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Electromagnetic Waves Chapter - Physics JEE Main

Last updated date: 01st Mar 2024
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Concepts for JEE Main Electromagnetic Waves Physics

This chapter deals with the concept of electromagnetic waves and lets us know about how these waves are produced and what are the properties of electromagnetic waves. Basically, the electromagnetic waves consist of an electric field vector and a magnetic vector field vector both are perpendicular to each other and as well as the direction of propagation of the wave.

The chapter includes the topics like Maxwell’s equations and Lorentz force formula which make all the basic laws of electromagnetism. It also contains the concept of displacement current which helps us to understand the continuity of the current between the plates of the capacitor. It includes the electromagnetic wave velocity concept and properties of electromagnetic waves like transverse nature of the wave, intensity of electromagnetic waves, momentum and radiation pressure of the waves.

Now, let's move on to the important concepts and formulae related to JEE Exam and JEE Main physics electromagnetic waves exam along with a few solved examples.

JEE Main Physics Chapters 2024

Topics of Electromagnetic Waves:

Electromagnetic Waves and Its Characteristics:

In JEE Main Physics, we dive into electromagnetic waves, like those carrying signals for phones and radios. They're like invisible messengers, traveling through space with electric and magnetic qualities.

Transverse Nature of Electromagnetic Waves:

These waves dance side to side, showing a transverse nature. It's like watching a wave move horizontally, with the electric and magnetic parts shaking at right angles to the direction they travel.

Electromagnetic Spectrum:

Picture a rainbow, but with waves instead of colors. This spectrum includes radio waves (for communication), microwaves (for heating), infrared (felt as heat), visible light, ultraviolet (sun rays), x-rays (for medical imaging), and gamma rays (from nuclear reactions).

Applications of Electromagnetic Waves:

From cooking with microwaves to medical scans with x-rays, electromagnetic waves power numerous applications. They help us communicate, cook, see, heal, and explore the universe.

Important Topics of Electromagnetic Waves Chapter

  • What are electromagnetic waves?

  • Displacement current

  • Continuity of current

  • Maxwell’s Equation  

  • Lorentz force

  • Velocity of electromagnetic waves

  • Properties of electromagnetic waves

  • Electromagnetic spectrum

Electromagnetic Waves Important Concept for JEE Main

Name of the Concept

Key Points of the Concepts

What are electromagnetic waves?

  • EM waves are another name for electromagnetic waves. Electromagnetic radiation is made up of electromagnetic waves that are generated when an electric field collides with a magnetic field. The combination of oscillating electric and magnetic fields can also be characterised as electromagnetic waves. Maxwell's equations, which are the fundamental equations of electrodynamics, are solved by electromagnetic waves. The various examples of electromagnetic waves are microwaves, radio waves, infrared etc. 

Wave Nature

  • Electromagnetic waves are produced by the oscillation of electric and magnetic fields. The changing electric field induces a magnetic field, which, in turn, generates an electric field, leading to a self-propagating wave. This self-sustaining interaction creates a continuous electromagnetic wave.

  • The speed of electromagnetic waves in a vacuum, denoted as 'c,' is approximately $3 \times 10^8$ meters per second. This is a fundamental constant of nature.

Transverse Nature

  • Electromagnetic waves are transverse in nature, meaning the oscillations of electric and magnetic fields occur perpendicular to the direction of wave propagation. This property distinguishes them from longitudinal waves where oscillations are parallel to the wave's path.

Displacement Current

  • The displacement current ($I_D$) is the current which comes into play in the region in which the electric field and the electric flux ($\phi_E$) is changing with time. This displacement current is given as;

$I_D=\epsilon_o \frac{\text{d}\phi_E}{\text{d}t}$

Continuity of Current

  • Maxwell pointed out that for consistency of Ampere’s circuital law, there must be displacement current ($I_D$) along with the conduction current ($I$) in the closed-loop as $I+I_D$ has the property of continuity although individually they may not be continuous. After Maxwell’s modification, Ampere’s law is written as,

$\oint\overrightarrow{B}.\overrightarrow{dl}=\mu_o (I+I_D)$

Maxwell’s Equation

  • In the absence of any dielectric or magnetic material, the four Maxwell’s equations are written as;

  1. $\oint\overrightarrow{E}.\overrightarrow{ds}=\dfrac{q}{\epsilon_o}$

This equation gives an electric field due to discrete charge or due to certain charge distribution.

