Have you ever wondered how your phone can pick up radio signals or how the sun can travel millions of kilometres to hit the Earth? The solution is in electromagnetic waves, which constitute the basis of Chapter 8 of Class 12 Physics. This chapter discusses the properties, behaviour and uses of the electromagnetic waves in real life, and the NCERT Solutions of Class 12 Physics Chapter 8 - Electromagnetic Waves provide these simple and step-by-step explanations, which help students to understand them well.
This Story also Contains
These NCERT Solutions for Class 12 Physics Chapter 8 - Electromagnetic Waves are very useful in the preparation of CBSE Class 12 board examination and competitive examinations such as JEE Main and NEET because they contain not only solved exercises, but also extra practice questions and an understanding of the concept. These NCERT solutions, which are designed by the subject experts, not only conserve the precious study time but also simplify the revision of the complex concepts. It is also free of cost, and the NCERT Solutions for Class 12 Physics Chapter 8 - Electromagnetic Waves PDF can be downloaded by the student, allowing one to revise anytime and anywhere conveniently. These solutions are well constructed, and thus learners are able to reinforce their fundamentals and build confidence in exams.
Also Read
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 solutions 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.
Electromagnetic Waves class 12 question answers of textbook exercises offer detailed and easy-to-understand answers to all in-text and exercise problems. These solutions help students grasp key concepts, practice effectively, and prepare confidently for board exams and competitive tests like JEE and NEET.
Answer:
Radius of the discs(r) = 12cm
Area of the discs(A) =
Permittivity,
Distance between the two discs = 5cm=0.05m
=
=
=8.003 pF
But
Therefore,
=
Answer:
The value of the displacement current will be the same as that of the conduction current, which is given to be 0.15A. Therefore displacement current is also 0.15A.
Answer:
Yes, Kirchhoff's First rule (junction rule) is valid at each plate of the capacitor. This might not seem like the case at first, but once we take into consideration both the conduction and displacement current, Kirchhoff's first rule will hold good.
Answer:
Capacitance(C) of the parallel plate capacitor = 100 pF
Voltage(V) = 230 V
Angular Frequency (ω) =
RMS value of conduction current is
Answer:
Yes, conduction current is equal to the displacement current. This will be the case because otherwise, we will get different values of the magnetic field at the same point by taking two different surfaces and applying Ampere–Maxwell Law.
Answer:
We know Ampere-Maxwell's Law,
Between the plates conduction current
For a loop of radius r smaller than the radius of the discs,
Since we have to find the amplitude of the magnetic field, we won't use the RMS value but the maximum value of the current.
The amplitude of B at a point 3.0 cm from the axis between the plates is
Answer:
The speed with which these electromagnetic waves travel in a vacuum will be the same and will be equal to
Answer:
Since the electromagnetic wave travels along the z-direction, its electric and magnetic field vectors are lying in the x-y plane as they are mutually perpendicular.
Frequency of wave = 30 MHz
Wavelength =
Answer:
Frequency range = 7.5 MHz to 12 MHz
Speed of light =
Wavelength corresponding to the frequency of 7.5 MHz
=
Wavelength corresponding to the frequency of 12 MHz =
The corresponding wavelength band is 25m to 40m.
Answer:
The frequency of the electromagnetic waves produced by the oscillation of a charged particle about a mean position is equal to the frequency of the oscillation of the charged particle. Therefore, the electromagnetic waves produced will have a frequency of
Answer:
Magnetic Field (
Speed of light(c) = 3
Electric Field =
= 510
= 153
Answer:
(a)
=
Angular frequency (
=3.14
Propagation constant(k)
=
=
=
=
=1.05 rad
Wavelength(
=
(b) Assuming the electromagnetic wave propagates in the positive z-direction, the Electric field vector will be in the positive x-direction and the magnetic field vector will be in the positive y-direction, as they are mutually perpendicular and
Answer:
Now substitute a different range of wavelengths in the electromagnetic spectrum to obtain the energy
EM wave | One wavelength is taken from the range | Energy in eV |
Radio | 1 m | |
Microwave | 1 mm | |
Infra-red | 1000 nm | 1.2375 |
Light | 500 nm | 2.475 |
Ultraviolet | 1nm | 1237.5 |
X-rays | 0.01 nm | 123750 |
Gamma rays | 0.0001 nm | 12375000 |
What is the amplitude of the oscillating magnetic field?
Answer:
The amplitude of the oscillating magnetic field(B0) =
Answer:
The average energy density of the Electric field(uE)
=
=
=
=
Therefore, the average energy density of the electric field is equal to the average energy density of the Magnetic field.
Electromagnetic Waves NCERT Solutions of additional questions offer additional work beyond school textbook questions to learners to allow them to solidify their concept knowledge, such as the propagation of waves, energy transfer, and uses of electromagnetic waves. These are solutions which are capable of raising exam preparedness and problem-solving abilities, both on the board, as well as competitive examinations.
