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NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves- If you're a Class 12 student concerned about NCERT solutions electromagnetic waves, you're in the right place on Careers360. On this NCERT solutions page, you'll find answers to all the exercises and additional exercise questions from one to fifteen. These class 12 chapter 8 physics NCERT solutions, authored by subject experts, are detailed and presented in easy-to-understand language
The Electromagnetic Waves NCERT solutions cover the question related to the concept of displacement current. Problems related to the Electromagnetic Spectrum are also discussed in NCERT Solutions for Class 12 Physics chapter 8 Electromagnetic Waves. The class 12 physics ch 8 NCERT solutions help in testing the knowledge about the concepts studied in a chapter. Students must refer to these solutions of NCERT to prepare for the examination effectively. The NCERT Solutions for Class 12 Physics Chapter 8 PDF download also helps in preparing for various entrance exams like JEE, NEET, etc.
The NCERT textbook is usually explicitly referenced in the questions that are asked in the CBSE EMW class 12 exams. One of the subjects that comes up frequently on the board test is electromagnetism. Therefore, it is advised that students use the electromagnetic waves NCERT solutions to get a firm understanding of both the chapter and the subject.
Free download electromagnetic waves class 12 ncert solutions PDF for CBSE exam.
Calculate the capacitance and the rate of change of potential
difference between the plates.
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,
=
Obtain the displacement current across the plates.
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.
Is Kirchhoff’s first rule (junction rule) valid at each plate of the
capacitor? Explain.
Answer:
Yes, Kirchoff'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 the Kirchoff's first rule will hold good.
What is the rms value of the conduction current?
Answer:
Capacitance(C) of the parallel plate capacitor = 100 pF
Voltage(V) = 230 V
Angular Frequency (ω) =
RMS value of conduction current is
Is the conduction current equal to the displacement current?
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.
Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Answer:
We know the Ampere - Maxwells 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 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 (speed of light in vacuum).
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 electromagnetic waves produced will have a frequency of 10 9 Hz.
Answer:
Magnetic Field (B 0 )=510 nT = 510 10 -9 T
Speed of light(c) = 3 10 8 ms -1
Electric Field = B 0 c
= 510 10 -9 T 3 10 8 ms -1
= 153 NC -1
Answer:
E 0 = 120 NC -1
(a)
=
Angular frequency ( ) = 2
=3.14 10 8 rad s -1
Propagation constant(k)
=
=
=
=
=1.05 rad m -1
Wavelength( ) =
=
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 gives the direction of propagation of the wave.
Answer:
Now substitute the different range of wavelength 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(B 0 ) =
Answer:
The average energy density of the Electric field(U E )
=
= (as E=Bc)
=
=U B
Therefore the average energy density of the electric field is equal to the average energy density of the Magnetic field.
Answer:
The electric field vector is in the negative x-direction and the wave propagates in the negative y-direction.
11 (b) Suppose that the electric field part of an electromagnetic wave in vacuum is
Answer:
From the equation of the wave given we can infer k = 1.8 rad m -1
11 (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( ) = 5.4 10 8 rad s -1
Frequency( ) =
=
=8.6 10 7 Hz
=86 MHz
11 (d) Suppose that the electric field part of an electromagnetic wave in vacuum is
What is the amplitude of the magnetic field part of the wave?
Answer:
From the given equation of the electric field, we can infer Electric field amplitude (E 0 ) =3.1 NC -1
=1.03 10 -7 T
11(e) Suppose that the electric field part of an electromagnetic wave in vacuum is
Write an expression for the magnetic field part of the wave.
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,
Assume that the radiation is emitted isotropically and neglect reflection.
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 Wm -2
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 10 -3 Wm -2
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 radiations in different parts of the spectrum
Answer:
Frequency( )=1057 MHz
Wavelength( )
=0.283 m
=28.3 cm
Radio waves
Answer:
Using formula,
= 0.107cm
Microwaves.
Answer:
E=14.4 keV
Wavelength( ) =
=
= 0.85 A 0
X-rays
Answer the following questions
15. (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
15. (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 to be 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 and 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
15. (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 radiations coming from the sun which are very harmful to humans.
Answer the following questions
Answer:
If the earth did not have an atmosphere its average surface temperature be lower than what it is now as in the absence of 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 reaching earth surface and it will also deplete the atmosphere and therefore stopping the greenhouse effect and thus doubling the cooling effect.