  1. $\oint\overrightarrow{B}.\overrightarrow{ds}=0$

This equation shows that the number of magnetic lines of force entering a closed surface is equal to the number of magnetic lines of force leaving it. It means magnetic line forces always form closed paths.

  1. $\oint\overrightarrow{E}.\overrightarrow{dl}=-\frac{\text{d}}{\text{d}t}\oint\overrightarrow{B}.\overrightarrow{ds}$

This equation represents Faraday's law of electromagnetic induction. This law shows that the line integral of the electric field around any closed path is equal to the time rate of change of magnetic flux through the surface bounded by the closed path.

  1. $\oint\overrightarrow{B}.\overrightarrow{dl}=\mu_o I+\mu_o \epsilon_o \frac{\text{d}}{\text{d}t}\oint\overrightarrow{E}.\overrightarrow{ds}$

This equation is a generalised form of Ampere’s Law as modified by Maxwell.

Lorentz Force

  • Suppose a charge $q$ travelling at velocity $v$ in the presence of both electric ($E$) and magnetic fields ($B$). Then it will experience the force due to electric field as well as magnetic field and this force is termed as Lorentz force. Mathematically, it is written as;


Velocity of Electromagnetic Waves

  • The velocity or speed of electromagnetic waves is always equal to speed of light ($c$) in vacuum and it is given as;

$c=\frac{1}{\sqrt{\mu_o \epsilon_o}}=3 \times 10^8\,m/s$

Properties of Electromagnetic Waves

  • The electromagnetic waves are produced by accelerated or oscillating charges and these waves don’t require any medium for their propagation.

  • In electromagnetic waves, the sinusoidal variation in both electric and magnetic field vectors occurs simultaneously. As a result, they attain maxima and minima at the same place and at the same time.

  • The velocity of these waves entirely depends on the electric and magnetic properties of the medium in which these waves are travelling.

  • These waves carry energy which is divided equally between electric field and magnetic field vectors. The average electric density ($u_E$) and magnetic energy density ($u_B$) due to static electric field ($E$) and magnetic field ($B$) which don’t vary with time are given as;

$u_E=\dfrac{1}{2} \epsilon_o E^2$

$u_B=\dfrac{1}{2} \dfrac{B^2}{\mu_o}$

  • Intensity of an electromagnetic wave ($I$) is defined as the energy crossing per second per unit area perpendicular to the direction of propagation of electromagnetic waves. Mathematically it is given as;

$I=\dfrac{1}{2} \dfrac{B_o^2}{\mu_o}c$

Here, $B_o$ is the maximum value magnetic field vector.

  • The electromagnetic waves carry energy and momentum. If a portion of electromagnetic wave of energy $U$ is propagating with speed $c$ then its linear momentum ($p$) will be,


  • Radiation pressure of an electromagnetic wave is defined as the force exerted by the wave on the unit area of the surface on which it is incident. We can write this as,

$\text{Radiation pressure}=\dfrac{\text{Force}}{\text{Area}}=\dfrac{\text{change in momentum}}{\text{area}\times \text{time}}$

Electromagnetic Spectrum

  • The orderly distribution of electromagnetic radiation according to its wavelength or frequency is called the electromagnetic spectrum.

  • The classification of the electromagnetic spectrum in increasing order of frequency of radiation is given below as;

  1. Radio Waves: These waves have frequencies ranging from $5\times 10^5\, Hz$ to $10^9\, Hz$. These waves are produced by oscillating electric circuits having an inductor and capacitor.

Properties of Radio Wave

  • They are present in radar, radio, and satellite communication.


  • Useful in marine and navigation.

  • Useful in long-range and long-distance communication.

  1.  Microwaves: The frequency range of these waves are from $1\, GHz$ to $300\, Hz$. They are produced by special vacuum tubes named klystron, magnetrons and Gunn diodes etc.

Properties of Microwaves

  • Seen in the reflection, refraction, and diffraction process.


  • Present in radar and telecommunication.

  • Analysis of fine details of molecular structure.

  1. Infrared Waves: These waves are discovered by Herschell and have a frequency range $3\times 10^{11}\, Hz$ to $4\times 10^{14}\, Hz$. Infrared waves are sometimes called heat waves and they are produced by hot bodies.

Properties of Infrared Waves

  • They have thermal effects also.

  • All properties are similar to those of light except wavelength.


  • Used in industry, medicine, and astronomy.

