
The electric field part of an electromagnetic wave in a medium is represented by
$
{E_x} = 0 \\
{E_y} = 2.5\dfrac{N}{C}\cos \left[ {(2\pi \times {{10}^6}\dfrac{{rad}}{s})t - (\pi \times {{10}^{ - 2}}\dfrac{{rad}}{m}x)} \right]; \\
{E_x} = 0. \\
$
The wave is:
(A) Moving along x direction with frequency $2\pi \times {10^6}Hz$ and the wavelength 200 m.
(B) Moving along y direction with frequency $2\pi \times {10^6}Hz$ and wavelength 200 m.
(C) Moving along x direction with frequency ${10^6}Hz$ and wavelength 100 m
(D) Moving along x direction with frequency ${10^6}Hz$ and wavelength 200 m.
Answer
133.2k+ views
Hint: To answer this question, we need to know the standard equation of an electromagnetic wave. The equation of the electromagnetic wave that is then given in the question should be compared with the standard equation. From the comparison we can obtain the quantities required for the answer. From the sign of the answer that we will obtain, we can find the coordinate to which it is belonging.
Complete step by step answer:
We know that the standard equation of electromagnetic wave is given by:
${E_y} = {E_0}\cos (wt - kx)$
The equation that is given is:
$
\\
{E_y} = 2.5\dfrac{N}{C}\cos \left[ {(2\pi \times {{10}^6}\dfrac{{rad}}{s})t - (\pi \times {{10}^{ - 2}}\dfrac{{rad}}{m}x)} \right] \\
\\
$
Now we have to compare the given equation with the standard equation.
On the comparison we get that:
$
w = 2\pi f \\
\Rightarrow w = 2\pi \times {10^6} \\
\Rightarrow \lambda f = {10^6}Hz \\
$
It is also known to us that:
$
\dfrac{{2\pi }}{\lambda } = k = \pi \times {10^{ - 2}}{m^{ - 1}} \\
\Rightarrow f = \lambda = 200m \\
$
Hence we can say that the wave is moving along the positive x direction with a frequency of ${10^6}Hz $and a wavelength of 200 m.
So option D is the correct answer.
Note: In this question we have come across the term electromagnetic wave. For better understanding we should be defining the electromagnetic waves. The waves that are created due to the occurrence of the vibrations between the electric and the magnetic field are known as electromagnetic waves.
The standard equation of the electromagnetic wave is described as a second order partially differentiated equation. The main aim of the equation is to describe the propagation of the electromagnetic waves through a medium or even through a vacuum.
The standard equation of the electromagnetic waves represents a three dimensional form of the wave equation.
Complete step by step answer:
We know that the standard equation of electromagnetic wave is given by:
${E_y} = {E_0}\cos (wt - kx)$
The equation that is given is:
$
\\
{E_y} = 2.5\dfrac{N}{C}\cos \left[ {(2\pi \times {{10}^6}\dfrac{{rad}}{s})t - (\pi \times {{10}^{ - 2}}\dfrac{{rad}}{m}x)} \right] \\
\\
$
Now we have to compare the given equation with the standard equation.
On the comparison we get that:
$
w = 2\pi f \\
\Rightarrow w = 2\pi \times {10^6} \\
\Rightarrow \lambda f = {10^6}Hz \\
$
It is also known to us that:
$
\dfrac{{2\pi }}{\lambda } = k = \pi \times {10^{ - 2}}{m^{ - 1}} \\
\Rightarrow f = \lambda = 200m \\
$
Hence we can say that the wave is moving along the positive x direction with a frequency of ${10^6}Hz $and a wavelength of 200 m.
So option D is the correct answer.
Note: In this question we have come across the term electromagnetic wave. For better understanding we should be defining the electromagnetic waves. The waves that are created due to the occurrence of the vibrations between the electric and the magnetic field are known as electromagnetic waves.
The standard equation of the electromagnetic wave is described as a second order partially differentiated equation. The main aim of the equation is to describe the propagation of the electromagnetic waves through a medium or even through a vacuum.
The standard equation of the electromagnetic waves represents a three dimensional form of the wave equation.
Recently Updated Pages
Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

JEE Main 2023 (April 11th Shift 2) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Wheatstone Bridge for JEE Main Physics 2025

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Current Loop as Magnetic Dipole and Its Derivation for JEE

Inertial and Non-Inertial Frame of Reference - JEE Important Topic

Write the value of charge in coulombs on the nucleus class 12 physics JEE_Main

Other Pages
Diffraction of Light - Young’s Single Slit Experiment

In Bohrs model of the hydrogen atom the radius of the class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Clemmenson and Wolff Kishner Reductions for JEE

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

CBSE Date Sheet 2025 Class 12 - Download Timetable PDF for FREE Now
