Chapter 8 Electromagnetic Waves

NCERT Solutions Class 12 Physics Chapter 8 - Electromagnetic Waves

The physics curriculum for class XII, chapter 8, titled "Electromagnetic Waves" is a culmination of all the topics learned prior. It is through electromagnetic waves that the link between electromagnetic theory and optical theory has been established due to the fact that electric and magnetic fields (that are continually changing) move through space in the form of waves. This chapter also begins with the resolution of a conflict within Ampere's Law, which led to the introduction of the "Displacement Current" concept, published by James Clerk Maxwell. This concept was central to providing complete representation of all classically defined electromagnetic events, in the seven equations. Understanding Chapter 8 of Class XII Physics is necessary to understand how radio, television, and mobile phones transmit information, and therefore is considered to be one of the most important areas of study for both the CBSE (Central Board of Secondary Education) Board Examination and for an array of different competitive examinations such as JEE (Joint Entrance Examination) and NEET (National Eligibility cum Entrance Test).

The solutions provided from the NCERT for Class XII Physics Chapter 8 ("Electromagnetic Waves") will assist students in visualising the transverse nature of the various kinds of electromagnetic waves via the breakdown of the Electromagnetic Spectrum from low wavelength (long wave) (i.e. radio waves), to high frequency (gamma rays) (high energy).

1.0Download NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves: Free PDF

In order to study electromagnetic waves, you must first develop a conceptual grasp of the wave's characteristics and how they carry energy. The following documents contain NCERT Solutions to the exercises of the NCERT books for the electromagnetic waves. The NCERT Solutions contain complete detailed information about the Electromagnetic Waves, including how to calculate their momentum and intensity and the relationship of their electric and magnetic fields. Students can save these solutions in PDF form to help them learn the different types of waves within the electromagnetic spectrum and to understand how to use them in various applications and how to create those types of waves.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

Available Soon

2.0Class 12 Physics Chapter 8 Electromagnetic Waves: Key Concepts

Class 12 Physics Chapter 8, Electromagnetic Waves, explains how electromagnetic waves are produced, their properties, and the different regions of the electromagnetic spectrum. The main concepts covered in this chapter include:

Displacement Current: Understand the concept of displacement current and why Maxwell introduced it while modifying Ampere’s circuital law for time-varying electric fields, such as those between the plates of a charging capacitor.
Maxwell’s Equations: Learn about Maxwell’s equations and their role in describing the basic laws of electricity and magnetism.
Nature of Electromagnetic Waves: Understand that electromagnetic waves are transverse waves and do not need a material medium to travel. In vacuum, they travel at the speed of light.
Energy and Radiation Pressure: Learn that electromagnetic waves carry energy and can exert pressure when they fall on a surface.
Electromagnetic Spectrum: Study the different regions of the electromagnetic spectrum, including radio waves, microwaves, infrared rays, visible light, ultraviolet rays, X-rays, and gamma rays.
Uses of Electromagnetic Waves: Understand the applications of different types of electromagnetic waves in areas such as communication, radar, microwave ovens, medical imaging, and other technologies.

3.0NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves : Detailed Solutions

SOLVED EXAMPLES

A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, $E = 6.3 j V / m$ . What is B at this point ? Sol. Using Eq. the magnitude of B is

$B = \frac{E}{c} = \frac{6.3 V / m}{3 \times 1 0 ^{8} m / s} = 2.1 \times 1 0^{- 8} T$

To find the direction, we note that $E$ is along y-direction and the wave propagates along x-axis. Therefore, $B$ should be in a direction perpendicular to both x and y-axes. Using vector algebra, $E \times B$ should be along x-direction. Since $(+ \hat{j}) \times (+ \hat{k}) = \hat{i}, B$ is along the z-direction. Thus, $B = 2.1 \times 1 0^{- 8} \hat{k} T$

The magnetic field in a plane electromagnetic wave is given by $B_{y} = (2 \times 1 0^{- 7}) T sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t)$ (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field.

Sol.

(a) Comparing the given equation with

$B_{y} = B_{0} sin [2 π (\frac{x}{λ} + \frac{t}{T})] We get, λ = \frac{2 π}{0.5 \times 1 0 ^{3}} m = 1.26 cm and \frac{1}{T} = v = \frac{1.5 \times 1 0 ^{11}}{2 π} = 23.9 GHz$

(b)

$E_{0} = B_{0} c = 2 \times 1 0^{- 7} T \times 3 \times 1 0^{8} m / s = 6 \times 10 V / m = 60 V / m$

The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along the z-axis is obtained as $E_{z} = 60 sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t) V / m$

EXERCISE QUESTIONS WITH SOLUTIONS

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.