  • Used from fog or haze photography.

  • Elucidating molecular structure.

  1. Visible Light: It is the narrow region which is detected by the human eye. It has a frequency range from $4\times 10^{14}\,Hz$ to $8\times 10^{14}\,Hz$. These are produced by atomic excitation.

Properties of Visible Light

  • Sensitive to the human eye.


  • Helpful in seeing objects.

  • Helpful in studying molecular structure.

  1. Ultraviolet Rays: These rays were discovered by Ritter in 1801 and their frequency ranges from $8\times 10^{14}\, Hz$ to $5\times 10^{16}\, Hz$. These rays are produced by sum and very hot bodies.

Properties of Ultraviolet rays

  • They are present in all properties of light.

  • Present in the photoelectric effect.


  • Helps in the detection of adulteration, writing, and signature.

  • Helpful in the sterilization of water due to its destructive action of bacteria.

  1. X-Rays: These rays are discovered by the German physicist W. Roentgen. Their frequency range is $10^{16}\, Hz$ to $3\times 10^{21}\, Hz$ and they are produced when high energy electrons are stopped suddenly on the metal of high atomic number.

Properties of X-rays

  • They have a low penetrating power.

  • Their other properties are similar to gamma rays except for their wavelength.


  • Helps in medical diagnosis and treatment.

  • Helps in the study of the crystal structure.

  • Useful in industrial radiography.

  1. Gamma Rays: These rays are the waves of frequency range $3\times 10^{18}\, Hz$ to $5\times 10^{22}\, Hz$. These are produced by radioactive substances.

Properties of Gamma rays

  • They have a high penetrating power.

  • They are uncharged.

  • They have low ionizing power.


  • Gives information on nuclear structure.

  • Help in medical treatment.

How are Electromagnetic waves formed?

Electric Field and Force:

  • An electric field is formed by a charged particle.

  • This electric field exerts a force on other charged particles.

  • Positive charges accelerate in the direction of the electric field, while negative charges accelerate in the opposite direction.

Magnetic Field and Force:

  • A magnetic field is created by a moving charged particle.

  • This magnetic field applies a force to other moving particles.

  • The force acts perpendicular to the direction of their motion, changing the direction of their speed.

Electromagnetic (EM) Field:

  • An accelerating charged particle generates an EM field.

  • Electromagnetic waves consist of electric and magnetic fields propagating through free space at the speed of light.

Oscillating Charged Particle:

  • Electromagnetic waves are produced when a charged particle oscillates around a central position.

  • The frequency of oscillation corresponds to the frequency of the resulting electromagnetic wave.

Frequency and Wavelength:

  • The frequency of the electromagnetic wave is denoted as "f."

  • The wavelength of the wave is represented as "λ."

  • The relationship between wavelength and frequency is given by λ = c/f, where "c" is the speed of light.

Propagation of Energy:

  • Electromagnetic waves transmit energy through space.

Sinusoidal Waveform:

  • Electromagnetic waves are represented by a sinusoidal (wave-like) graph.

  • This graph illustrates time-varying electric and magnetic fields.

  • The fields are both perpendicular to each other and to the direction of wave propagation.

Diagonal Nature:

  • Electromagnetic waves are described as diagonal in nature, indicating their perpendicularity and transverse characteristics.

Types of Electromagnetic Waves

Electromagnetic waves encompass a spectrum that includes various types, each characterized by distinct properties and applications.

  1. Radio Waves: These have the longest wavelengths and are utilized in communication systems like radios and broadcasting.

  2. Microwaves: With shorter wavelengths than radio waves, microwaves power microwave ovens, communication technology, and radar systems.

  3. Infrared Waves: Beyond microwaves, they're used in night vision technology, remote controls, and thermal imaging.

  4. Visible Light: The only segment perceivable by the human eye, enabling vision and serving diverse applications from photography to optical communications.

  5. Ultraviolet Waves: Just beyond visible light, used in sterilization, fluorescent lighting, and skin tanning.

  6. X-Rays: Higher-energy waves used in medical imaging (X-rays) and security screening.

  7. Gamma Rays: The shortest wavelength and highest energy, employed in cancer treatment and exploring nuclear reactions.

What are the Source of Electromagnetic Waves?