Answer:
The electric field vector is in the negative x-direction, and the wave propagates in the negative y-direction.
Answer:
From the equation of the wave given, we can infer k = 1.8 rad
Q.1 (c) Suppose that the electric field part of an electromagnetic wave in vacuum is
Answer:
From the given equation of the electric field, we can infer angular frequency(
Frequency(
=
=8.6
=86 MHz
Answer:
From the given equation of the electric field, we can infer Electric field amplitude (E 0 ) =3.1
=1.03
Answer:
As the electric field vector is directed along the negative x-direction and the electromagnetic wave propagates along the negative y-direction, the magnetic field vector must be directed along the negative z-direction. (
Therefore,
Answer:
Total power which is converted into visible radiation = 5% of 100W = 5W
The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.
Intensity for a sphere of radius 1m
=0.398
Answer:
Total power which is converted into visible radiation = 5% of 100W = 5W
The above means 5J of energy is passing through the surface of a concentric sphere(with the bulb at its centre) per second.
Intensity for a sphere of radius 10 m
=3.98
Answer:
EM wave | One wavelength is taken from the range | Temperature |
Radio | 100 cm | |
Microwave | 0.1cm | 2.9 K |
Infra-red | 100000ncm | 2900K |
Light | 50000 ncm | 5800K |
Ultraviolet | 100ncm | |
X-rays | 1 ncm | |
Gamma rays | 0.01 nm |
These numbers indicate the temperature ranges required for obtaining radiation in different parts of the spectrum
Answer:
Frequency(
Wavelength(
=0.283 m
=28.3 cm
Radio waves
Answer:
Using the formula,
= 0.107cm
Microwaves.
Answer:
E=14.4 keV
Wavelength(
=
= 0.85
X-rays
Answer the following questions
Q5. (a) Long-distance radio broadcasts use short-wave bands. Why?
Answer:
Long-distance radio broadcasts use short-wave bands as these are refracted by the ionosphere.
Answer the following questions
Q5. (b) It is necessary to use satellites for long-distance TV transmission. Why?
Answer:
As TV signals are of high frequencies, they are not reflected by the ionosphere and therefore satellites are used to reflect them.
Answer the following questions
Answer:
X-rays are absorbed by the atmosphere, and therefore, the source of X-rays must lie outside the atmosphere to carry out X-ray astronomy. Therefore, satellites orbiting the earth are necessary, but radio waves and visible light can penetrate through the atmosphere, and therefore optical and radio telescopes can be built on the ground.
Answer the following questions
Q5. (d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
Answer:
The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs the ultraviolet radiation coming from the sun, which is very harmful to humans.
Answer the following questions
Answer:
If the Earth did not have an atmosphere, its average surface temperature would be lower than what it is now, as in the absence of an atmosphere, there would be no greenhouse effect.
Answer the following questions
Answer:
The use of nuclear weapons would cause the formation of smoke clouds, preventing the light from the sun from reaching Earth's surface, and it would also deplete the atmosphere and therefore stop the greenhouse effect and thus doubling the cooling effect.
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:
Or
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
But
Here
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
Q4: The magnetic field of a plane electromagnetic wave is given by :
where
The rms value of the force (in Newton) experienced by a stationary charge
Answer:
Amplitude of resultant magnetic field
Q5: Two light waves are given by,
Answer:
As we learned
Intensity of EM wave -
where
so
Class 12 Physics Chapter 8 - Electromagnetic Waves is to explain the oscillation of electric and magnetic fields in order to transmit the energy across the space. The chapter identifies the source, characteristics and the uses of electromagnetic waves, which is the backbone of the modern communication systems. These topics are not just some good preparation in 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
The electromagnetic waves are the foundation of the modern world of communication, medical science and even 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 application in real life rather than memorization. The systematic approach will assist in addressing the conceptual as well as the numerical issues without any problems.
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 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.
Frequently Asked Questions (FAQs)
Experts at Careers360 have created the chapter 8 physics class 12 ncert solutions after completing extensive research on each concept. To assist students perform well on class assignments and board exams, every small element is well addressed. Additionally, it enables students to successfully complete their tasks on time.
Before starting the preparation of NCERT class 12 chapter 8 it is better to have an idea of chapters 1 to 7. Chapter 1 to 7 is interconnected.
One question from chapter 8 electromagnetic waves can be expected for the NEET exam. The questions can be theoretical or numerical.
From the chapter on electromagnetic waves, one question can be expected for JEE Main. More than one question can also be asked.
3 to 5 marks questions are asked from class 12 chapter electromagnetic waves for CBSE board exams.
On Question asked by student community
Hello,
The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.
Hope this information is useful to you.
Hello,
Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.
Hope this information is useful to you.
Hello Pruthvi,
Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.
The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.
For more details about the KCET Exam preparation,
CLICK HERE.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.
Hello
For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
As per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters