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.
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.
(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.
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.
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:
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.
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.
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.