Understanding the diverse sources and types of electromagnetic waves is fundamental to comprehending their pervasive influence on our technological landscape. These waves, stemming from natural phenomena and human-engineered devices, power various communication systems, medical technologies, and scientific advancements. Delving into their origins illuminates the breadth of applications and the intricate interplay between nature and technology in harnessing these waves for human benefit.

Electromagnetic waves originate from various natural and artificial sources:

  1. Natural Sources: Celestial bodies, such as stars, including our sun, emit electromagnetic waves across the spectrum. Earth's natural phenomena like lightning also generate radio waves.

  2. Artificial Sources: Technology drives many artificial sources, including:

  3. Antennas: Used in broadcasting, antennas emit radio waves.

  4. Microwaves: Devices like microwave ovens and radars produce and utilize microwaves.

  5. Light Bulbs: Emit visible light waves for illumination.

  6. X-ray Machines: Create X-rays for medical imaging.

  7. Satellites and Communication Devices: Generate and receive various waves for communication.

  8. Lasers: Produce coherent and focused light waves for diverse applications.

Electromagnetic Spectrum of Wavelengths

Physicists has categorized electromagnetic waves into certain groups based on their frequencies f. The wavelength of the electromagnetic wave is equal to λ = c/f. ~400 nm to ~700 nm is the wavelength of visible light whereas ~400 nm is the wavelength of violet light. Its frequency is ~7.5*1014 Hz. 

In the full electromagnetic spectrum, visible light has a small part. The following table shows the frequency range and wavelength range of certain electromagnetic waves.

Type of Radiation

Frequency Range (Hz)

Wavelength Range


1020 - 1024

< 10-12 m


1017 - 1020

1 nm - 1 pm


1015 - 1017

400 nm - 1 nm


4 - 7.5*1014

750 nm - 400 nm


1*1014 - 4*1014

2.5 μm - 750 nm


1013 - 1014

25 μm - 2.5 μm


3*1011 - 1013

1 mm - 25 μm

radio waves

< 3*1011

> 1 mm


Properties of electromagnetic waves

  • They are transverse.

  • They have the oscillating electric and also magnetic vectors always at right angles which are in the direction of the propagation of the waves.

  • They can travel well in a vacuum because they do not require a medium to travel.

  • They travel at the speed of light.

  • They are produced by the oscillating of the electric charges and also by the transition of electrons between the energy levels from the atom.


Applications of the Electromagnetic Spectrum 

  • Radio waves are transmitted through the air which does not cause damage even if it is absorbed by the human body.

  • Microwaves are used in cooking. What happens is when the high-frequency microwaves get absorbed by the food molecules the heating is caused by the increase of the internal energy of the molecules.

  • Fiber optic communication uses visible light.

  • The sterilization of water using ultraviolet rays.

Relation Between Wavelength and Frequency 

Wavelength and frequency of a wave are inversely proportional. As the wavelength increases, frequency decreases and as the frequency increases wavelength decreases. As we know –

c = λ

Where c = speed of light 

λ = wavelength

 v = frequency

 So, we can write 

\[\frac{1}{v}\]= λ

Thus, λ ∝\[\frac{1}{v}\]

The relationship of frequency and wavelength has been shown above in the figure of the electromagnetic spectrum as well. 

Relation Between Frequency and Energy

As the frequency increases, energy also increases. Thus, frequency and energy are directly proportional. It can be expressed as –

E = hv

Where E = energy 

h = Plank’s constant which is equal to 6.6 × 10⁻³⁴ J

v = frequency

The relationship of frequency and energy has been shown above in the figure of the electromagnetic spectrum as well. 

This ends our coverage on the topic “Electromagnetic Radiation: Wave Nature”. We hope you enjoyed learning and were able to grasp the concepts. We hope after reading this article you will be able to solve problems based on the topic. If you are looking for solutions to NCERT Textbook problems based on this topic, then log on to the Vedantu website or download the Vedantu Learning App. By doing so, you will be able to access free PDFs of NCERT Solutions as well as Revision notes, Mock Tests and much more.

Types of EMF Exposure

Radiation is present in the so-called electromagnetic spectrum. These rays range from high intensity (so-called high-frequency) on one side of the spectrum, to very low (or low-frequency) intensity on the other side.