Charging parallel-plate capacitor with two circular plates, showing the charging current between the plates.

(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. Sol. Here; $r = 12 cm = 12 \times 1 0^{- 2} m$ ,

$d = 5 mm = 5 \times 1 0^{- 3} m,$

Charging current $(I_{c}) = 0.15 A$ (a) $C = \frac{ε _{0} A}{d}$ here $A = π r^{2}$

$C = \frac{8.854 \times 1 0 ^{- 12} \times 3.14 \times ( 12 \times 1 0 ^{- 2} ) ^{2}}{5 \times 1 0 ^{- 3}}$

$C = 80 pF$

$q = CV \Rightarrow \frac{dq}{dt} = C \frac{dV}{dt}$

⇒

$I_{c} = C \frac{d V}{d t} \Rightarrow \frac{d V}{d t} = \frac{I _{c}}{C}$

$\frac{dV}{dt} = \frac{0.15}{80 \times 1 0 ^{- 12}} = 1.87 \times 1 0^{9} V / s$ (b) Displacement current,

$I_{d} = ε_{0} \frac{d ϕ _{E}}{d t} here, ϕ_{E} = E A I_{d} = ε_{0} A \frac{d E}{d t} \Rightarrow I_{d} = ε_{o} A \frac{d}{d t} (\frac{V}{d}) \Rightarrow I_{d} = \frac{ε _{o} A}{d} (\frac{d V}{d t}) \Rightarrow I_{d} = C (\frac{d V}{d t}) \Rightarrow I_{d} = I_{c} = 0.15 A$

(c) Yes, the conduction current (charging current) $I_{c}$ is not continuous across the capacitor gap as no charge is transferred across this gap. But the sum $(I_{c} + I_{d})$ is continuous as $I_{c} = I_{d}$ . Thus, Kirchoff's first rule (which states the current, is continuous) is valid.

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 $300 rad s^{- 1}$ .

Parallel-plate capacitor with circular plates connected to an AC supply, illustrating displacement current between the plates

(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. Here;

$C = 100 pF = 100 \times 1 0^{- 12} F = 1 0^{- 10} F, V_{rms} = 230 V, ω = 300 rad / s$

(a) $V_{rms} = I_{rms} X_{C}$

$\Rightarrow I_{r m s} = \frac{V _{r m s}}{X _{C}}$

(b) Yes, the conduction current is equal to the displacement current even in case of ac.

This value of B is peak value as we have

considered the peak displacement current

while calculating it.

What physical quantity is the same for X-raysof wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m?

Sol. The speed in vacuum for X-rays, red light andradiowaves (though all of different wavelengths) is the same, i.e, c = 3 × 108 m/s.

A plane electromagnetic wave travels in vaccum 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. E and B in x-y plane and mutually perpendicular to each other

A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? Sol. A radio can tune to minimum frequency,

$v_{1} = 7.5 MHz = 7.5 \times 1 0^{6} Hz$

Maximum frequency,

$v_{2} = 12 MHz = 12 \times I 0^{6} Hz$

Speed of light, $c = 3 \times 1 0^{8} m / s$ Corresponding wavelength for $v_{1}$ can be calculated as :

$λ_{1} = \frac{c}{v _{1}} = \frac{3 \times 1 0 ^{8}}{7.5 \times 1 0 ^{6}} = 40 m$

Corresponding wavelength for $v_{2}$ can be calculated as :

$λ_{2} = \frac{c}{v _{2}} = \frac{3 \times 1 0 ^{8}}{12 \times 1 0 ^{6}} = 25 m$

Thus, the wavelength band of the radio is 40 m to 25 m.

