Class 12 Physics Chapter 8: Electromagnetic Waves - NCERT Solutions

This page focuses on the complete NCERT solutions for Class 12 Physics Chapter 8: Electromagnetic Waves. This section provides detailed, easy-to-understand solutions for all the questions from this chapter. These Electromagnetic Waves question answers will offer you valuable insights and explanations.

This eighth chapter in Class 12 Physics is inclusive of three different topics. All of these three topics; displacement current, electromagnetic waves, and the electromagnetic spectrum are interrelated and based on the common concept of magnetism. The chapter comprises the different electromagnetic waves which include microwaves, x-rays, radio waves and also their usage in practical life, Maxwell’s Equation, the frequency of the different electromagnetic waves and their wavelengths, etc.

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Exercise 1

Q1	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.
Ans:	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 = $πr squared$ 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. (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.

Q3	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?
Ans:	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.

Q4	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?
Ans:	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:

Q5	A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Ans:	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.

Q6	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?
Ans:	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.

Q7	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?
Ans:	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.

Q15	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?
Ans:	(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.