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# JEE Important Chapter - Electromagnetic Waves

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## Important Concepts For JEE Electromagnetic Waves Chapter

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

### 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 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 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.

## JEE Top Colleges

View all JEE Main 2022 Top Colleges
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.

## JEE Main Upcoming Dates

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