Hello Students. Are you Searching for NCERT Solutions for Class 12 Physics Chapter 8? If yes then you are most welcome to NCERTian. Here we have provided you with the complete Question and Answers of Chapter 8: Electromagnetic Waves. These solutions are written by expert teachers and faculties keeping the need of students in mind.
| Chapter | 8. Electromagnetic Waves |
| Subject | Physics |
| Textbook | Physics I |
| Class | Twelve |
| Category | NCERT Solutions for Class 12 |
Class 12 Physics NCERT Solutions are the excellent practice set for students to ace Physics with better understandings and good marks. With the NCERT Solution for Class 12 Physics, students can confidently answer their exam questions and enjoy the best possible marks in the subject. These solutions are prepared keeping in view the latest syllabus and curriculum. In addition, the solutions are designed keeping in mind the requirements of students. Therefore, students can use these solutions for revision or to score high marks in the final examinations. Let’s get started with the solutions of Chapter 8: Electromagnetic Waves.
NCERT Solutions for Class 12 Physics Chapter 8
Electromagnetic Waves Solutions
Exercise
Q 8.1) Two circular plates having radius of 12 cm each and separated by 5 cm are used to make a capacitor as shown in the Figure 8.6.
An external source charges this capacitor. 0.15 A is the charging current which remains constant.
(a) Determine the capacitance and the rate of charge of potential difference between the two capacitiveplates.
(b) Calculate the displacement current across the capacitive plates.
(c) Kirchhoff’s first rule (junction rule) is applicable to each plate of the capacitor. Yes or No. Give Reasons.
Answer)
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
Answer a) The capacitance between the two plates can be calculated as follows:
where,
A = Area of each plate = 
= 8.0032 × 10−12F
= 80.032 pF
The charge on each plate is given by,
q = CV
where,
V is the 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.
Answer b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, id is 0.15 A.
Answer 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.
Q 8.2) A parallel plate capacitor (Fig. 8.7) 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 a point 3.0 cm from the axis between the plates.
Answer)
Radius of each circular plate, R = 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C = 100 pF = 100 × 10−12 F
Supply voltage, V = 230 V
Angular frequency, ω = 300 rad s−1
Answer a)
Rms value of conduction current, I = 
Where,
XC = Capacitive reactance
∴ I = V × ωC
= 230 × 300 × 100 × 10−12
= 6.9 × 10−6 A
= 6.9 μA
Hence, the rms value of conduction current is 6.9 μA.
Answer b) Yes, conduction current is equal to displacement current.
Answer c) Magnetic field is given as:
Where,
μ0 = Free space permeability = 4 π × 10−7NA−2
I0 = Maximum value of current = 
r = Distance between the plates from the axis = 3.0 cm = 0.03 m
= 1.63 × 10−11 T
Hence, the magnetic field at that point is 1.63 × 10−11 T.
Q 8.3) 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?
Answer) 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 8.4) 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?
Answer)
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. 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 8.5) A radio can tune in to any station in the 7.5 MHz to 12 MHz band.
What is the corresponding wavelength band?
Answer)
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 8.6) 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?
Answer) 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 8.7) 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?
Answer)
Amplitude of magnetic field of an electromagnetic wave in a vacuum,
B0=510nT=510×10−9T
Speed of light in vacuum, c = 3×108m/s
Amplitude of electric field of an electromagnetic wave is given by the relation,
E=cB0=3×108×510×10−9=153N/C
Therefore, the electric field part of the wave is 153 N/C.
Q 8.8) Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.0 MHz.
(a) Determine, B0, ω, k, and λ. (b) Find expressions for E and B.
Answer)
Electric field amplitude, E0 = 120 N/C
Frequency of source, ν = 50.0 MHz = 50 × 106 Hz
Speed of light, c = 3 × 108 m/s
Answer a)
Magnitude of magnetic field strength is given as:
Angular frequency of source is given by:
ω = 2πν
= 2π × 50 × 106
= 3.14 × 108 rad/s
Propagation constant is given as:
Wavelength of wave is given as:
Answer b) Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular.
Equation of electric field vector is given as:
And, magnetic field vector is given as:
Q 8.9) The terminology of different parts of the electromagnetic spectrum is given in the text.
Use the formula E = hν (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?
Answer)
Energy of a photon is given as:
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
If the wavelength λ is in metre and the energy is in Joule, then by dividing E by 1.6 × 10-19 will convert the energy into eV.
(a) For Gamma rays, the wavelength ranges from 10-10 to 10-14 m, therefore the photon energy can be calculated as follows:
(b) The wavelength for X-rays ranges between 10-8 m to 10-13 m
For λ = 10-8,
For λ = 10-13 m, energy = 107 eV
(c) For ultraviolet radiation, the wavelength ranges from 4 × 10-7 m to 6 × 10-7 m.
