{r}}}$

$\vec{E}(z, t)=\hat{a}_{x} 5 \cos \left(2 \pi \times 10^{9} t+\beta z\right)$ $+\hat{a}_{y} 3 \cos \left(2 \pi \times 10^{9} t+\beta z-\frac{\pi}{2}\right)$ $\Rightarrow$ Wave is travelling in $-\hat{a}_{z}$ direction. $\Rightarrow$ Wave has orthogonal components with unequal amplitudes and checking the time trace. $\therefore$ Wave is left hand elliptically polarized.

$\vec{E}(x, t)=\hat{a}_{y} 24 \pi \cos \left(\omega t-K_{o} x\right) V / m$ Power density $\vec{P}=\frac{1}{2} \frac{|E|^{2}}{\eta}=\frac{1}{2} \frac{(24 \pi)^{2}}{120 \pi} \hat{a}_{x}$ Time-averaged power

$\Rightarrow$ Wave is travelling in $-\hat{a}_{z}$ direction. $\Rightarrow$ Wave has orthogonal components with unequal amplitudes and checking the time trace. $\therefore$ Wave is left hand elliptically polarized.

Material has $\mu_{R}$ and $\epsilon_{R}$ complex. $\eta \text { of material }=\sqrt{\frac{j \omega \mu}{\sigma+j \omega \epsilon}}=\sqrt{\frac{\mu_{o} \mu_{R}}{\epsilon_{o} \epsilon_{R}}}=120 \pi$ where, $\sigma=0 \quad$ (dielectric) No reflections at the boundary.

$\vec{E}(z, t)=\hat{a}_{y} 2 \cos \left(10^{8} t-\frac{z}{\sqrt{2}}\right) \mathrm{V} / \mathrm{m}$ General form of $\vec{E}(z, t)=\hat{a}_{y} E_{o} \cos (\omega t-\beta z) \vee / m$ Comparing both, we get $\beta=\frac{1}{\sqrt{2}}$ since, $\beta=\frac{2 \pi}{\lambda}$ $\Rightarrow \quad \lambda=\frac{2 \pi}{\beta}=\frac{2 \pi}{\frac{1}{\sqrt{2}}}=2 \sqrt{2} \pi m$ $\Rightarrow \quad \lambda=8.88 \mathrm{m}$

Since, the TEM wave or plane wave is travelling in positive x-direction, then conditions should be

So, the option (B) satisfies the condition.

$\vec{E}(z, t)=3 \hat{x} \cos \left(\omega t-k z+30^{\circ}\right)-4 \hat{y} \sin \left(\omega t-k z+45^{\circ}\right)$ The plane wave is travelling in +z direction

At z=0, tracing E locus for various times

At $t=\frac{T}{6}$ $E_{x}=3 \cos \left(\frac{2 \pi}{T} \times \frac{T}{6}+30^{\circ}\right)=0$ $E_{y}=-4 \sin \left(60^{\circ}+45^{\circ}\right)=-3.86$ This is left elliptical polarized plane-wave.

For reflected wave. $E_{r}$ propagation direction changes without change in angle $45^{\circ}$ $E_{r} \text { propagation direction }=\frac{\hat{a}_{x}-\hat{a}_{z}}{\sqrt{2}}$ $\Gamma$ reflection coefficient $\Gamma=-0.23$ $E_{r}$ field direction suitably changes satisfying $E \cdot P=0$