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NCERT Class 12 Physics Chapter 8 Electromagnetic Waves
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Electromagnetic Waves
Chapter: 8
| Part – I |
EXERCISE
1. Figure 8.5 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 0.15A.
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
Ans: C = Σ0 × A/d
(Σ0 = 8.854 × 10−12)
r = 12 cm = 0.12 m
d = 5.0 cm = 0.05 m.
A = 𝜋r2 = 𝜋 (0.12)2 = 𝜋 × 0.0144
= 0.0454m2
C = Σ0A/ d = (8.854 × 10-12 × 0.0454) / 0.05
= 4.021 × 10-13 / 0.05
= 8.42 × 10-12 F
= 8.04 pF.
q = CV
Ⅰ = dq / dt = e × dV / dt
= dV/dt
= 1/c = 0.15 / 80.1 × 10-12
= 1.875 × 109 Vs-1.
(b) Obtain the displacement current across the plates.
Ans: In a capacitor, the displacement current is equal to the conduction current.
Id = I
Id = 0.15 A.
(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
Ans: As the sum continuous, so is Kirchhoff’s first law at the capacitor plate if the current in the law is the sum of conduction and displacement currents.
2. A parallel plate capacitor (Fig. 8.6) 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?
Ans: Given:
R = 6.0 cm = 0.06 m
C = 100 pF = 100 × 10-12 F
V = 230 V
ω = 300 rad/s
r = 3.0 cm = 0.03 m
Here:
Ⅰrms = ωCVrms
= 300 × 100 × 10-12 × 230
= 6.9 × 10-6 A
= 6.9 μA.
(b) Is the conduction current equal to the displacement current?
Ans: Yes, the conduction current is equal to the displacement current even if Ⅰ is oscillating in time.
(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Ans: B = μ0Ⅰrmsr / 2𝜋 × R2
We know that μ0 = (4𝜋 × 10-7)
B = 4𝜋 × 10-7 × 6.9 × 10-6 × 0.03 / 2𝜋 × (0.06)2
= 1.15 × 10-11 T
= 11.5 pT.
3. What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radio waves of wavelength 500m?
Ans: The wave speed in vacuum is the same for all radiations: C = 3 × 108 ms-1.
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?
Ans: Given:
𝑓 = 30 MHz = 30 × 106 Hz.
c = 3 × 108 m/s
λ = c/𝑓
= (3 × 108 m/s) / 30 × 106 Hz
= 10 m.
5. A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Ans: Given:
𝑓 = 7.5 MHz = 7.5 × 106 Hz.
c = 3 × 108 m/s
λ = c/𝑓
Wavelength for 7.5 MHz:
= (3 × 108 m/s) / (7.5 × 106 Hz) = 40m.
Wavelength for 12 MHz:
= (3 × 108 m/s) / (12 × 106 Hz) = 25 m.
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?
Ans: The frequency of the electromagnetic waves produced by the oscillator is 109 Hz.
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?
Ans: Given:
E0 = ?
c = 3 × 108 m/s
B0 = 510 nT = 510 × 10-9 T
E0 = c × B0
= 3 × 108 m/s × 510 × 10-9 T
= 3 × 510 × 10-1 V/m
= 153 V/m.
8. Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz.
(a) Determine, B0 ,ω, k, and l.
Ans: Given:
E0 = 120 N/C
𝑓 = 50.0 MHz = 50.0 × 106 Hz.
Determine B0, ω, k and λ.
ω = 2𝜋𝑓
= 2𝜋 × 50.0 × 106
= 3.14 × 108 rad/s.
Wavelength λ:
(c = 3.00 × 108 m/s)
λ = c/f
= 3.00 × 108 / 50.0 × 106
= 6.00 m.
Calculate k:
k = 2𝜋 / λ
= 2𝜋 / 6.00
= 1.05 rad/m.
Magnetic field:
E0 = cB0
B0 = E0/c
= 120 / 3.00 × 108
= 4.00 × 10-7 T.
(b) Find expressions for E and B.
Ans:
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?
Ans: Given:
E = hv
= (h = 6.626 × 10-19 J)
E(eV) = E(J) / 1.602 × 10-19.
(i) Radio waves:
Frequency: 1 GHz = 1 × 109 Hz.
E = hv = 6.626 × 10-34 × 1 × 109 J
= 6.626 × 10-25 J.
E(eV) = 6.626 × 10-25 / 1.602 × 10-19
= 4.14 × 10-6 eV.
(ii) Microwaves:
Frequency: 1 THz = 1 × 1012 Hz
E = hv = 6.626 × 10-34 × 1 × 1011 J
= 6.626 × 10-22 J.
E(eV) = 6.626 × 10-22 / 1.602 × 10-19
= 0.414 eV.
(iii) Infrared:
Frequency: 1 THz = 1 × 1012 Hz.
E = hv = 6.626 × 10-34 × 1 × 1012 J
= 6.626 × 10-22 J
E(eV) = 6.626 × 10-22 / 1.602 × 10-19
= 0.414 eV.
(iv) Visible Light:
Frequency: 5 Hz = 5 × 1015 Hz.
E = hv = 6.626 × 10-34 × 5 × 1014 J
= 3.313 × 10-19 J
E(eV) = 3.313 × 10-19 / 1.602 × 10-19
= 2.07 eV.
(v) Ultraviolet:
Frequency: 1 PHz = 1 × 1016 Hz
E = hv = 6.626 × 10-34 × 1 × 1016 J
= 6.626 × 10-8 J.
E(eV) = 6.626 × 10-18 / 1.602 × 10-19
= 41.4 eV.
(vi) X – rays:
Frequency: 1 EHz = 1 × 1018 Hz.
E = hv = 6.626 × 10-34 × 1 × 1018 J
= 6.626 × 10-16 J.
E(eV) = 6.626 × 10-16 / 1.602 × 10-19
= 4.14 KeV.
(vii) Gamma Rays:
1 ZHz = 1 × 1021 Hz.
E = hv = 6.626 × 10-34 × 1 × 1021 J
= 6.626 × 10-13 J.
E(eV) = 6.626 × 10-13 / 1.602 × 10-19
= 4.14 MeV.
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?
Ans: Given:
𝑓 = 2.0 × 1010 Hz
E0 = 48 V/m
c = 3 × 108 m/s
λ = c/𝑓
= 3 × 108 m/s / 2.0 × 1010 Hz.
= 0.015 m = 1.5 cm.
(b) What is the amplitude of the oscillating magnetic field?
Ans: B0 = E0 /c
= B0 = 48 V/m / 3 × 108 m/s
= 1.6 × 10-7 T.
(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.]
Ans:
