# NCERT 12 Physics Electromagnetic Waves Chapter 8 Exercise

Q.1. shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 mm. 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 change of potential difference between the plates.

(b) Obtain the displacement current across the plates.

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

Given ∊0 = 8.85 ⨯10-12 C2 N-1 m-2  now(a) We know that capacity of a parallel plate capacitor is given by (c) Yes, Kirchhoff’s first rule is valid at each plate of the capacitor provided wo take the current to be the sum of the conduction and displacement currents.

Q.2. A parallel plate capacitor made of circular plates each of radius R = 6. Cm. has a capacitance C = 100  F. The capacitor is connected to a 230 V a. c. supply with an angular frequency of 300 rad. S-1.

(a) What is the r. m. s. value of the conduction current ?

(b) Is the conduction current equal to the displacement current ?

(c) Determine the magnitude of B at a point 3.0 cm from the axis between the plates.    Q.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 ?
Sol. The speed in vacuum is same for all the given wavelengths, which is 3 ⨯ 108 ms-1
Q.4. A plane electromagnetic wave travels in vacuum along Z- direction. What can you say about the direction of its electric and magnetic field vectors ? If the frequency of the wave is 30 MHz, what is its wavelength ?

Sol. In electromagnetic wave, the electric field vector  and magnetic field vector show their variations perpendicular to the direction of propagation o wave as well as perpendicular to each other. As the electromagnetic wave is travelling along Z- direction, hence  show their variation in x-y plane. Q.5. A radio can tune to any station the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band ? Q.6. A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of electromagnetic waves produced by the oscillator ?
Sol. The frequency of electromagnetic wave is the same as that of oscillating charged particle about its equilibrium position ; which is 109 Hz.
Q.7. The amplitude of the magnetic fild part of a harmonic electromagnetic wave in vacuum is B0 = 510 n T. What is the amplitude of the electric field part of the wave ?
Sol. Here,  B0 = 510 n T = 510 ⨯ 10-9 T, E0 = c B0 = 3 ⨯ 108 ⨯510⨯10-9 = 153 NC-1

Q.8. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and its frequency is  v = 50.0 MHz. (a) Determine B0, , k and λ, (b) Find expressions for  and . Q.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 e. m. spectrum. In what why are the different scales of photon energies that you obtain related to the sources of electromagnetic tradition. Where h = 6.6 × 10-34 j. The energy of photon of different parts of electromagnetic spectrum in joules and eV are shown in table below, along with their sources of origin.

Q.10. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 ⨯ 1010 Hz and amplitude 48 Vm-1. (a) What is the wavelength of the wave ? (b) What is the amplitude of oscillating magnetic field ?  (c) Show that the total average energy density of the electric field equals the average energy density of magnetic field. (c = 3⨯ 108 m/s). (c) For average energy density, see point 10.

Q.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 ⨯ 108 rad/s)t]

(a) What is the direction of motion ?

(b) What is the wavelength λ ?

(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.

Sol. (a) From the given equation, it clear that the direction of motion of e. m. wave is along negative y direction I, e., along

(b) Comparing the given equation with the equation

E = E0 cos(ky+ wt), we have Q.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 emitted isotropiclly and neglect reflection. Q.13. Use the formula  K to obtain the characteristic temperature ranges for different parts of the e. m. spectrum. What do the numbers that you obtain tell you ?

Sol. We know, every body at a given temperature T, emits radiations of all wavelengths in certain rabge. For a black body, the wavelength corresponding to maximum intensity of radiation at a given temperature T is given, according to Wein’s law, by the relation  : Temperature for other wavelengths can be similarly found. These numbers tell us the temperature ranges required for obtaining radiation in different parts of the e. m. spectrum. Thus to obtain It is to be noted that, a body at lower temperature will also produce this wavelength but not with maximum intensity.

Q.14. Given below are some famous numbers associated with electromagnetic radiation in different contexts in Physics. State the part of the e. m. 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) 5896 Å – (double lines of sodium).

(e) 14.4 keV (energy of a particular transition in Fe57 nucleus associated with a famous high resolution spectroscopic method Mossbauer spectroscopy)].

Sol. (a) This wavelength corresponds to radio waves (short wavelength or high frequency end).

(b) This frequency also corresponds to radio waves (short wavelength or high frequency end). This wavelength corresponds to microwave region of the electromagnetic spectrum.

(d) This wavelength lies in the visible region (yellow) of the electromagnetic spectrum. This frequency lies in the X-ray region of the electromagnetic spectrum.

Q.15. Answer the following questions :

(b) It is necessary to use satellites for long distance T.V. 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 atmosphere is crucial for human survival. Why ?

(e) If the earth did not have atmosphere, would its average surface temperature be higher of 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 earth. What might be the basis of this prediction.

Sol. (a) It is so because ionosphere reflects the waves in these bands.

(b) Yes, television signals being of high frequency are not reflected by the ionosphere. Therefore, to reflect them satellites are needed. That is why, satellites are used for long distance T. V.  transmission.

(c) Optical and radio waves can penetrate the atmosphere whereas X- rays being of much smaller wavelength are absorbed by the atmosphere. That is why we can work with optical and radio telescopes on earth’s surface but X-ray astronomical telescopes much be used on the satellite orbiting above the earth’s atmosphere.

(d) The small ozone layer present on the top of the stratosphere absorbs most of the ultraviolet radiations form the sun which are dangerous and cause genetic damage to the living cells, prevents them form reaching the earth’ surface and earth helps in the survival of the life.

(e) The temperature of the earth would be lover because the Green House effect to the atmosphere would be absent.

(f) The clouds by a global nuclear war would perhaps cover most parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘nuclear winter’.

Updated: October 14, 2020 — 10:47 am