NCERT Solutions for Class 12 Physics Chapter 8 - Electromagnetic Waves
NCERT Solutions for Class 12 Physics Chapter 8 Free PDF Download
Please Click on Free PDF Download link to Download the NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves
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 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 charge 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.
Sol. Given, radius of plates r = 12 cm = 12 × 10–2m
Separation between circular plates
d = 5 cm = 5 × 10–2m
Current I = 0.15A
(a) Capacitance
$$\text{C}=\frac{\epsilon _{0}\text{A}}{d}\\\text{where, A is the area of plates.}\\\text{C}=\frac{8.854×10^{\normalsize-12}×3.14(12×10^{2})^{2}}{5×10^{\normalsize-2}}\\\text{C}=\frac{8.854×3.14×144×10^{\normalsize-12-4+2}}{5}$$
C = 800.6 × 10–14F = 8.01pF
Charge on the plates of the capacitor
q = CV
$$\frac{dq}{dt}=\text{C}.\frac{dV}{dt}\\\text{I}=\text{C}.\frac{dV}{dt}\qquad\bigg[\because\frac{dq}{dt}=1\bigg]\\\frac{\text{dV}}{\text{dt}}=\frac{1}{\text{C}}=\frac{0.15}{8.01×10^{\normalsize-12}}=18.7×10^{9}\text{V/S}$$
Thus, the rate of change of potential
is 18.7 × 109 V/s.
(b) The displacement current is equal to the conduction current Id = 0.15A.
(c) Yes, Kirchhoff’s first rule is valid because we take the current to be the sum of conduction currents and the displacement currents.
2. A parallel plate capacitor (shown in figure) 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.
(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.
Sol. Given radius of plates R = 6 cm = 6 × 10–2 m
Capacitance
C= 100 pF = 100 × 10–12 F = 10–10F
Voltage of capacitor V = 230V
Frequency, w = 300 rad/s
(a) The rms value of current
$$\text{I}_{rms}=\frac{\text{V}_{rms}}{\text{X}_{c}}\\\therefore\space\text{X}_{c}=\frac{1}{\omega C}=\frac{1}{300×10^{\normalsize-10}}=\frac{10^{10}}{300}Ω\\\therefore\space\text{I}_{rms}=\frac{230×300}{10^{10}}=3×23×1000×10^{\normalsize-10}$$
= 69 × 10–7 = 6.9 × 10–6A
Irms = 6.9µA
(b) Yes, the conduction current is same as displacement current
$$\text{I}_{d}=\epsilon_{0}\frac{d\phi_{E}}{dt}\\\text{(By the definition of displacement current)}\\\text{I}_{d}=\epsilon_{0}\frac{d}{dt}\text{(EA)}\space(\phi_{E}=\text{EA})\\\text{I}_{d}=\epsilon_{0}\text{A}\frac{\text{dE}}{\text{dt}}\space\bigg(\text{E}=\frac{\sigma}{\epsilon_{0}}=\frac{\text{Q}}{\epsilon_{0}\text{A}}\bigg)\\\text{I}_{d}=\epsilon_{0}\text{A}\frac{d}{dt}\bigg(\frac{Q}{\epsilon_{0}A}\bigg)\\\text{I}_{d}=\epsilon_{0}\text{A}.\frac{1}{\epsilon_{0}A}.\frac{dQ}{dt}=\frac{dQ}{dt}=\text{I}$$
Id = I
(c) We have the distance of point from the axis between the plates
r = 3 cm = 3 × 10–2m
Radius of plates R = 6 cm = 6 × 10–2m
The magnetic field at a point between the plates
$$\text{B}=\frac{\mu_{0}}{2\pi R^{2}}.r\text{I}_{d}\\\text{B}=\frac{\mu_{0}r}{2\pi R^{2}}.\text{I}\space(I_{d}=1)\\\text{If I = I}_0, \text{maximum value of current}\\\text{then}\space\text{I}=\sqrt{2}\text{I}_{rms}$$
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. Here, X-rays, red light and radiowaves all are known as the electromagnetic waves and all the electromagnetic waves travel with the same speed as speed of light c. Thus, the speed is same for X-rays, red light and radiowaves.