Class 12 physics ch 8 NCERT solutions is an essential resource for students preparing for both board exams and competitive exams like JEE. This chapter can be considered moderately challenging but also offers scoring potential. These solutions provide a clear understanding of electromagnetic waves, making complex concepts manageable, and play a crucial role in helping students excel in both board exams and JEE, contributing significantly to their success. There are 15 questions in the Electromagnetic Waves NCERT solutions. The NCERT solution for Class 12 Physics chapter 8 covers theoretical as well as numerical problems.
NCERT solutions for class 12 physics chapter-wise
Gauss’ law of electricity:
Gauss’s law of magnetism:
Faraday’s laws of Electromagnetic induction:
Since emf can be defined as the line integral of the electric field, the above relation can be expressed as
Ampere-Maxwell’s Circuital law:
Where
Gamma Rays:
Wavelength range: 1 x 10¹⁴ to 1 x 10⁻¹⁰m
Frequency range: 3 x 10²² to 3 x 10¹⁸Hz
Production: By transition of atomic nuclei and decay of certain elementary particles.
X- rays:
Wavelength range: 1 x 10⁻¹¹ to 3 x 10⁻⁸m
Frequency range: 3 x 10¹⁹ to 1 x 10¹⁶Hz
Production: By sudden deceleration of high-speed electrons at a high-atomic number target, and also by the electronic transition among the innermost orbits of atoms.
Ultraviolet- Rays:
Wavelength range: 1 x 10⁻⁸ to 3 x 10⁻⁸m
Frequency range: 3 x 10¹⁶ to 1 x 10¹⁴Hz
Production: By sun, arc, vacuum spark, and ionized gases.
Visible Light:
Wavelength range: 1 x 10⁻⁷ to 1 x 10⁻⁷m
Frequency range: 7 x 10¹⁴ to 4 x 10¹⁴Hz
Production: Radiated by excited atoms in gases and incandescent bodies.
Infrared Radiation
Wavelength range: 8 x 10⁻⁷ to 5 x 10⁻³m
Frequency range: 4x 10¹⁴ to 6 x 10¹⁰ Hz
Production: From hot bodies, and by rotational and vibrational transitions in molecules
Radio waves
Wavelength range: 1 x 10⁻¹ to 1 x 10⁴m
Frequency range: 3 x 10⁹ to 3 x 10⁴ Hz
Production: By oscillating electric circuits.
Also, check
Electromagnetic Wave Class 12 Physics NCERT: Topics
The following are the main headings covered in the Chapter 8 Physics Class 12-
For the CBSE board exam, 4% of questions are expected from chapter 8 physics class 12.
NCERT solutions for class 12 will give a clear idea about how to use the formulas studied in the NCERT chapter 8 Physics Class 12.
For exams like NEET and JEE Main one or two questions are asked from the Electromagnetic Waves NCERT.
Questions based on the wave equation are expected from the chapter for the CBSE board exam. The relation between electric field, magnetic field and speed of light is also important.
The electromagnetic Waves Class 12 pdf download button is available for the ease of students to prepare offline using Electromagnetic Waves Class 12 NCERT solutions.
Comprehensive Coverage: These ncert solutions electromagnetic waves comprehensively cover all the topics and questions presented in the Class 12 Physics chapter on electromagnetic waves.
Detailed Explanations: Each class 12 chapter 8 physics ncert solutions provides detailed step-by-step explanations, simplifying complex concepts for students.
Clarity and Simplicity: The solutions are presented in clear and simple language, ensuring ease of understanding.
Exam Preparation: These solutions are crucial for board exam preparation and provide valuable support for competitive exams like JEE.
Balanced Complexity: The chapter balances complexity with scoring potential, making it accessible for students aiming to score well.
Excluded Content:
Certain portions have been omitted, which include:
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.
Admit Card Date:04 October,2024 - 29 November,2024
Admit Card Date:04 October,2024 - 29 November,2024
Application Date:07 October,2024 - 22 November,2024
Application Correction Date:08 October,2024 - 27 November,2024
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
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hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
Hello Akash,
If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.
You can get the Previous Year Questions (PYQs) on the official website of the respective board.
I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.
Thank you and wishing you all the best for your bright future.
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