Examples of High-Energy Radiation include:

  • x-rays

  • Gamma rays

  • Some high-intensity ultraviolet (UV) rays

These are ionizing radiation, which means that these forces can affect atomic-level cells by releasing an electron atom, or "ionizing" them. Ionizing radiation can damage the DNA and cells of the body, which in turn may contribute to genetic mutations and cancer. On the other side of the spectrum are very low-frequency (ELF) rays. This is a type of non-ionizing radiation. It can move atoms in the body or cause them to vibrate, but most researchers agree that it is not enough to damage DNA or cells.Among the ELF rays and the strongest rays in the spectrum there are other types of non-ionizing radiation, such as:

  • Radiofrequency (RF) radiation.

  • Visible light

  • Infrared

Electric and magnetic fields join as one field in many types of radiation. The result is called the electromagnetic field (EMF).

In Short, Here are Two Types of EMF Exposure:

  1. High-frequency EMFs. This is a type of ionizing radiation. Scientific literature acknowledges that excessive exposure can damage the DNA or cells of a reliable source. Medical devices such as X-ray imaging and CT scans produce low levels of this radiation. Other sources include gamma rays from radiation and UV rays from sunburn or sun beds.

  2. Low- to mid-frequency EMFs. This is a type of ionizing radiation. Gentle and thought to be harmless to humAns: Household appliances such as microwave ovens, cell phones, hair dryers, and washing machines, as well as power cords and MRIs, emit this type of radiation. This category of EMFs comprises very low EMF (EMF-EMFs) and radiofrequency EMFs (RF-EMFs).

Magnetic Field Lines

When a magnetic field is created in an area, the magnetic dipoles create some particular force field lines. These lines are called magnetic field lines. Michael Faraday discovered the concept of magnetic field lines. Magnetic field lines show the direction of produced current and its strength. There are several laws about the characters and properties of magnetic field lines.

Direction of Field Lines

As the magnetic field is a vector quantity, magnetic field lines also have particular directions. The directions of magnetic field lines are different for inside and outside of the magnet. The magnetic field lines are directed from the north pole to the south pole outside the magnet. The direction of magnetic field lines inside the magnet is from the south pole to the north pole.

Strength of Magnetic Field Lines

Now, we are going to discuss the strength of magnetic field lines. Magnetic field strength is the intensity of the magnetic field. The magnetic field strength is proportional to the number of the magnetic field lines and their closeness. The closeness of the magnetic field lines is called the areal density. As the magnetic field lines don’t cross each other, the strength of the field is unique at any point of the magnetic field.

Properties of Magnetic Field Lines

Magnetic field lines create the visualized format of the magnetic field. The magnetic field lines have some particular properties, which are mentioned below.

  • The magnetic field lines are proportional to the number of lines and their density.

  • The strength of magnetic field lines is proportional to the number of lines and their density.

  • Magnetic field lines cannot cross each other.

  • Magnetic field lines are continuous loops.

  • The direction of the field lines is indicated by arrows at any point (south pole to north pole inside the magnet and north pole to south pole outside the magnet).

  • The density of the field lines is high near the poles.

Laws About Magnetic Field

There are several laws depicting the characteristics, properties, and directions of the magnetic field. Biot Savart discovered the concept of the magnetic field and gave a law about the creation of the magnetic field. Hendrik A. Lorentz gave a law about calculating the force of a magnetic field. Michael Faraday discovered the concept of magnetic field lines.

List of Important Formulas for Electromagnetic Waves Chapter


Name of the Concept


Displacement current

  • The expression of displacement current is;

$I_D=\epsilon_o \frac{\text{d}\phi_E}{\text{d}t}$

Continuity of current

  • The mathematical form of modified Ampere’s circuital law is;

$\oint\overrightarrow{B}.\overrightarrow{dl}=\mu_o (I+I_D)$

Maxwell’s Equation 

  • The expression of four Maxwell equations are;

  1. $\oint\overrightarrow{E}.\overrightarrow{ds}=\dfrac{q}{\epsilon_o}$

  2. $\oint\overrightarrow{B}.\overrightarrow{ds}=0$

  3. $\oint\overrightarrow{E}.\overrightarrow{dl}=-\frac{\text{d}}{\text{d}t}\oint\overrightarrow{B}.\overrightarrow{ds}$

  4. $\oint\overrightarrow{B}.\overrightarrow{dl}=\mu_o I+\mu_o \epsilon_o \frac{\text{d}}{\text{d}t}\oint\overrightarrow{E}.\overrightarrow{ds}$


Lorentz force

  • The expression of Lorentz force is given as,



Velocity of electromagnetic waves

  • The velocity of electromagnetic wave is given as;

$c=\frac{1}{\sqrt{\mu_o \epsilon_o}}=3 \times 10^8\,m/s$


Average electric field density

  • The expression of average electric field density is,

$u_E=\dfrac{1}{2} \epsilon_o E^2$


Average magnetic field density

  • The expression of average magnetic field density is,

$u_B=\dfrac{1}{2} \dfrac{B^2}{\mu_o}$


Intensity of an electromagnetic wave

  • The expression of intensity of an electromagnetic wave is,

$I=\dfrac{1}{2} \dfrac{B_o^2}{\mu_o}c$

Here, $B_o$ is the maximum value magnetic field vector.


