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# NCERT Solution for Class 12 Physics Chapter 8

Welcome to the NCERT Solutions for Class 12 Physics Chapter 8, provided by Swastik Classes. In this chapter, you will explore the topic of Electromagnetic Waves.

You will learn about the properties of electromagnetic waves, including their speed, wavelength, frequency, and energy. The chapter covers the various types of electromagnetic waves, such as radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

Additionally, the chapter discusses the electromagnetic spectrum and the applications of electromagnetic waves in various fields, such as communication, medical imaging, and spectroscopy. The chapter also covers the phenomenon of reflection, refraction, and polarization of electromagnetic waves.

Our NCERT Solutions for Class 12 Physics Chapter 8 provide step-by-step explanations and solutions to all the questions in the textbook. With our solutions, you can easily understand the concepts covered in the chapter and develop a deeper understanding of the subject.

## Answers of Physics NCERT solutions for class 12 Chapter 8 – Electromagnetic Waves

Page No 285:

Question 8.1:

Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.

(a) Calculate the capacitance and the rate of charge of potential difference between the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

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, = 8.85 × 10−12 C2 N−1 m−2

(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.

Page No 286:

Question 8.2:

A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius = 6.0 cm has a capacitance = 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.

Question 8.3:

What physical quantity is the same for X-rays of wavelength 10−10 m, red light of wavelength 6800 Å and radiowaves 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.

Question 8.4:

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 xy 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:

Question 8.5:

A radio can tune in to 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.

Question 8.6:

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.

Question 8.7:

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 = 510 nT = 510 × 10−9 T

Speed of light in a vacuum, c = 3 × 108 m/s

Amplitude of electric field of the electromagnetic wave is given by the relation,

E = cB0

= 3 × 10× 510 × 10−9 = 153 N/C

Therefore, the electric field part of the wave is 153 N/C.

Question 8.8:

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, B0ω, 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 × 10m/s

(a) Magnitude of magnetic field strength is given as:

Angular frequency of source is given as:

ω = 2πν

= 2π × 50 × 106

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:

Question 8.9:

The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula  (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

The given table lists the photon energies for different parts of an electromagnetic spectrum for differentλ.

The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.

Question 8.10:

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. [= 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,

… (2)

Putting equation (2) in equation (1), we get

Squaring both sides, we get

Page No 287:

Question 8.11:

Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) + (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 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,  = 3.490 m

(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:

Question 8.12:

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:

Question 8.13:

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;  and so on.

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.

Question 8.14:

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

(Mössbauer spectroscopy)].

(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

λis 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 =

Where,

= Planck’s constant = 6.6 × 10−34 Js

Energy, = 14.4 K eV

This corresponds to X-rays.

Question 8.15:

(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 form reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.

## NCERT Solutions for Class 12 Physics Chapter 8 – Electromagnetic Waves

As observed in Chapters 4 and 6, electricity and magnetism are intertwined. Faraday’s experiments proved that a relative motion between a magnet and an electric conductor induced an emf in the conductor. A physicist named James Clerk Maxwell (1831-1879) argued that the converse was also true. He suggested that if electromagnetic induction existed, then it was imperative that a varying electric field would generate a magnetic field.

Maxwell devised a series of equations using electric and magnetic fields, as well as the charge and current densities, as their origins. Maxwell’s equations are the name for these equations. In the NCERT Class 12 Physics Chapter 8, we will discuss the displacement current and electromagnetic waves. A broader spectrum of these waves will be covered later in Chapter 15 of the book.

### Topics to study in Class 12 Physics Chapter 8 Electromagnetic Waves

1. Introduction
2. Displacement Current
3. Electromagnetic Waves
1. Sources of electromagnetic waves
2. Nature of electromagnetic waves
4. Electromagnetic Spectrum
2. Microwaves
3. Infrared waves
4. Visible rays
5. Ultraviolet rays
6. X-rays
7. Gamma rays

### Why do Students prefer SWC’s NCERT Solutions for Class 12 Physics Chapter 7?

At SWC, students directly learn from IITians and experienced faculties. We provide quality study materials to prepare and follow a LEAP model. We teach not only for exams but for lifelong applications

## FAQs on Physics NCERT Class 12 Chapter 7

### What are the uses of electromagnetic waves?

Used for telecommunication, televisions, radio transmission, ultraviolet disinfecting techniques, and more.

### What are the advantages of electromagnetic waves?

Electromagnet waves cause no harm to the human tissue if absorbed and can refract and reflected to conveniently change the direction of transmission.

### How many electromagnetic waves are there?

There are seven different electromagnetic waves – radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays, and gamma rays.

## Summary of NCERT Solutions for Class 12 Physics Chapter 7

1- Because all electromagnetic waves travel through a vacuum at the same speed, the fundamental distinction between them is their wavelengths or frequencies. As a result, the waves’ modes of contact with matter varied significantly.

1. Electromagnetic waves are emitted by accelerated charged particles. The wavelength of an electromagnetic wave is frequently linked to the size of the system that emits it. Gamma radiation, with wavelengths ranging from 10–14 m to 10–15 m, is commonly emitted by an atomic nucleus. Heavy atoms release X-rays as a form of radiation. Electrons in a circuit are accelerated to produce radio waves. Waves with a wavelength of about the same size as the antenna can be transmitted most efficiently by a transmitting antenna. The wavelength of visible radiation emitted by atoms, on the other hand, is significantly longer than the atomic size.
1. An electromagnetic wave’s oscillating fields can accelerate charges and produce oscillating currents. As a result, an electromagnetic wave detection device is based on this fact. This is how Hertz’s original receiver functioned. Practically all modern receiving devices use the same basic idea. Other methods for detecting high-frequency electromagnetic waves are based on the physical effects they have when interacting with materials.
1. With frequencies lower than visible light, infrared waves vibrate not only electrons but entire atoms or molecules of a substance. This vibration raises the substance’s internal energy and, as a result, its temperature. Infrared waves are often referred to as heat waves because of this.
1. The center of our eyes’ sensitivity corresponds to the center of the sun’s wavelength dispersion. It’s because people have evolved visions that are most sensitive to the sun’s strongest wavelengths.

Conclusion

The NCERT Solutions for Class 12 Physics Chapter 8 provided by Swastik Classes offer a comprehensive understanding of the concepts of Electromagnetic Waves. The solutions provide step-by-step explanations of all the questions given in the textbook, making it easier for students to grasp the concepts and solve the problems with ease.

Our solutions cover all the important topics of the chapter, such as properties of electromagnetic waves, types of electromagnetic waves, electromagnetic spectrum, and their applications in various fields. Additionally, the solutions provide a detailed explanation of the phenomenon of reflection, refraction, and polarization of electromagnetic waves.

With these solutions, students can develop a better understanding of the subject and excel in their exams. The solutions are designed to help students gain a deeper understanding of the concepts covered in the chapter and provide them with the necessary tools to solve complex problems.

Overall, the NCERT Solutions for Class 12 Physics Chapter 8 are an excellent resource for students who want to develop a strong foundation in the subject of Electromagnetic Waves. With these solutions, students can improve their problem-solving skills and achieve academic success.