A charged particle oscillates about its mean equilibrium position with a frequency of $1 0^{9} Hz$ . What is the frequency of the electromagnetic waves produced by the oscillator? Sol. 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., $1 0^{9} Hz$ .
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_{0} = 510 nT$ . What is the amplitude of the electric field part of the wave? Sol. Amplitude of magnetic field of an electromagnetic wave in a vacuum,

$B_{0} = 510 nT = 510 \times 1 0^{- 9} T$

Speed of light in a vacuum,

$c = 3 \times 1 0^{8} m / s$

Amplitude of electric field of the electromagnetic wave,

$E_{0} = c B_{0} = 3 \times 1 0^{8} \times 510 \times 1 0^{- 9} = 153 N / C$

Suppose that the electric field amplitude of an electromagnetic wave is $E_{0} = 120 N / C$ and that its frequency is $v = 50.0 MHz$ . (a) Determine, $B_{0}, ω, k$ and $λ$ . (b) Find expressions for E and B. Sol. Given,

$E_{0} = 120 N / C v = 50.0 MHz = 50 \times 1 0^{6} Hz .$

(a)

$\frac{E _{0}}{B _{0}} = c \Rightarrow B_{0} = \frac{E _{0}}{c} = \frac{120}{3 \times 1 0 ^{8}} = 4 \times 1 0^{- 7} = 400 nT$

$ω k λ = 2 π v = 2 \times 3.14 (50 \times 1 0^{6}) = 3.14 \times 1 0^{8} rad / s = \frac{ω}{c} = \frac{3.14 \times 1 0 ^{8}}{3 \times 1 0 ^{8}} = 1.05 rad / m = \frac{c}{v} = \frac{3 \times 1 0 ^{8}}{50 \times 1 0 ^{6}} = 6.00 m$

(b) If we take the wave to be propagating along X-axis,

$E = E_{0} sin (kx - ω t)$

$\Rightarrow E = (120 N / C) sin [1.05 x - (3.14 \times 1 0^{8}) t]$

$B = B_{0} sin (kx - ω t)$

$\Rightarrow B = (400 nT) sin [1.05 x - (3.14 \times 1 0^{8}) t]$

The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $E = h ν$ (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. Energy of a photon is given as, $E = h ν = \frac{hc}{λ}$

$\Rightarrow E = \frac{6.6 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{λ} = \frac{19.8 \times 1 0 ^{- 26}}{λ} J = \frac{19.8 \times 1 0 ^{- 26}}{λ \times 1.6 \times 1 0 ^{- 19}} = \frac{12.375 \times 1 0 ^{- 7}}{λ} eV$

The given table lists the photon energies for different parts of an electromagnetic spectrum for different $λ$ .

$λ (m)$	$1 0^{3}$	1	$1 0^{- 3}$	$1 0^{- 6}$	$1 0^{- 8}$	$1 0^{- 10}$	$1 0^{- 12}$
E(eV)	12.375	12.375	12.375	12.375	12.375	12.375	12.375
	$\times 1 0^{- 10}$	$\times 1 0^{- 7}$	$\times 1 0^{- 4}$	$\times 1 0^{- 1}$	$\times 1 0^{1}$	$\times 1 0^{3}$	$\times 1 0^{5}$

The portion energies for different parts of the spectrum of source indicate the spacing of the relevant energy levels of the source.

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 1 0^{10} Hz$ 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.

$[c = 3 \times 1 0^{8} m s^{- 1}]$

Sol. Here; $E_{0} = 48 V / m$ ,

$v = 2.0 \times 1 0^{10} Hz$

and $c = 3 \times 1 0^{8} m / s$ (a) $λ = \frac{c}{v} = \frac{3 \times 1 0 ^{8}}{2 \times 1 0 ^{10}} = 1.5 \times 1 0^{- 2} m$

(b)

$B_{0} = \frac{E _{0}}{c} = \frac{48}{3 \times 1 0 ^{8}} = 1.6 \times 1 0^{- 7} T$

$u_{E} = \frac{1}{4} ε_{0} E_{0}^{2}$

Energy density in B,

$u_{B} = \frac{1}{4} \frac{B _{0}^{2}}{μ _{0}}$

Here $E_{0} = B_{0} c = B_{0} (\frac{1}{μ _{0} ε _{0}})$

$u_{E} = \frac{1}{4} ε_{0} B_{0}^{2} c^{2} = \frac{1}{4} ε_{0} B_{0}^{2} (\frac{1}{μ _{0} ε _{0}})^{2} = \frac{1}{4} ε_{0} B_{0}^{2} \times (\frac{1}{μ _{0} ε _{0}})$