Linear momentum of electromagnetic waves

  • The expression of linear momentum of electromagnetic wave is,


JEE Main Electromagnetic Waves Solved Examples 

1. The cross-sectional area of a $26\,mW$ laser beam is $10\,mm^2$. Then the  greatest electric field magnitude in this electromagnetic wave will be?(Given permittivity of space $\epsilon_o = 9 \times 10^{-12}$ SI units, speed of light $c = 3 \times 10^8\,m/s$)


Given, cross sectional area, $A= 10\,mm^2= 10 \times 10^{-6}\,m$

Power of the wave, $P=26\,mW=26 \times 10^{-3}\,W$

To find: Magnitude of the greatest electric field, $E=?$

In order to find the value of the greatest magnitude of electric field, we have to use the relation of intensity for electromagnetic waves as it is the ratio of power to area of cross section.

The expression of intensity for electromagnetic wave is,

$I=\dfrac{\text{Power (P)}}{\text{Area(A)}}=\dfrac{1}{2}\epsilon_o E^2 c$

Now using the above relation we can write;

$E=\sqrt{\dfrac{2P}{A \epsilon_o c}}$

After putting the values of known quantities, we get;

$E=\sqrt{\dfrac{2\times 26 \times 10^{-3}}{10 \times 10^{-6} \times 9 \times 10^{-12} \times 3 \times 10^8 }}$

$E= \sqrt{\dfrac{52\times 10^{-3}}{270 \times 10^{-10}}}$

$E=\sqrt{0.1925 \times 10^7}$

$E=\sqrt{1.925 \times 10^6}$

$E=1.387 \times 10^3\,V/m$


Hence, the greatest magnitude of the electric field is 1.387 kV/m.

Key point: The expression of intensity for electromagnetic waves is essential to solve such a type of problem.

2. If the maximum value of the electric field produced by sunlight is 620 N/C then calculate the overall average energy density of electromagnetic waves.


Given that, 

Maximum value of electric field vector, $E_o=620\,N/C$

To find: Overall average energy density

In order to find the overall average energy density we have to use the expression of total energy density i.e, the energy density due to electric field component and energy density due to magnetic field component. 

Therefore, the total energy density ($U$) of the wave is given as;

$U=\dfrac{1}{2}\epsilon_o E_o^2+ \dfrac{1}{2}\dfrac{B_o^2}{\mu_o}$.........(1)

As $B_o$ can be written in terms of $E_o$ as;


Now after putting the value of $B_o$ in the equation (1), we get;

$U=\dfrac{1}{2}\epsilon_o E_o^2+ \dfrac{1}{2 \mu_o}\dfrac{E_o^2}{c^2}$

$U=\dfrac{1}{2}\epsilon_o E_o^2+ \dfrac{1}{2 \mu_o}\dfrac{E_o^2}{\epsilon_o \mu_o}$..........(As $\dfrac{1}{c^2}=\epsilon_o \mu_o$)

On further solving, we get;

$U=\epsilon_o E_o^2$ 

Now after putting the values of $E_o$ in the above expression, we obtain;

$U=8.85 \times 10^{-12} \times (620)^2$

$U= 3.40 \times 10^{-6}\,J/m$

Hence, the overall average density of the electromagnetic wave is $3.40 \times 10^{-6}\,J/m$.

Key point: The expressions of average density due electric field component and magnetic field are important to solve such types of problems.

