NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT Solutions for Class 12 Physics Chapter 8 - Electromagnetic Waves

Welcome to the Chapter 8 - Electromagnetic Waves, Class 12 Physics NCERT Solutions page. Here, we provide detailed question answers for Chapter 8 - Electromagnetic Waves. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Electromagnetic Waves and excel in their exams. By going through these Electromagnetic Waves question answers, you can strengthen your foundation and improve your performance in Class 12 Physics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Exercise 1

Q 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.

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,

Where, A = Area of each plate = $πr squared$

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, i_d is 0.15 A.

Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.

Q 15:

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

Q 3:

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

Q 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 x-yplane. 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:

Q 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.

Q 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.

Q 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 × 10⁸ × 510 × 10⁻⁹ = 153 N/C

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

Frequently Asked Questions about Electromagnetic Waves - Class 12 Physics

- 1. How many questions are covered in Electromagnetic Waves solutions?
- All questions from Electromagnetic Waves are covered with detailed step-by-step solutions including exercise questions, additional questions, and examples.
- 2. Are the solutions for Electromagnetic Waves helpful for exam preparation?
- Yes, the solutions provide comprehensive explanations that help students understand concepts clearly and prepare effectively for both board and competitive exams.
- 3. Can I find solutions to all exercises in Electromagnetic Waves?
- Yes, we provide solutions to all exercises, examples, and additional questions from Electromagnetic Waves with detailed explanations.
- 4. How do these solutions help in understanding Electromagnetic Waves concepts?
- Our solutions break down complex problems into simple steps, provide clear explanations, and include relevant examples to help students grasp the concepts easily.
- 5. Are there any tips for studying Electromagnetic Waves effectively?
- Yes, practice regularly, understand the concepts before memorizing, solve additional problems, and refer to our step-by-step solutions for better understanding.

Exam Preparation Tips for Electromagnetic Waves

The Electromagnetic Waves is an important chapter of 12 Physics. This chapter’s important topics like Electromagnetic Waves are often featured in board exams. Practicing the question answers from this chapter will help you rank high in your board exams.

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