NCERT CLASS 12 PHYSICS

CHAPTER 8 – ELECTROMAGNETIC WAVES

PART 1: IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS

1. What are electromagnetic waves?

Electromagnetic waves are waves consisting of oscillating electric and magnetic fields that travel through space.

The electric and magnetic fields are perpendicular to each other and also perpendicular to the direction of propagation.

Examples include radio waves, microwaves, infrared rays, visible light, ultraviolet rays, X-rays and gamma rays.

2. Who predicted the existence of electromagnetic waves?

James Clerk Maxwell predicted the existence of electromagnetic waves theoretically.

He showed that changing electric and magnetic fields can produce each other and propagate through space as electromagnetic waves.

3. What is displacement current?

Displacement current is the current associated with a changing electric field.

It was introduced by James Clerk Maxwell to modify Ampere's circuital law.

It is given by:

Iᵈ = ε₀ (dΦᴇ/dt)

Where:

Iᵈ = Displacement current
ε₀ = Permittivity of free space
Φᴇ = Electric flux

4. Why did Maxwell introduce the concept of displacement current?

Maxwell introduced displacement current to remove the inconsistency in Ampere's circuital law when applied to a charging capacitor.

In a charging capacitor, conduction current flows through the connecting wires, but no charge actually crosses the gap between the plates.

The changing electric field between the plates produces displacement current.

This makes the current continuous throughout the circuit.

5. What is the difference between conduction current and displacement current?

Conduction current:

It is produced by the actual flow of electric charges.
It flows through conducting materials such as wires.
It is represented by I꜀.

Displacement current:

It is associated with a changing electric field.
It can exist in the space between capacitor plates.
It is represented by Iᵈ.

6. Write the expression for displacement current.

The displacement current is given by:

Iᵈ = ε₀ (dΦᴇ/dt)

For a parallel-plate capacitor with a uniform electric field:

Iᵈ = ε₀ A (dE/dt)

Where:

A = Area of capacitor plate
E = Electric field

7. Is displacement current equal to conduction current in a charging capacitor?

Yes. For an ideal charging capacitor, the displacement current between the plates is equal to the conduction current in the connecting wires.

Therefore:

Iᵈ = I꜀

This equality holds instantaneously during charging.

8. What is Ampere-Maxwell law?

Ampere-Maxwell law is the modified form of Ampere's circuital law.

It states that the magnetic field is produced by both conduction current and displacement current.

Its integral form in vacuum is:

∮ B⃗ · dl⃗ = μ₀ (I꜀ + Iᵈ)

Where:

μ₀ = Permeability of free space
I꜀ = Conduction current
Iᵈ = Displacement current

9. What are Maxwell's equations?

Maxwell's equations are four fundamental equations that describe the relationship between electric fields, magnetic fields and electric charges and currents.

They are:

Gauss's law for electricity:
∮ E⃗ · dA⃗ = Q/ε₀
Gauss's law for magnetism:
∮ B⃗ · dA⃗ = 0
Faraday's law of electromagnetic induction:
∮ E⃗ · dl⃗ = −dΦᴮ/dt
Ampere-Maxwell law:
∮ B⃗ · dl⃗ = μ₀(I꜀ + Iᵈ)

These equations form the foundation of classical electromagnetic theory.

10. What is the speed of electromagnetic waves in vacuum?

The speed of electromagnetic waves in vacuum is:

c = 3 × 10⁸ m/s

It is equal to the speed of light in vacuum.

It is given by:

c = 1/√(μ₀ε₀)

11. How are electromagnetic waves produced?

Electromagnetic waves are produced by accelerating electric charges.

An accelerating charge produces changing electric and magnetic fields, which travel through space in the form of electromagnetic waves.

12. Why do stationary charges not produce electromagnetic waves?

A stationary charge produces an electric field that does not change with time.

Since there is no acceleration or changing electromagnetic field associated with the stationary charge, it does not radiate electromagnetic waves.

13. Are electromagnetic waves transverse or longitudinal?

Electromagnetic waves are transverse waves.

In these waves:

The electric field oscillates perpendicular to the direction of propagation.
The magnetic field oscillates perpendicular to the direction of propagation.
The electric and magnetic fields are perpendicular to each other.

14. Do electromagnetic waves require a material medium for propagation?

No. Electromagnetic waves do not require a material medium for propagation.

They can travel through vacuum.

For example, sunlight travels through the vacuum of space and reaches the Earth.

15. What is the relationship between electric and magnetic fields in an electromagnetic wave?

In vacuum, the amplitudes of the electric and magnetic fields are related by:

E₀/B₀ = c

Therefore:

E₀ = cB₀

Where:

E₀ = Amplitude of electric field
B₀ = Amplitude of magnetic field
c = Speed of light in vacuum

16. What is the electromagnetic spectrum?

The electromagnetic spectrum is the complete range of electromagnetic waves arranged according to their frequencies or wavelengths.

