NCERT solutions for class 12 physics chapter 8 Electromagnetic Waves

NCERT Solutions for Class 12 Physics Chapter 8 - Electromagnetic Waves: Download PDF

The Class 12 Physics Chapter 8 - Electromagnetic Waves question answers help students gain a clear understanding of the concepts and reinforce their knowledge by giving out detailed and stepwise answers to all the questions present in the chapter. The Class 12 physics chapter 8 Electromagnetic Waves questions answers are crafted according to the new NCERT syllabus and are also very beneficial in preparation for the board exams and competitive exams like JEE and NEET. You may as well download the free PDF to read continuously and revise successfully.

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Class 12 Physics Chapter 8 - Electromagnetic Waves: Exercise Questions

Electromagnetic Waves class 12 question answers of textbook exercises offer detailed and easy-to-understand answers to all in-text and exercise problems. These Electromagnetic Waves class 12 question answers help students grasp key concepts, practice effectively, and prepare confidently for board exams and competitive tests like JEE and NEET.

Class 12 Physics Chapter 8 - Electromagnetic Waves: Higher Order Thinking Skills (HOTS) Questions

Chapter 8 Higher Order Thinking Skills (HOTS) Questions provide students with an opportunity to think beyond ordinary exercises and think critically in understanding electromagnetic waves. These Electromagnetic Waves class 12 question answers also assist students to apply concepts in real life and prepare themselves well for both board exams as well as competitive ones like JEE and NEET.

Q1. A red LED emits light at 0.2 watts uniformly around it. The amplitude of the electric field of the light at a distance of 3 m from the diode is ___ V/m.

Answer:

$I=P / 4 \pi r^2$ and $p_{avg}=\frac{1}{2} \varepsilon_0 E_{o}^2$
$\therefore P / 4 \pi r^2=\left(\frac{1}{2}\right) \mathcal{E}_0 E_0^2 C$

Or $E_0=\sqrt{\frac{2 P}{4 \pi \varepsilon_0 r^2 C}}$

$\begin{aligned} & \text { Here } p=0.2 \mathrm{w}, r=3 \mathrm{~m}, c=3 \times 10^8 \mathrm{~ms}^{-2} \\ & \frac{1}{4} \pi \varepsilon_0=9 \times 10^9\end{aligned}$

$\begin{aligned} & E_0=\sqrt{\frac{2 \times 0.2 \times 9 \times 10^9}{9 \times 3 \times 10^8}} \\ & E_0=\sqrt{\frac{0.4 \times 9 \times 10^9}{27 \times 10^8}} \\ & E_0=\sqrt{\frac{36}{27}}\end{aligned}$

$E_0=1.15 \mathrm{~v} / \mathrm{m}$

Q2. The following figure shows a capacitor made of two circular plates. The capacitor is being charged by an external source which supplies a constant current equal to 0.15 A. What is the displacement current (in amperes) across plates?

Answer:

As we know,

Displacement current

$I_d=\frac{\varepsilon_0 d \phi_c}{d t}$

But $\phi_E=E A=\frac{q}{A \varepsilon_0} A=\frac{q}{\varepsilon_0}$

$\therefore \quad I_d=\varepsilon_0 \frac{d}{d t}\left(\frac{q}{\varepsilon_0}\right)=\frac{\varepsilon_0}{\varepsilon_0} \frac{d q}{d t}=\frac{d q}{d t}=I$

Here $I=0.15 \mathrm{~A} \quad \therefore I_d=0.15 \mathrm{~A}$

Q3. A radiation of 200 W is incident on a surface which is 60% reflecting and 40% absorbing. The total force on the surface is

Answer:
Force exerted by incident radiation is $F=\frac{P}{C}$ where $P$ is the power of radiation and $C$ is the velocity of light. For absorbed radiations are $40 \%$ then exerted force is $F_{a b s}=\frac{0.4 P}{c}$ and $60 \%$ part of radiations is reflected by the surface, so force exerted by theses radiations is $$ F_{r e f}=2 \times \frac{0.6 P}{C}=\frac{1.2 P}{C} $$ Then total force exerted by radiations on the surface is $$ \begin{aligned} & F_{\text {total }}=F_{\text {ref }}+F_{\text {abs }} \\ & =\frac{1.2 P}{c}+\frac{0.4 P}{c}=\frac{1.6 P}{c} \\ & =\frac{1.6 \times 200}{3 \times 10^8}=1.07 \times 10^{-6} \mathrm{~N} \end{aligned} $$