$u_{E} = \frac{1}{4} \frac{B _{0}^{2}}{μ _{0}}$

$\Rightarrow u_{E} = u_{B}$

4.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Physics chapter-wise, with exercise answers, formulas, and step-by-step solutions to learn concepts and solve questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

5.0Key Features of NCERT Solutions for Class 12 Physics Chapter 8

Clear Classification of the Electromagnetic Spectrum: The solutions compare different regions of the electromagnetic spectrum based on their wavelength, frequency, source, and common uses.
Electric and Magnetic Field Relationship: The solutions explain the relationship between the amplitudes of electric and magnetic fields in an electromagnetic wave and show how to use E/B=cE/B = c in numerical questions.
Easy Understanding of Displacement Current: The concept of displacement current is explained using the example of a charging capacitor, making it easier to understand why Maxwell introduced it.
Step-by-Step Wave Equations: The solutions explain the equations for sinusoidally varying electric and magnetic fields and help students identify the direction, frequency, and other properties of an electromagnetic wave.
Simple Answers to Theory Questions: Important descriptive questions are answered in a clear and concise manner, covering the key points students need to include in CBSE board examination answers.

Frequently Asked Questions

NCERT Solutions Class 12 Physics Chapter 8 - Electromagnetic Waves

1.0Download NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves: Free PDF

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

Available Soon

2.0Class 12 Physics Chapter 8 Electromagnetic Waves: Key Concepts

Displacement Current: Understand the concept of displacement current and why Maxwell introduced it while modifying Ampere’s circuital law for time-varying electric fields, such as those between the plates of a charging capacitor.
Maxwell’s Equations: Learn about Maxwell’s equations and their role in describing the basic laws of electricity and magnetism.
Nature of Electromagnetic Waves: Understand that electromagnetic waves are transverse waves and do not need a material medium to travel. In vacuum, they travel at the speed of light.
Energy and Radiation Pressure: Learn that electromagnetic waves carry energy and can exert pressure when they fall on a surface.
Electromagnetic Spectrum: Study the different regions of the electromagnetic spectrum, including radio waves, microwaves, infrared rays, visible light, ultraviolet rays, X-rays, and gamma rays.
Uses of Electromagnetic Waves: Understand the applications of different types of electromagnetic waves in areas such as communication, radar, microwave ovens, medical imaging, and other technologies.

3.0NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves : Detailed Solutions

SOLVED EXAMPLES

A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, $E = 6.3 j V / m$ . What is B at this point ? Sol. Using Eq. the magnitude of B is

$B = \frac{E}{c} = \frac{6.3 V / m}{3 \times 1 0 ^{8} m / s} = 2.1 \times 1 0^{- 8} T$

The magnetic field in a plane electromagnetic wave is given by $B_{y} = (2 \times 1 0^{- 7}) T sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t)$ (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field.

Sol.

(a) Comparing the given equation with

$B_{y} = B_{0} sin [2 π (\frac{x}{λ} + \frac{t}{T})] We get, λ = \frac{2 π}{0.5 \times 1 0 ^{3}} m = 1.26 cm and \frac{1}{T} = v = \frac{1.5 \times 1 0 ^{11}}{2 π} = 23.9 GHz$

(b)

$E_{0} = B_{0} c = 2 \times 1 0^{- 7} T \times 3 \times 1 0^{8} m / s = 6 \times 10 V / m = 60 V / m$

EXERCISE QUESTIONS WITH SOLUTIONS

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.

$d = 5 mm = 5 \times 1 0^{- 3} m,$

Charging current $(I_{c}) = 0.15 A$ (a) $C = \frac{ε _{0} A}{d}$ here $A = π r^{2}$

$C = \frac{8.854 \times 1 0 ^{- 12} \times 3.14 \times ( 12 \times 1 0 ^{- 2} ) ^{2}}{5 \times 1 0 ^{- 3}}$

$C = 80 pF$

$q = CV \Rightarrow \frac{dq}{dt} = C \frac{dV}{dt}$

⇒

$I_{c} = C \frac{d V}{d t} \Rightarrow \frac{d V}{d t} = \frac{I _{c}}{C}$

$\frac{dV}{dt} = \frac{0.15}{80 \times 1 0 ^{- 12}} = 1.87 \times 1 0^{9} V / s$ (b) Displacement current,

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 $300 rad s^{- 1}$ .