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?
Sol. The direction of electromagnetic wave is perpendicular to both electric and magnetic fields. As electromagnetic wave is travelling in Z-direction, then electric and magnetic fields are in X-Y direction and are perpendicular to each other. f = 30
MHz = 30 × 106 Hz.
Speed, c = 3 × 108 m/s
We know, c = fλ
Wavelength of e-m waves
$$\lambda=\frac{c}{f}=\frac{3×10^{8}}{30×10^{6}}=\frac{300}{30}=10\text{m}$$
Thus, the wavelength of electromagnetic waves is 10m.
5. A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Sol. Given, frequency f1= 7.5 MHz
Frequency f2 = 12MHz
Speed of e-m wave c = 3 × 108 m/s
Wavelength corresponding to frequency f1
$$\lambda_{1}=\frac{c}{f_{1}}=\frac{3×10^{8}}{7.5×10^{8}}=\frac{300}{7.5}=40\text{m}$$
Wavelength corresponding to frequency f2
$$\lambda_{2}=\frac{c}{f_{2}}=\frac{3×10^{8}}{12×10^{6}}=\frac{300}{12}=25\text{m}$$
Thus, the corresponding wavelength band is 25m to 40m.
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. According to the question,
Frequency of the electromagnetic waves = 109 Hz.
The frequency of electromagnetic waves produced by the oscillator is equal to the frequency of the oscillating charged particle (in equilibrium position).
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?
Sol. Given, magnetic field part of harmonic electromagnetic wave
B0 = 510 nT
$$\text{Speed of light in vacuum c =}=\frac{\text{E}_{0}}{\text{B}_{0}}$$
where, E0 is the electric part of the wave
$$3×10^{8}=\frac{\text{E}_{0}}{510×10^{\normalsize-9}}$$
or E0 = 153 N/C
Thus, the amplitude of the electric field part of wave is 153 N/C.
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,w, k and λ. (b) Find expressions for E and B.
Sol. (a) Speed of light (in vacuum)
$$c=\frac{\text{E}_{0}}{\text{B}_{0}}\\\text{B}_{0}=\frac{\text{E}_{0}}{c}=\frac{120}{3×10^{8}}=40×10^{\normalsize-8}$$
or B0 = 400 × 10–9T = 400 nT
Angular frequency
w = 2πf = 2 × 3.14 × 50 × 106
w = 3.14 × 108 rad/s
Wave number of e-m waves
$$\text{K}=\frac{\omega}{c}=\frac{3.14×10^{8}}{3×10^{8}}=1.05\space\text{rad/m}$$
$$\lambda=\frac{c}{f}=\frac{3×10^{8}}{50×10^{6}}=6.00\text{m}$$
(b) The electric field E= E0 sin (kx - wt)
E = 120 sin (1.05 x – 3.14 × 108 t)
The magnetic field B
B = B0 sin (kx – wt)
B = 4 × 10–7 sin (1.05x – 3.14 × 108t)
9. The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (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?
Sol. Given, energy of photon E = hγ
For γ-rays
Frequency of γ-rays n = 3 × 1020 Hz
Energy E = hv = 6.6 × 10–34 × 3 × 1020 = 19.8 ×10–14 J
$$\text{or}\space\text{E}=\frac{19.8×10^{\normalsize-14}}{1.6×10^{\normalsize-19}}=1.24×10^{6}\space\text{eV}$$
The source of g-rays is nuclear origin.
For X-rays
Frequency of X-rays v = 3 × 1018 Hz
Energy of X-rays E = hv = 6.6 × 10–34 × 3 × 1018
= 19.8 × 10–16J
$$\text{or}\space\text{E}=\frac{19.8×10^{-16}}{1.6×10^{-19}}=1.24×10^{4}\text{eV}$$
The retardation of high energy electron produces X-rays.