Q4. The magnetic field of a plane electromagnetic wave is given by :

$
\vec{B}=B_0[\cos (k z-\omega t)] \hat{i}+B_1 \cos (k z+\omega t) \hat{j}
$

where $B_0=3 \times 10^{-5} T$ and $B_1=2 \times 10^{-6} \mathrm{~T}$.
The rms value of the force (in Newton) experienced by a stationary charge $Q=10^{-4} C$ at $Z=0$ is closest to (up to one decimal).

Answer:

$
\begin{aligned}
& B=B_0 \cos (k x-\omega t) \hat{i}+B_1 \cos (k z+\omega t) \hat{j} \\
& B_0=3 \times 10^{-5} T \\
& B_1=2 \times 10^{-6} T
\end{aligned}
$

Amplitude of resultant magnetic field $=B^{\prime}=\sqrt{B_o^2+B_1^2}$

$
\begin{aligned}
Q & =10^{-4} C \\
F_{r m s} & =Q E_{r m s}=Q\left(c B_{r m s}^{\prime}\right) \\
F_{r m s} & =Q \sqrt{\left(\frac{C B_0}{\sqrt{2}}\right)^2+\left(\frac{C B_1}{\sqrt{2}}\right)^2} \\
& =10^{-4} \times \frac{3 \times 10^8}{\sqrt{2}} \sqrt{\left(3 \times 10^{-5}\right)^2+\left(2 \times 10^{-6}\right)^2}
\end{aligned}
$

$
\begin{aligned}
& =10^{-4} \times \frac{3 \times 10^8}{\sqrt{2}} \sqrt{9 \times 10^{-10}+4 \times 10^{-12}} \\
& =10^{-4} \times \frac{3 \times 10^8}{\sqrt{2}} \sqrt{900 \times 10^{-12}+4 \times 10^{-12}} \\
& =\frac{3 \times 10^4}{\sqrt{2}} \sqrt{904} \times 10^{-6} \\
& =0.6 \mathrm{~N}
\end{aligned}
$

Q5. Two light waves are given by, $E_1=2 \sin (100 \pi t-k x+30)$ and $E_2=3 \cos (200 \pi t-k x+60)$. The ratio of the intensity of the first wave to that of the second wave is :
Answer:
As we learned
Intensity of EM wave -

$
I=\frac{1}{2} \epsilon_o E_o^2 c
$

where
$\epsilon_o=$ Permittivity of free space
$E_o=$ Electric field amplitude
$\mathrm{c}=$ Speed of light in vacuum
so

$
I \propto A^2 \therefore \frac{I_1}{I_2}=\frac{2^2}{3^2}=4 / 9
$

Class 12 Physics Chapter 8 - Electromagnetic Waves: Topics

Class 12 Physics Chapter 8 - Electromagnetic Waves is to explain the oscillation of electric and magnetic fields in order to transmit the energy across space. The chapter identifies the source, characteristics and uses of electromagnetic waves, which are the backbone of modern communication systems. These topics are not just some good preparation for board exams, but also a good way of learning the concepts that are helpful in JEE and NEET.

8.1 Introduction
8.2 Displacement current
8.3 Electromagnetic waves
8.3.1 Sources of electromagnetic waves
8.3.2 Nature of electromagnetic waves
8.4 Electromagnetic spectrum
8.4.1 Radio waves
8.4.2 Microwaves
8.4.3 Infrared waves
8.4.4 Visible rays
8.4.5 Ultraviolet rays
8.4.6 X-rays
8.4.7 Gamma rays

Class 12 Physics Chapter 8 - Electromagnetic Waves: Important Formulae

Class 12 Physics Chapter 8 Electromagnetic Waves explains how electric and magnetic fields combine to form waves that travel through space without needing any medium. This chapter introduces students to Maxwell’s equations, displacement current, and the nature of the electromagnetic spectrum. Learning the important formulas from this chapter helps students solve numerical problems quickly and understand the relationship between wavelength, frequency, and energy in different types of electromagnetic waves.

