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JEE Important Chapter - Electromagnetic Waves

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Important Concepts For JEE Electromagnetic Waves Chapter

Important Concepts For JEE Electromagnetic Waves Chapter

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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.


Important Topics of Electromagnetic Waves

  • What are electromagnetic waves?

  • Displacement current

  • Continuity of current

  • Maxwell’s Equation  

  • Lorentz force

  • Velocity of electromagnetic waves

  • Properties of electromagnetic waves

  • Electromagnetic spectrum


Important Concept of Electromagnetic Waves for JEE

Name of the Concept

Key Points of the Concepts

1. 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. 

2. 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;

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

3. 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,

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

4. 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.

5. 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;

$\overrightarrow{F}=q(\overrightarrow{E}+\overrightarrow{v}\times\overrightarrow{B})$

6. 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;

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

7. 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,

$p=\dfrac{U}{c}$

  • 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,

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

8. 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.

  2.  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.

  3. 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.

  4. 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.

  5. 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.

  6. 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.

  7. 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.


List of Important Electromagnetic Waves Formulas  

S.No.

Name of the Concept

Formula


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

      4. 

Lorentz force

  • The expression of Lorentz force is given as,

$\overrightarrow{F}=q(\overrightarrow{E}+\overrightarrow{v}\times\overrightarrow{B})$

      5.  

Velocity of electromagnetic waves

  • The velocity of electromagnetic wave is given as;

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

    6.  

Average electric field density

  • The expression of average electric field density is,

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

    7. 

Average magnetic field density

  • The expression of average magnetic field density is,

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

    8. 

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.

    9.

Linear momentum of electromagnetic waves

  • The expression of linear momentum of electromagnetic wave is,

$p=\dfrac{U}{c}$


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


Conclusion

In this article, we have studied electromagnetic waves and their properties. We came to know what an electromagnetic wave is and discussed the various electromagnetic wave examples. We have also discussed the displacement current which is responsible for the flow of current between the plates of capacitors. We discussed the four Maxwell equations which are the backbone of electromagnetism in physics. We have also talked about several concepts like velocity, energy density and intensity of electromagnetic waves.

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JEE Main 2022 Book Solutions and PDF Download

JEE Main 2022 Book Solutions and PDF Download

View all JEE Main Important Books
In order to prepare for JEE Main 2022, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2022 exam so that they can grab the top rank in the all India entrance exam.
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Maths
NCERT Book for Class 12 Maths
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Physics
NCERT Book for Class 12 Physics
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JEE Main Mock Tests

JEE Main Mock Tests

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JEE Main 2022 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2022 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.
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JEE MAIN MOCK TEST - 1
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 3
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 2
3 hr  • 75 questions • OBJECTIVE
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JEE Main 2022 Cut-Off

JEE Main 2022 Cut-Off

JEE Main Cut Off
NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
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JEE Main 2022 Results

JEE Main 2022 Results

JEE Main 2022 June and July Session Result - NTA has announced JEE Main result on their website. To download the Scorecard for JEE Main 2022 June and July Session, visit the official website of JEE Main NTA.
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Rank List
Counselling
Cutoff
JEE Main 2022 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in NITs and CFTIs colleges. JEE Main 2022 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2022 state rank list on the official website or on our site.
The NTA will conduct JEE Main 2022 counselling at https://josaa.nic.in/. There will be two rounds of counselling for admission under All India Quota (AIQ), deemed and central universities, NITs and CFTIs. A mop-up round of JEE Main counselling will be conducted excluding 15% AIQ seats, while the dates of JEE Main 2022 June and July session counselling for 85% state quota seats will be announced by the respective state authorities.
NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
Want to know which Engineering colleges in India accept the JEE Main 2022 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.
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Counselling

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FAQs on JEE Important Chapter - Electromagnetic Waves

FAQ

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.