## 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

### 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

## 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.

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

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

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

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

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

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

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:

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.

Artificial Sources: Technology drives many artificial sources, including:

Antennas: Used in broadcasting, antennas emit radio waves.

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

Light Bulbs: Emit visible light waves for illumination.

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

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

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.

### 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:

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.

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

### 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 $c = 3 \times 10^8\,m/s$)

Sol:

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;

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

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

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

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

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

$E=1.387\,kV/m$

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.

Sol:

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;

$B_o=\dfrac{E_o}{c}$..........(2)

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. $30^\circ C$

Sol:

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;

$v^2=4c^2$

After taking square root on both sides of the equation,

$v=2c$

$\dfrac{c}{v}=\dfrac{1}{2}$

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

Sol:

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

Here, you'll find a comprehensive collection of study resources for Electromagnetic Waves designed to help you excel in your JEE Main preparation. These materials cover various topics, providing you with a range of valuable content to support your studies. Simply click on the links below to access the study materials of Electromagnetic Waves and enhance your preparation for this challenging exam.

## JEE Main Physics Study and Practice Materials

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.

### Conclusion

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.