Previous Year Questions from JEE Paper

1. The critical angle of medium for a specific wavelength, if the medium has relative permittivity $3$ and relative permeability $\dfrac{4}{3}$ for this wavelength, will be (JEE Main 2020)

a. $60^\circ C$

b. $45^\circ C$

c. $15^\circ C$

d. $30^\circ C$


Given that,

Relative permittivity of the medium = 3 i.e, $\epsilon= 3\epsilon_o$........(3)

Relative permeability of the medium= $\dfrac{4}{3}$ i.e, $\mu=\dfrac{4}{3}\mu_o$..........(4)

To find: Critical angle, $\theta_c=?$

To find the critical angle for a specific wavelength, first we have to multiply the equation (3) and (4) and then obtain the ratio of speed of light in vacuum to the speed of light in medium which is equal to the sine of critical angle. 

Now after multiplying the equation (3) with (4), we obtain;

$\epsilon \mu =3\epsilon_o  \dfrac{4}{3}\mu_o$

$\epsilon \mu = 4\epsilon_o \mu_o$

As speed of light in medium is $v^2=\epsilon \mu$ and speed of light in vacuum is $c^2=\epsilon_o \mu_o$, therefore after putting these values in the above equation we get;


After taking square root on both sides of the equation,



As $\dfrac{c}{v}$= refractive index of the medium which in terms equals to the sine of critical angle, therefore using this we get;

$\sin \theta_c=\dfrac{1}{2}$

$\sin \theta_c= \sin 30^\circ$

$\theta_c= 30^\circ$

Hence, the value of critical angle for medium is $30^\circ$ and thus option d is the correct answer.

Key point: The knowledge of the ratio of speed of light in vacuum to speed of light in medium is equal to the sine of the critical angle is important to solve this problem.

2. A plane electromagnetic wave of frequency $25\,GHz$ is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by $\overrightarrow{B}=5 \times 10^{-8}\hat{j}\,T$. The corresponding electric field $\overrightarrow{B}$  is (speed of light $c=3 \times 10^8\,m/s$)  (JEE MAIN 2020)

a. $-1.66\times 10^{-16}\,\hat{i}\,V/m$

b. $1.66\times 10^{-16}\,\hat{i}\,V/m$

c. $-15 \hat{i}\,V/m$

d. $15 \hat{i}\,V/m$


Given that,

Magnetic field, $\overrightarrow{B}=5 \times 10^{-8}\hat{j}\,T$

As the wave is travelling in z-direction therefore $c=3 \times 10^8\,\hat{k}\,m/s$.

To find: Electric field, $\overrightarrow{E}=?$

To solve this problem, we have to use the relation between electric field and magnetic field. According to this relation, the electric field vector is equal to the $c$ (speed of light) times the magnetic field vector.

The relation between electric field and magnetic field is,

$\overrightarrow{E}=\overrightarrow{B}. \overrightarrow{c}$

$\overrightarrow{E}= 5 \times 10^{-8}\hat{j}.3 \times 10^8\,\hat{k}$

$\overrightarrow{E}= 5 \times 10^{-8}\times 3 \times 10^8\, \hat{j}.\hat{k}$

$\overrightarrow{E}= 15\hat{j}\, V/m$

Hence the value of the electric field vector is $15\hat{j}\, V/m$. Therefore, option d is the correct answer.

Key point: The relation between the electric field vector and the magnetic field vector is essential to solve this type of problem.

Practice Questions

1. In empty space, a planar electromagnetic wave with frequency $\nu= 23.9\,GHz$ propagates along the positive z-direction. If the electric field's maximum value is $60\,V/m$ then obtain the magnetic field vector.

(Ans: $\overrightarrow{B}=2 \times 10^{-7} \sin(0.5 \times 10^3z-1.5\times 10^{11}t)\hat{i}$)

2. What would be the electric field expression if the magnetic field in a plane electromagnetic wave is provided by $\overrightarrow{B}=3 \times 10^{-8} \sin(1.6 \times 10^8 x+48\times 10^{10}t)\hat{j}\,T$?

(Ans: $\overrightarrow{E}=9 \sin(1.6 \times 10^8 x+48\times 10^{10}t)\hat{k}\,V/m$)

JEE Main Physics Electromagnetic Waves Study Materials

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Explore an array of resources in the JEE Main Physics Study and Practice Materials section. Our practice materials offer a wide variety of questions, comprehensive solutions, and a realistic test experience to elevate your preparation for the JEE Main exam. These tools are indispensable for self-assessment, boosting confidence, and refining problem-solving abilities, guaranteeing your readiness for the test. Explore the links below to enrich your Physics preparation.