For ultraviolet rays
Frequency n = 1015 Hz
Energy E = hv = 6.6 × 10–34 × 1015 = 6.6 × 10–19J
$$\text{or}\space\text{E}=\frac{6.6×10^{\normalsize-19}}{1.6×10^{-19}}=4.125\space\text{eV}$$
It originates by the excitation of atoms.
For visible rays
Frequency v = 6 × 1014 Hz
Energy E = hv = 6.6 × 10–34 × 6 × 1014 = 39.6 × 10–20J
$$\text{or}\space\text{E}=\frac{39.6×10^{\normalsize-20}}{1.6×10^{\normalsize-19}}=2.475\space\text{eV}$$
The visible rays produce by the excitation of valance electrons.
For infrared rays
Frequency rays v = 1013 Hz
Energy E = hv = 6.6 × 10–34 × 1013 = 6.6 × 10–21J
$$\text{or}\space\text{E}=\frac{6.6×10^{\normalsize-21}}{1.6×10^{\normalsize-19}}=4.125×10^{\normalsize-2}\space\text{eV}$$
They originate by the excitation of atoms and molecules.
For microwaves
Frequency v = 1010 Hz
Energy of microwaves E = hv = 6.6 × 10–34 × 1010 = 6.6 × 10–24J
$$\text{or}\space\text{E}=\frac{6.6×10^{\normalsize-24}}{1.6×10^{\normalsize-19}}=4.125×10^{\normalsize-5}\text{eV}$$
They originate by the oscillating current in vacuum tubes.
For radiowaves
Frequency v = 3 × 108 Hz
Energy E = hv = 6.6 × 10–34 × 3 × 108 = 19.8 × 10–26J
$$\text{or}\space\text{E}=\frac{19.8×10^{\normalsize-26}}{1.6×10^{\normalsize-19}}=4.125×10^{\normalsize-5}\space\text{eV}$$
They produced by oscillating current.
| Types of radiation | Photon energy |
| γ-rays | 1.24 × 106 eV |
| X-rays | 1.24 × 104 eV |
| Ultraviolet rays | 4.12 eV |
| Visible waves | 2.475 eV |
| infrared waves | 4.125 × 10–2 eV |
| Microwaves | 4.125 × 10–5 eV |
| Radiowaves | 1.24 × 10–6 eV |
10. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V/m.
(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]
Sol. Frequency = 2 × 1010 Hz
Given, c = 3 × 106 m/s
Electric field amplitude E0 = 48 V/m
$$\text{(a) Wavelength} \space\lambda =\frac{c}{f}=\frac{3×10^{8}}{2×10^{10}}$$
= 1.5 × 10–2 m
$$\text{(b) Using formula, c =}\frac{\text{E}_{0}}{\text{B}_{0}}$$
The oscillating magnetic field amplitude
$$\text{B}_{0}=\frac{\text{E}_{0}}{c}=\frac{48}{3×10^{8}}=1.6×10^{\normalsize-7}\text{T}$$
(c) The average energy density of electric field
$$\text{u}_{\text{E}}=\frac{1}{4}\epsilon_{0}\text{E}_{0}^{2}\qquad...\text{(i)}\\\text{We know that}\space\frac{\text{E}_{0}}{\text{B}_{0}}=c\\\text{Putting in Eq. (i),}\\\therefore\space u_{\text{E}}=\frac{1}{4}\epsilon_{0}c^{2}\text{B}_{0}^{2}\qquad...(\text{ii})$$
Speed of electromagnetic waves,
$$c=\frac{1}{\sqrt{\mu}_{0}\epsilon_{0}}\\\text{Putting in Eq. (ii), we get}\\\mu_{\text{E}}=\frac{1}{4}.\epsilon_{0}\text{B}_{0}^{2}.\frac{1}{\mu_{0}\epsilon_{0}}\\\text{u}_{E}=\frac{1}{4}.\frac{\text{B}_{0}^{2}}{\mu_{0}}\\=\frac{\text{B}_{0}^{2}}{2\mu_{0}}=\mu_{0}$$
Thus, the average energy density of the E field equals the average energy density of B field.
Physics Most Likely Question Bank
CBSE Class 12 for 2025 Exam
Share page on