It includes:

Radio waves
Microwaves
Infrared rays
Visible light
Ultraviolet rays
X-rays
Gamma rays

17. Arrange electromagnetic waves in increasing order of frequency.

The increasing order of frequency is:

Radio waves < Microwaves < Infrared rays < Visible light < Ultraviolet rays < X-rays < Gamma rays

18. Arrange electromagnetic waves in increasing order of wavelength.

The increasing order of wavelength is:

Gamma rays < X-rays < Ultraviolet rays < Visible light < Infrared rays < Microwaves < Radio waves

19. What are radio waves?

Radio waves are electromagnetic waves having relatively long wavelengths and low frequencies.

They are produced by accelerated charges in conducting aerials.

Uses:

Radio broadcasting
Television communication
Wireless communication

20. What are microwaves?

Microwaves are electromagnetic waves having wavelengths shorter than radio waves.

They are commonly produced by special electronic devices.

Uses:

Radar systems
Satellite communication
Microwave ovens
Wireless communication

21. What are infrared rays?

Infrared rays are electromagnetic waves having wavelengths longer than visible red light.

They are commonly associated with thermal radiation.

Uses:

Remote controls
Thermal imaging
Night-vision equipment
Physiotherapy and heat treatment

22. What is visible light?

Visible light is the part of the electromagnetic spectrum that can be detected by the human eye.

Its wavelength range is approximately 400 nm to 700 nm.

It enables us to see objects.

23. What are ultraviolet rays?

Ultraviolet rays are electromagnetic waves having frequencies higher than visible violet light.

They are produced by very hot bodies and certain electronic and atomic processes.

Uses:

Sterilisation of medical equipment
Detection of forged documents
Fluorescence applications
Scientific research

Excessive exposure to ultraviolet radiation can be harmful to living tissues.

24. What are X-rays?

X-rays are high-frequency electromagnetic waves with very short wavelengths.

They are commonly produced when high-speed electrons are suddenly decelerated or when inner-shell atomic transitions occur.

Uses:

Medical imaging
Detection of fractures
Security scanning
Study of crystal structures

25. What are gamma rays?

Gamma rays are electromagnetic waves having the highest frequencies and shortest wavelengths in the electromagnetic spectrum.

They are generally emitted during nuclear transitions and radioactive processes.

Uses:

Cancer radiotherapy
Sterilisation of medical equipment
Nuclear research
Scientific investigations

26. Which electromagnetic waves are used in radar systems?

Microwaves are commonly used in radar systems.

They are useful for detecting the position, speed and distance of objects.

27. Which electromagnetic waves are used in satellite communication?

Microwaves are widely used in satellite communication because they can pass through the atmosphere in suitable frequency bands with relatively low attenuation.

28. Which electromagnetic waves are used in TV and radio broadcasting?

Radio waves are used in radio and television broadcasting.

Different frequency bands are used for transmitting audio and video signals.

29. Which electromagnetic waves are used in night-vision devices?

Infrared rays are used in night-vision and thermal imaging devices because warm objects emit infrared radiation.

30. Which electromagnetic waves are used to detect bone fractures?

X-rays are used to detect bone fractures because they pass through soft tissues more easily than through bones.

31. Which electromagnetic waves are used for sterilisation?

Ultraviolet radiation is used for sterilisation and disinfection of air, water and suitable surfaces.

It can damage the genetic material of microorganisms.

32. Which electromagnetic waves are used in cancer treatment?

High-energy gamma rays and X-rays are used in radiotherapy to destroy cancer cells.

33. Why are electromagnetic waves not deflected by electric or magnetic fields?

Electromagnetic waves consist of oscillating electric and magnetic fields and do not carry a net electric charge.

Therefore, they are not deflected like charged particles when passing through external electric or magnetic fields.

34. Do all electromagnetic waves travel with the same speed in vacuum?

Yes. All electromagnetic waves travel with the same speed in vacuum.

Their speed is:

c = 3 × 10⁸ m/s

However, their frequencies and wavelengths are different.

35. What happens to the wavelength of an electromagnetic wave when its frequency increases?

The wavelength decreases when frequency increases, provided the wave travels in the same medium.

The relation is:

c = νλ

Therefore:

λ = c/ν

36. What is the relation between the energy of a photon and its frequency?

The energy of a photon is directly proportional to its frequency.

E = hν

Where:

E = Energy of photon
h = Planck's constant
ν = Frequency

37. Which electromagnetic waves have the highest photon energy?

Gamma rays have the highest photon energy because they have the highest frequency.

According to:

E = hν

Higher frequency means higher photon energy.

38. Which electromagnetic waves have the lowest photon energy?

Radio waves have the lowest photon energy because they have the lowest frequency in the electromagnetic spectrum.

39. What is the importance of electromagnetic waves in communication?

Electromagnetic waves are used to transmit information over long distances.

Radio waves and microwaves are widely used in radio broadcasting, television, mobile communication, radar and satellite communication.

40. What is the importance of electromagnetic waves in daily life?

Electromagnetic waves are important in many areas of daily life, including:

Communication
Medical diagnosis
Medical treatment
Cooking
Remote controls
Weather forecasting
Scientific research
Satellite navigation

PART 2: NCERT EXERCISE QUESTIONS AND ANSWERS

QUESTION 8.1

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 0.15 A.