For 4 × 10-7 m,
For 6 × 10-7 m, the energy is equal to 103 eV.
The energy of the ultraviolet radiation varies between 1010 to 103 eV.
(d) For visible light, the wavelength ranges from 4 × 10-7 m to 7 × 10-7 m.
For 4 × 10-7, the energy is the same as above, that is 1010 eV
For 7 × 10-7 m, the energy is 100 eV
(e) For infrared radiation, the wavelength ranges between 7 × 10-7 m to 7 × 10-14 m.
The energy for 7 × 10-7 m is 100 eV
The energy for 7 × 10-14 m is 10-3 eV
(f) For microwaves, the wavelength ranges from 1 mm to 0.3 m.
For 1 mm, the energy is 10-3 eV.
For 0.3 m, the energy is 10-6 eV.
(g) For radio waves, the wavelength ranges from 1 m to few km.
For 1 m, the energy is 10-6 eV.
The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source
Q 8.10) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 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 × 108 m s−1.]
Answer)
Frequency of the electromagnetic wave, ν = 2.0 × 1010 Hz
Electric field amplitude, E0 = 48 V m−1
Speed of light, c = 3 × 108 m/s
Answer a) Wavelength of a wave is given as:
Answer b) Magnetic field strength is given as:
Answer c) Energy density of the electric field is given as:
And, energy density of the magnetic field is given as:
Where,
∈0 = Permittivity of free space
μ0 = Permeability of free space
We have the relation connecting E and B as:
E = cB … (1)
Where,
Putting equation (2) in equation (1), we get
Squaring both sides, we get
Q 8.11) 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 × 106 rad/s)t} ˆi .
(a) What is the direction of propagation?
(b) What is the wavelength λ?
(c) What is the frequency ν?
(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.
Answer a) The direction of motion is along the negative y-direction. i.e., along -j.
Answer b) The given equation is compared with the equation,
E = E0 cos (ky + ωt)
⇒ k = 1.8 rad/s
ω = 5.4 x 106 rad/s
λ = 2π/k = (2 x 3.14)/1.8 = 3.492 m
Answer c) Frequency, ν = ω/2π = 5.4 x 106/(2 x 3.14) = 0.859 x 106 Hz
Answer d) Amplitude of the magnetic field, B0 = E0/c
= 3.1/(3 x 108) = 1.03 x 10-8 T= 10.3 x 10-9 T= 10.3 nT
Answer e) Bz = B0 cos (ky + ωt)ˆk ={(10.3 nT) cos[(1.8 rad/m)y + (5.4 × 106 rad/s)t]} kˆ
Q 8.12) 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 1m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
Answer)
Power rating of bulb, P = 100 W
It is given that about 5% of its power is converted into visible radiation.
Power of visible radiation,
Hence, the power of visible radiation is 5W.
Answer a) Distance of a point from the bulb, d = 1 m
Hence, intensity of radiation at that point is given as:
Answer b)
Distance of a point from the bulb, d1 = 10 m
Hence, intensity of radiation at that point is given as:
Q 8.13) Use the formula λ m T = 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum.
What do the numbers that you obtain tell you?
Answer)
We have the equation, λ m T = 0.29 cm K
⇒ T = (0.29/λ m )cm K
Here, T is the temperature
λ m is the maximum wavelength of the wave
For λ m = 10-4 cm
T = (0.29/10-4)cm K = 2900 K
For the visible light, λ m = 5 x 10-5 cm
T = (0.29/ 5 x 10-5 )cm K ≈ 6000 K
Note: a lower temperature will also produce wavelength but not with maximum intensity.
Q 8.14) 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 Å – 5896 Å [double lines of sodium]
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus associated with a famous high-resolution spectroscopic method (Mössbauer spectroscopy)].
Answer a) Radio waves (short-wavelength end)
Answer b) Radio waves (short-wavelength end)
Answer c) Microwave
Answer d) Visible light (Yellow)
Answer e) X-rays (or soft γ-rays) region
Q 8.15) 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?
Answer a) Ionosphere reflects waves in the shortwave bands.
Answer b) Television signals have high frequency and high energy. Therefore, it is not properly reflected by the ionosphere. Satellites are used to reflect the TV signals.
Answer c) Atmosphere absorbs X-rays, while visible and radio waves can penetrate it.
Answer d) Ozone layer absorbs the ultraviolet radiations from the sunlight and prevents it from reaching the surface of the earth and causing damage to life.
Answer e) If the atmosphere is not present, there would be no greenhouse effect. As a result, the temperature of the earth would decrease.
Answer f) The smoke clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’.
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