# Uniform Plane Waves

 Question 1
The electric field of a plane electromagnetic wave is

$E=a_{x} C_{1 x} \cos (\omega t-\beta z)+a_{y} C_{1 y} \cos (\omega t-\beta z+\theta) \mathrm{V} / \mathrm{m}$

Which of the following combination(s) will give rise to a left handed elliptically polarized (LHEP) wave?
 A $C_{1 x}=1, C_{1 y}=1, \theta=\pi / 4$ B $C_{1 x}=2, C_{1 y}=1, \theta=\pi / 2$ C $C_{1 x}=1, C_{1 y}=2, \theta=3 \pi / 2$ D $C_{1 x}=2, C_{1 y}=1, \theta=3 \pi / 4$
GATE EC 2023   Electromagnetics
Question 1 Explanation:
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.
 Question 2
Consider the following wave equation,
$\frac{\partial^2 f(x,t)}{\partial t^2}=10000\frac{\partial^2 f(x,t)}{\partial x^2}$
Which of the given options is/are solution(s) to the given wave equation?
 A $f(x,t)=e^{-(x-100t)^2}+e^{-(x+100t)^2}$ B $f(x,t)=e^{-(x-100t)}+0.5e^{-(x+1000t)}$ C $f(x,t)=e^{-(x-100t)}+\sin (x+100t)$ D $f(x,t)=e^{j100 \pi(-100x+t)}+e^{j100 \pi(100x+t)}$
GATE EC 2022   Electromagnetics
Question 2 Explanation:
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.

 Question 3
The magnetic field of a uniform plane wave in vacuum is given by

$\vec{H}(x,y,z,t)=(\hat{a}_x+2\hat{a}_y+b\hat{a}_z) \cos (\omega t+3x-y-z)$

The value of b is _____
 A 0 B 1 C -1 D -2
GATE EC 2020   Electromagnetics
Question 3 Explanation:
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
$\hat{a}_{H }=\frac{1\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z}}{\sqrt{1^{2}+2^{2}+b^{2}}}$
$\hat{a}_{\rho }=\frac{-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z}}{\sqrt{3^{2}+1^{2}+1^{2}}}$
$\hat{a}_{H}\cdot \hat{a}_{\rho }=0 (\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z})\cdot (-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z})=0$
$-3+2+b=0$
$b=1$
 Question 4
A uniform plane wave traveling in free space and having the electric field
$\vec{E}=(\sqrt{2}\hat{a}_{x}-\hat{a}_{z})cos[6\sqrt{3}\pi \times 10^{8}t-2\pi (x+\sqrt{2}z)]V/m$
is incident on a dielectric medium (relative permittivity $\gt$1, relative permeability = 1) as shown in the figure and there is no reflected wave.

The relative permittivity (correct to two decimal places) of the dielectric medium is___________.
 A 2 B 2.9 C 1 D 3
GATE EC 2018   Electromagnetics
Question 4 Explanation:
The wave has $E_{i}$ direction and propagation direction both in same plane (ZX).
The wave is plane of incidence (P) polarized.
\begin{aligned} \beta_{x} &=1, \quad \beta_{2}=\sqrt{2} \\ \tan \theta_{i} &=\frac{\beta_{z}}{\beta_{x}}=\sqrt{2} \end{aligned}
No reflection in P polarized wave under Brewster's angle of incidence.
\begin{aligned} \theta_{i} &=\theta_{B} \\ \tan \theta_{B} &=\sqrt{2}=\sqrt{\epsilon_{r}} \\ \epsilon_{r} &=2 \end{aligned}
 Question 5
The distance (in meters) a wave has to propagate in a medium having a skin depth of 0.1 m so that the amplitude of the wave attenuates by 20 dB, is
 A 0.12 B 0.23 C 0.46 D 2.3
GATE EC 2018   Electromagnetics
Question 5 Explanation:
Attenuation constant,
\begin{aligned} \alpha &=\frac{1}{\text { skin depth }}=10 \mathrm{Np} / \mathrm{m} \\ \text { 20log }_{10}\left(\frac{E_{0}}{E_{x}}\right) &=20 \mathrm{dB} \\ \frac{E_{o}}{E_{x}} &=10 \Rightarrow E_{x}=\frac{E_{0}}{10} \\ E_{x} &=E_{0} e^{-\alpha x}=E_{0} e^{-10 x}=\frac{E_{0}}{10} \\ e^{-10 x} &=\frac{1}{10} \\ x &=\frac{1}{10} \ln (10)=0.23 \mathrm{m} \end{aligned}

There are 5 questions to complete.