$C = 100 pF = 100 \times 1 0^{- 12} F = 1 0^{- 10} F, V_{rms} = 230 V, ω = 300 rad / s$

(a) $V_{rms} = I_{rms} X_{C}$

$\Rightarrow I_{r m s} = \frac{V _{r m s}}{X _{C}}$

(b) Yes, the conduction current is equal to the displacement current even in case of ac.

This value of B is peak value as we have

considered the peak displacement current

while calculating it.

What physical quantity is the same for X-raysof wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m?

Sol. The speed in vacuum for X-rays, red light andradiowaves (though all of different wavelengths) is the same, i.e, c = 3 × 108 m/s.

A plane electromagnetic wave travels in vaccum 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. E and B in x-y plane and mutually perpendicular to each other

A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? Sol. A radio can tune to minimum frequency,

$v_{1} = 7.5 MHz = 7.5 \times 1 0^{6} Hz$

Maximum frequency,

$v_{2} = 12 MHz = 12 \times I 0^{6} Hz$

Speed of light, $c = 3 \times 1 0^{8} m / s$ Corresponding wavelength for $v_{1}$ can be calculated as :

$λ_{1} = \frac{c}{v _{1}} = \frac{3 \times 1 0 ^{8}}{7.5 \times 1 0 ^{6}} = 40 m$

Corresponding wavelength for $v_{2}$ can be calculated as :

$λ_{2} = \frac{c}{v _{2}} = \frac{3 \times 1 0 ^{8}}{12 \times 1 0 ^{6}} = 25 m$

Thus, the wavelength band of the radio is 40 m to 25 m.

A charged particle oscillates about its mean equilibrium position with a frequency of $1 0^{9} Hz$ . What is the frequency of the electromagnetic waves produced by the oscillator? Sol. 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., $1 0^{9} Hz$ .
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_{0} = 510 nT$ . What is the amplitude of the electric field part of the wave? Sol. Amplitude of magnetic field of an electromagnetic wave in a vacuum,

$B_{0} = 510 nT = 510 \times 1 0^{- 9} T$

Speed of light in a vacuum,

$c = 3 \times 1 0^{8} m / s$

Amplitude of electric field of the electromagnetic wave,

$E_{0} = c B_{0} = 3 \times 1 0^{8} \times 510 \times 1 0^{- 9} = 153 N / C$

Suppose that the electric field amplitude of an electromagnetic wave is $E_{0} = 120 N / C$ and that its frequency is $v = 50.0 MHz$ . (a) Determine, $B_{0}, ω, k$ and $λ$ . (b) Find expressions for E and B. Sol. Given,

$E_{0} = 120 N / C v = 50.0 MHz = 50 \times 1 0^{6} Hz .$

(a)

$\frac{E _{0}}{B _{0}} = c \Rightarrow B_{0} = \frac{E _{0}}{c} = \frac{120}{3 \times 1 0 ^{8}} = 4 \times 1 0^{- 7} = 400 nT$

(b) If we take the wave to be propagating along X-axis,

$E = E_{0} sin (kx - ω t)$

$\Rightarrow E = (120 N / C) sin [1.05 x - (3.14 \times 1 0^{8}) t]$

$B = B_{0} sin (kx - ω t)$

$\Rightarrow B = (400 nT) sin [1.05 x - (3.14 \times 1 0^{8}) t]$

The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $E = h ν$ (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. Energy of a photon is given as, $E = h ν = \frac{hc}{λ}$

The given table lists the photon energies for different parts of an electromagnetic spectrum for different $λ$ .

$λ (m)$	$1 0^{3}$	1	$1 0^{- 3}$	$1 0^{- 6}$	$1 0^{- 8}$	$1 0^{- 10}$	$1 0^{- 12}$
E(eV)	12.375	12.375	12.375	12.375	12.375	12.375	12.375
	$\times 1 0^{- 10}$	$\times 1 0^{- 7}$	$\times 1 0^{- 4}$	$\times 1 0^{- 1}$	$\times 1 0^{1}$	$\times 1 0^{3}$	$\times 1 0^{5}$

The portion energies for different parts of the spectrum of source indicate the spacing of the relevant energy levels of the source.

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 1 0^{10} Hz$ 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.