Displacement current:

$
I_d=\varepsilon_0 \frac{d \Phi_E}{d t}
$

Maxwell's equations (free space):

$
\begin{aligned}
& \nabla \cdot \vec{E}=\frac{\rho}{\varepsilon_0} \\
& \nabla \cdot \vec{B}=0 \\
& \nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t} \\
& \nabla \times \vec{B}=\mu_0 \vec{J}+\mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}
\end{aligned}
$

Speed of light:

In vacuum: $c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$
In medium: $v=\frac{c}{n}$
Angular frequency: $\omega=2 \pi f$
Wave number: $k=\frac{2 \pi}{\lambda}$
Energy densities:

Electric: $u_E=\frac{1}{2} \varepsilon_0 E^2$
Magnetic: $u_B=\frac{1}{2} \frac{B^2}{\mu_0}$
Total: $u=u_E+u_B=\varepsilon_0 E^2=\frac{B^2}{\mu_0}$

Approach to Solve Questions of Class 12 Physics Chapter 8 - Electromagnetic Waves

The electromagnetic waves are the foundation of the modern world of communication, medical science and even the transfer of energy in nature. During the process of answering questions in this chapter, the students are required to concentrate on the properties of waves and their equations and their application in real life, rather than memorisation. The systematic approach will assist in addressing the conceptual as well as the numerical issues without any problems.

• Revise Basics of Maxwell's Equations - Learn how electric and magnetic fields that vary with time produce electromagnetic waves.
• Memorise Important Formulas -Speed of EM waves, relationship between electric field (E) and magnetic field (B), energy density.
• Know the Spectrum - Learn the frequency, wavelength ranges, and other uses of various components of the electromagnetic spectrum (radio, microwave, infrared, visible, UV, X-ray, gamma).
• Concentrate on Wave Properties: Polarisation, reflection, refraction, and interaction of EM waves with objects.
• Solve Conceptual Problems -Use reasoning to solve problems in the real world, such as communication devices, satellites, and medical imaging.
• Step-by-Step approach to solving Numerical Problems - It is always important to begin with the known quantities (E, B, c, f) and use relations in a systematic manner.
• Practice Past Year questions -Work on JEE/ NEET and CBSE board questions to be acquainted with concepts that are commonly asked.
• Make Short Notes - Summarise formulas, spectrum chart, and key concepts for quick revision.

How Can NCERT Solutions for Class 12 Physics Chapter 8 Help in Exam Preparation?

Electromagnetic Waves class 12 question answers can be of great help to the students who want to perform well in their board examinations. In this chapter, the connection between magnetic and electric fields is established, and how electromagnetic waves are generated by changing fields is explained, which is a fundamental concept in the field of physics in the modern world. These Class 12 physics Electromagnetic Waves question answers can help students to have a clear idea of major concepts like the electromagnetic spectrum, characteristics of various waves, as well as their applications in various technologies such as radio, television, and satellites. Conceptual, numerical and application questions are included in these solutions as well, which intensify the problem-solving and analytical thinking. All the answers are presented in simple and exam-oriented language to assist students in writing accurate and comprehensive answers whenever undertaking exams. Moreover, revising by NCERT solutions provides total coverage of the syllabus and enables students to remember key points easily during the last-minute revision, hence one can be confident about answering questions in the board as well as in competitive exams in a more effective manner.

What Extra Should Students Study Beyond NCERT For JEE/NEET?

Beyond NCERT, students preparing for JEE/NEET should study additional concepts like Energy Density and Intensity of Electromagnetic Waves, which explain how energy is distributed in electric and magnetic fields and how it propagates through space. Understanding these topics enhances conceptual clarity and equips students to solve advanced numerical and application-based problems in competitive exams.

NCERT Solutions for Class 12 Physics Chapter-Wise

NCERT Solutions for Class 12 Physics Chapter-wise links provide a complete guide to all chapters, making it easier for students to access solved exercises, important formulas, and key concepts in one place. These solutions are designed to help students prepare effectively for board exams, JEE, NEET, and other competitive tests.

Also Check NCERT Books and NCERT Syllabus here:

• NCERT Books Class 12 Physics
• NCERT Syllabus Class 12 Physics
• NCERT Books Class 12
• NCERT Syllabus Class 12

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• NCERT solutions for Class 12 Mathematics
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