NCERT Solution for Class 12 Physics Chapter 8: Electromagnetic Waves deals with the generation and propagation of electromagnetic waves, which is a basic topic in the physic syllabus with the package of modern physics. This chapter is called 'Electromagnetic Waves,' offering a clear insight into how they are produced by accelerating charges and how such a wave travels through space. These are the equations that describe the behavior of electric and magnetic fields and their interaction in forming electromagnetic waves. The vast detail of explanations relating to the same can be found in the detailed pdf notes of Class 12 Physics Chapter 8. This chapter has been majorly emphasized in the deduction of the wave equation for electromagnetic waves and the examination of their properties like speed, wavelength, and frequency. The Class 12 Electromagnetic Waves chapter talks about the electromagnetic spectrum, ranging from radio waves and microwaves to infrared, visible light, ultraviolet, X-rays, and gamma rays. Students learn through solutions in Class 12 Physics Chapter 8 PDF about their practical applications in communication systems, medical imaging, and other technologies.
The NCERT Solution for Class 12 Physics Chapter 8: Electromagnetic Waves are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2025-2026. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Students can access the NCERT Solution for Class 12 Physics Chapter 8: Electromagnetic Waves. Curated by experts according to the CBSE syllabus for 2025–2026, these step-by-step solutions make Physics much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?
Amplitude of magnetic field of an electromagnetic wave in a vacuum,
B0=510nT=510×10−9T
Speed of light in vacuum, c = 3×108m/s
Amplitude of electric field of an electromagnetic wave is given by the relation,
E=cB0=3×108×510×10−9=153N/C
Therefore, the electric field part of the wave is 153 N/C.
A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109 Hz.
Two circular plates having a radius of 12 cm each and separated by 5 cm are used to make a capacitor as shown in Figure 8.6. An external source charges this capacitor. 0.15 A is the charging current which remains constant.
(a) Determine the capacitance and the rate of charge of potential difference between the two capacitor plates.
(b) Calculate the displacement current across the capacitive plates.
(c) Kirchhoff’s first rule (junction rule) is applicable to each plate of the capacitor. Yes or No. Give Reasons.
Radius of each circular plate, r = 12 cm = 0.12 m
Distance between the plates, d = 5 cm = 0.05 m
Charging current, I = 0.15 A
Permittivity of free space, 
(a) Capacitance between the two plates is given by the relation,
C
Where,
A = Area of each plate
Charge on each plate, q = CV
Where,
V = Potential difference across the plates
Differentiation on both sides with respect to time (t) gives:
Therefore, the change in potential difference between the plates is 1.87 ×109 V/s.
(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A.
(c) Yes
Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.
A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s−1.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10−12 F
Supply voltage, V = 230 V
Angular frequency, ω = 300 rad s−1
(a) Rms value of conduction current, I
Where,
XC = Capacitive reactance
∴ I = V × ωC
= 230 × 300 × 100 × 10−12
= 6.9 × 10−6 A
= 6.9 μA
Hence, the rms value of conduction current is 6.9 μA.
(b) Yes, conduction current is equal to displacement current.
(c) Magnetic field is given as:
B
Where,
μ0 = Free space permeability
I0 = Maximum value of current =
r = Distance between the plates from the axis = 3.0 cm = 0.03 m
∴B
= 1.63 × 10−11 T
Hence, the magnetic field at that point is 1.63 × 10−11 T.
What physical quantity is the same for X-rays of wavelength 10−10 m, red light of wavelength 6800 Å and radio waves of wavelength 500 m?
The speed of light (3 × 108 m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular.
Frequency of the wave, ν = 30 MHz = 30 × 106 s−1
Speed of light in a vacuum, c = 3 × 108 m/s
Wavelength of a wave is given as:
A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
A radio can tune to minimum frequency, ν1 = 7.5 MHz= 7.5 × 106 Hz
Maximum frequency, ν2 = 12 MHz = 12 × 106 Hz
Speed of light, c = 3 × 108 m/s
Corresponding wavelength for ν1 can be calculated as:
Corresponding wavelength for ν2 can be calculated as:
Thus, the wavelength band of the radio is 40 m to 25 m.
Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B0, ω, k, and λ. (b) Find expressions for E and B.
Electric field amplitude, E0 = 120 N/C
Frequency of source, ν = 50.0 MHz = 50 × 106 Hz
Speed of light, c = 3 × 108 m/s
(a) Magnitude of magnetic field strength is given as:
Angular frequency of source is given as:
ω = 2πν
= 2π × 50 × 106
= 3.14 × 108 rad/s
Propagation constant is given as:
Wavelength of wave is given as:
(b) Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular.
Equation of electric field vector is given as:
And, magnetic field vector is given as:
The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
Energy of a photon is given as:
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
λ = Wavelength of radiation
The given table lists the photon energies for different parts of an electromagnetic spectrum for differentλ.
|
λ (m) |
103 |
1 |
10−3 |
10−6 |
10−8 |
10−10 |
10−12 |
|
E (eV) |
12.375 × 10−10 |
12.375 × 10−7 |
12.375 × 10−4 |
12.375 × 10−1 |
12.375 × 101 |
12.375 × 103 |
12.375 × 105 |
The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m−1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s−1.]
Frequency of the electromagnetic wave, ν = 2.0 × 1010 Hz
Electric field amplitude, E0 = 48 V m−1
Speed of light, c = 3 × 108 m/s
(a) Wavelength of a wave is given as:
(b) Magnetic field strength is given as:
(c) Energy density of the electric field is given as:
And, energy density of the magnetic field is given as:
Where,
∈0 = Permittivity of free space
μ0 = Permeability of free space
We have the relation connecting E and B as:
E = cB … (1)
Where,
Putting equation (2) in equation (1), we get
Squaring both sides, we get
Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]}
(a) What is the direction of propagation?
(b) What is the wavelength λ?
(c) What is the frequency ν?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.
(a) From the given electric field vector, it can be inferred that the electric field is directed along the negative x direction. Hence, the direction of motion is along the negative y direction i.e., 
(b) It is given that,
The general equation for the electric field vector in the positive x direction can be written as:
On comparing equations (1) and (2), we get
Electric field amplitude, E0 = 3.1 N/C
Angular frequency, ω = 5.4 × 108 rad/s
Wave number, k = 1.8 rad/m
Wavelength,
(c) Frequency of wave is given as:
(d) Magnetic field strength is given as:
Where,
c = Speed of light = 3 × 108 m/s
(e) On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
Power rating of bulb, P = 100 W
It is given that about 5% of its power is converted into visible radiation.
Hence, the power of visible radiation is 5W.
(a) Distance of a point from the bulb, d = 1 m
Hence, intensity of radiation at that point is given as:
(b) Distance of a point from the bulb, d1 = 10 m
Hence, intensity of radiation at that point is given as:
Use the formula λm T= 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
A body at a particular temperature produces a continous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck’s law. It can be given by the relation,
Where,
λm = maximum wavelength
T = temperature
Thus, the temperature for different wavelengths can be obtained as:
For λm = 10−4 cm;
For λm = 5 ×10−5 cm;
For λm = 10−6 cm;
The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
(c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe].
(d) 5890 Å – 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method
(a) Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum.
(b) Radio waves; it belongs to the short wavelength end.
(c) Temperature, T = 2.7 °K
λm is given by Planck’s law as:
This wavelength corresponds to microwaves.
(d) This is the yellow light of the visible spectrum.
(e) Transition energy is given by the relation,
E = hν
Where,
h = Planck’s constant = 6.6 × 10−34 Js
ν = Frequency of radiation
Energy, E = 14.4 K eV
This corresponds to X-rays.
Answer the following questions:
(a) Long distance radio broadcasts use short-wave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?
(a) Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.
(b) It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions.
(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.
(d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.
(e) In theabsenceof an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.
(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light from reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.
The NCERT solution for Class 12 Chapter 8: Physics Electromagnetic Waves is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 12 Chapter 8: Physics Electromagnetic Waves is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 12 Physics Chapter 8 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 12 Physics Chapter 8: Physics Electromagnetic Waves it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 12 Physics Chapter 8 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.