Benefits of Using Vedantu for JEE Main Exam 2024 - Physics Electromagnetic Waves

Embark on a journey of comprehensive JEE Main preparation for the "Electromagnetic Waves" chapter with Vedantu's tailored approach. From expert guidance to interactive sessions, Vedantu ensures clarity, practical applications, and personalised study plans, fostering a deep understanding for success in JEE Main.

  1. Topic-Wise Clarity for Electromagnetic Waves: Vedantu ensures crystal-clear explanations, fostering a solid understanding of each topic within the Electromagnetic Waves chapter, vital for JEE Main preparation.

  2. Expert Guidance on Electromagnetic Waves: Receive expert guidance from Vedantu's instructors to navigate challenging concepts specific to Electromagnetic Waves, clarifying doubts for a stronger grasp of the subject in preparation for JEE Main.

  3. Interactive Learning Sessions for Electromagnetic Waves: Engage in interactive learning sessions tailored for Electromagnetic Waves, enhancing the effectiveness of your study and ensuring a more engaging exploration of the chapter.

  4. Customized Study Plans for Electromagnetic Waves: Vedantu offers personalized study plans for Electromagnetic Waves, aligning with your pace and enabling a comfortable, thorough understanding of this JEE Main-focused chapter.

  5. Live Problem Solving for Electromagnetic Waves: Participate in live problem-solving sessions specific to Electromagnetic Waves, honing your application of concepts and improving problem-solving skills for JEE Main preparation.

  6. Exam Preparation Support for Electromagnetic Waves: Access valuable resources and tips designed for efficient preparation specifically for JEE Main, ensuring confidence and readiness on exam day, particularly for the Electromagnetic Waves chapter.

  7. Real-Life Applications of Electromagnetic Waves: Vedantu connects theoretical concepts of Electromagnetic Waves to real-life applications, making the study more relatable and practical, aiding deeper understanding for JEE Main aspirants.

  8. Flexible Learning Schedule for Electromagnetic Waves: Vedantu's platform allows flexibility in learning schedules, accommodating various study preferences and time commitments, catering to the unique needs of JEE Main students focusing on Electromagnetic Waves.

  9. Accessible Study Materials for Electromagnetic Waves: Utilise Vedantu's comprehensive study materials specifically tailored for the Electromagnetic Waves chapter, making learning accessible and convenient for JEE Main aspirants.

  10. Regular Assessments for Electromagnetic Waves: Benefit from regular assessments and quizzes tailored for the Electromagnetic Waves chapter, tracking progress and identifying areas for improvement, crucial for effective JEE Main preparation.


In this article, we'll dive into the fascinating world of Electromagnetic Waves in physics. We'll explore the fundamental concepts and tackle questions related to this topic in your JEE Main preparation. You'll get a comprehensive understanding of electromagnetic waves, their properties, and how they play a crucial role in the world of physics. We've gathered everything you need in one place, including free downloadable PDFs with in-depth explanations. This resource will be your go-to guide for mastering Electromagnetic Waves, ensuring you excel in your exams.

FAQs on Electromagnetic Waves Chapter - Physics JEE Main

1. What is the weightage of the Electromagnetic waves in the JEE exam?

This chapter includes at least 1-2 questions in each year which ultimately lead to the weightage of approximately 2-3% in the exam.

2. What is the degree of difficulty of the Electromagnetic waves chapter questions?

The difficulty level of the questions asked in this chapter is easy to moderate. Therefore it is important to study this chapter as this chapter’s question can help in scoring good marks.

3. Is it really beneficial to review the Electromagnetic waves chapter questions from last year's papers?

We must practice the previous year's questions in order to score well and become familiar with the exam's difficulty level. It increases our self-esteem while also pointing us to areas where we may improve. Solving question papers from the last ten to fifteen years will help you better comprehend a subject and can also show you how many times a concept or topic will be repeated in the test. It is also beneficial to practise the previous year's issues in order to prepare for the Electromagnetic waves jee notes.

4. How does Vedantu support understanding the Electromagnetic Waves Chapter - Physics JEE Main?

Vedantu provides topic-wise clarity, expert guidance, and interactive learning sessions to ensure a solid understanding of the Electromagnetic Waves chapter for JEE Main.

5. What resources are available on Vedantu for preparing for the Electromagnetic Waves Chapter - Physics JEE Main?

Vedantu offers personalised study plans, live problem-solving sessions, and comprehensive study materials with real-life applications, facilitating flexible learning schedules and efficient exam preparation for JEE Main.