NCERT Solutions for class 12th Physics Chapter 8 Electromagnetic Waves


Question 1. Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being changed by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.

(a) Calculate the capacitance and the rate of change 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.

Note: Conduction current coming into the left plate is same (in magnitude) as the displacement current leaving it.

Question 2. A parallel plate capacitor (Fig.) 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 point 3.0 cm from the axis between the plates.

Question 3. What physical quantity is the same for X-rays of wavelength 10-10m, red light of wavelength 6800 A and radio waves of wavelength 500m?

Sol. The speed in vacuum is same for all of them. It is given by c = 3 x 108 m/s.

Question 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?

Question 5. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?

Question 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?

Sol.109 Hz. The electromagnetic waves will have same frequency as that of the oscillating charged particle.

Question 7. The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = SlOnT. What is the amplitude of the electric field part of the wave?

Question 8. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is v = 50.0 MHz.

(a) Determine, B0,w, k, and A.
(b) Find expressions for E and B.

Question 9. The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (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?

Question 10. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 x 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 x 108 m s-1.]


Question 11. 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 x 106 rad/s)t]} i.

(a) What is the direction of propagation?
(b) What is the wavelength A?
(c) What is the frequency v?
(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.

Question 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 lm from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.

Question 13. Use the formula Am T = 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum.

Question 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 A – 5896 A [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high resolution spectroscopic method (Mossbauer spectroscopy)].

Sol. (a) Radio waves (short wavelength end)
(b) Radio waves (short wavelength end)
(c) Microwave
(d) Visible (yellow)
(e) X-rays (or softy-rays) region.

Question 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 radiotelescopes 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 bya severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?

Sol. (a) Ionosphere reflects waves in these bands.
(b) Television signals are not properly reflected by the ionosphere. Therefore, reflection is effected by satellites.
(c) Atmosphere absorbs X-rays, while visible and radio waves can penetrate it.
(d) It absorbs ultraviolet radiations from the sun and prevents it from reaching the earth’s surface and causing damage to life.
(e) The temperature of the earth would be lower because the Greenhouse effect of the atmosphere would be absent.
(f) The clouds produced by global nuclear war would perhaps cover sustantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’.

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