$[c = 3 \times 1 0^{8} m s^{- 1}]$

Sol. Here; $E_{0} = 48 V / m$ ,

$v = 2.0 \times 1 0^{10} Hz$

and $c = 3 \times 1 0^{8} m / s$ (a) $λ = \frac{c}{v} = \frac{3 \times 1 0 ^{8}}{2 \times 1 0 ^{10}} = 1.5 \times 1 0^{- 2} m$

(b)

$B_{0} = \frac{E _{0}}{c} = \frac{48}{3 \times 1 0 ^{8}} = 1.6 \times 1 0^{- 7} T$

$u_{E} = \frac{1}{4} ε_{0} E_{0}^{2}$

Energy density in B,

$u_{B} = \frac{1}{4} \frac{B _{0}^{2}}{μ _{0}}$

Here $E_{0} = B_{0} c = B_{0} (\frac{1}{μ _{0} ε _{0}})$

$u_{E} = \frac{1}{4} ε_{0} B_{0}^{2} c^{2} = \frac{1}{4} ε_{0} B_{0}^{2} (\frac{1}{μ _{0} ε _{0}})^{2} = \frac{1}{4} ε_{0} B_{0}^{2} \times (\frac{1}{μ _{0} ε _{0}})$

$u_{E} = \frac{1}{4} \frac{B _{0}^{2}}{μ _{0}}$

$\Rightarrow u_{E} = u_{B}$

4.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Physics chapter-wise, with exercise answers, formulas, and step-by-step solutions to learn concepts and solve questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

5.0Key Features of NCERT Solutions for Class 12 Physics Chapter 8

Clear Classification of the Electromagnetic Spectrum: The solutions compare different regions of the electromagnetic spectrum based on their wavelength, frequency, source, and common uses.
Electric and Magnetic Field Relationship: The solutions explain the relationship between the amplitudes of electric and magnetic fields in an electromagnetic wave and show how to use E/B=cE/B = c in numerical questions.
Easy Understanding of Displacement Current: The concept of displacement current is explained using the example of a charging capacitor, making it easier to understand why Maxwell introduced it.
Step-by-Step Wave Equations: The solutions explain the equations for sinusoidally varying electric and magnetic fields and help students identify the direction, frequency, and other properties of an electromagnetic wave.
Simple Answers to Theory Questions: Important descriptive questions are answered in a clear and concise manner, covering the key points students need to include in CBSE board examination answers.

Frequently Asked Questions

What will I find in NCERT Solutions Class 12 Physics Chapter 8?

Why did Maxwell introduce displacement current in NCERT Solutions Class 12 Physics Chapter 8?

Does NCERT Solutions Class 12 Physics Chapter 8 explain how electromagnetic waves travel?

Which relation between electric and magnetic fields is used in NCERT Solutions Class 12 Physics Chapter 8?

How can I differentiate radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays using NCERT Solutions Class 12 Physics Chapter 8?

What are the practical uses of electromagnetic waves covered in NCERT Solutions Class 12 Physics Chapter 8?

Are NCERT Solutions Class 12 Physics Chapter 8 useful for competitive exams?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 8 - Electromagnetic Waves

1.0Download NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves: Free PDF

2.0Class 12 Physics Chapter 8 Electromagnetic Waves: Key Concepts

3.0NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

4.0Class 12 Physics NCERT Solutions – Chapter-wise Links

5.0Key Features of NCERT Solutions for Class 12 Physics Chapter 8

Table of Contents

Frequently Asked Questions

What will I find in NCERT Solutions Class 12 Physics Chapter 8?

Why did Maxwell introduce displacement current in NCERT Solutions Class 12 Physics Chapter 8?

Does NCERT Solutions Class 12 Physics Chapter 8 explain how electromagnetic waves travel?

Which relation between electric and magnetic fields is used in NCERT Solutions Class 12 Physics Chapter 8?

How can I differentiate radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays using NCERT Solutions Class 12 Physics Chapter 8?

What are the practical uses of electromagnetic waves covered in NCERT Solutions Class 12 Physics Chapter 8?

Are NCERT Solutions Class 12 Physics Chapter 8 useful for competitive exams?

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NCERT Solutions Class 12 Physics Chapter 8 - Electromagnetic Waves

1.0Download NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves: Free PDF

2.0Class 12 Physics Chapter 8 Electromagnetic Waves: Key Concepts

3.0NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

4.0Class 12 Physics NCERT Solutions – Chapter-wise Links

5.0Key Features of NCERT Solutions for Class 12 Physics Chapter 8

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