(a) Calculate the capacitance and the rate of change of potential difference between the plates.

(b) Obtain the displacement current across the plates.

Solution:

Given:

Radius of each plate:

r = 12 cm = 0.12 m

Distance between plates:

d = 5 cm = 0.05 m

Charging current:

I = 0.15 A

Permittivity of free space:

ε₀ = 8.85 × 10⁻¹² F/m

(a) Calculate capacitance and rate of change of potential difference.

Step 1: Calculate the area of the circular plate.

A = πr²

A = 3.14 × (0.12)²

A = 3.14 × 0.0144

A = 0.0452 m²

Step 2: Calculate capacitance.

For a parallel-plate capacitor:

C = ε₀A/d

Substituting:

C = (8.85 × 10⁻¹² × 0.0452)/0.05

C ≈ 8.0 × 10⁻¹² F

C = 8.0 pF

Step 3: Calculate the rate of change of potential difference.

We know:

Q = CV

Differentiating with respect to time:

dQ/dt = C(dV/dt)

Since:

I = dQ/dt

Therefore:

I = C(dV/dt)

dV/dt = I/C

Substituting:

dV/dt = 0.15/(8.0 × 10⁻¹²)

dV/dt = 1.875 × 10¹⁰ V/s

(b) Obtain the displacement current.

For a charging capacitor:

Displacement current = Conduction current

Therefore:

Iᵈ = I

Iᵈ = 0.15 A

Displacement current = 0.15 A

(c) Is Kirchhoff's first rule valid?

Yes. Kirchhoff's first rule is valid when both conduction current and displacement current are considered.

The conduction current flows through the connecting wires, while displacement current exists in the space between the capacitor plates.

The displacement current is equal to the conduction current.

Therefore, the total current remains continuous.

Final Answer:

(a) Capacitance = 8.0 pF

Rate of change of potential difference = 1.875 × 10¹⁰ V/s

(b) Displacement current = 0.15 A

QUESTION 8.2

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?

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

Solution:

Given:

Radius of plates:

R = 6.0 cm = 0.06 m

Capacitance:

C = 100 pF = 100 × 10⁻¹² F

RMS voltage:

V = 230 V

Angular frequency:

ω = 300 rad/s

Distance from axis:

r = 3.0 cm = 0.03 m

(a) Calculate RMS conduction current.

For a capacitor:

Capacitive reactance:

X꜀ = 1/ωC

RMS current:

I = V/X꜀

Therefore:

I = VωC

Substituting:

I = 230 × 300 × 100 × 10⁻¹²

I = 6.9 × 10⁻⁶ A

I = 6.9 μA

(b) Is conduction current equal to displacement current?

Yes.

For a charging or discharging capacitor, displacement current is equal to conduction current at every instant.

Therefore:

Iᵈ = I꜀

Yes, both currents are equal at every instant.

(c) Determine the amplitude of magnetic field.

According to Ampere-Maxwell law:

∮ B⃗ · dl⃗ = μ₀ Iᵈ

For a circular path of radius r between the plates:

B(2πr) = μ₀ Iᵈ (r²/R²)

Therefore:

B = μ₀ Iᵈr/(2πR²)

The given current is RMS current. First calculate the peak displacement current.

Iₘ = √2 Iᵣₘₛ

Iₘ = 1.414 × 6.9 × 10⁻⁶

Iₘ = 9.76 × 10⁻⁶ A

Now:

B₀ = μ₀ Iₘr/(2πR²)

Using:

μ₀/(2π) = 2 × 10⁻⁷ T m/A

Substituting:

B₀ = (2 × 10⁻⁷ × 9.76 × 10⁻⁶ × 0.03)/(0.06)²

B₀ = 1.63 × 10⁻¹¹ T

Final Answer:

(a) RMS conduction current = 6.9 μA

(b) Yes, conduction current equals displacement current at every instant.

QUESTION 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:

All three are electromagnetic waves.

The physical quantity that is the same for all three in vacuum is their speed.

The speed of electromagnetic waves in vacuum is:

c = 3 × 10⁸ m/s

Although their wavelengths and frequencies are different, their speed in vacuum is the same.

Final Answer:

Speed in vacuum = 3 × 10⁸ m/s

QUESTION 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?

Solution:

Given:

Direction of propagation = z-direction

Frequency:

ν = 30 MHz

ν = 30 × 10⁶ Hz

Speed of light:

c = 3 × 10⁸ m/s

Part 1: Directions of electric and magnetic fields

Electromagnetic waves are transverse waves.

Therefore:

Electric field is perpendicular to the direction of propagation.
Magnetic field is perpendicular to the direction of propagation.
Electric and magnetic fields are perpendicular to each other.

Since the wave travels along the z-direction, both fields lie in the x-y plane.

Their directions are mutually perpendicular.

Part 2: Calculate wavelength.

We know:

c = νλ

Therefore:

λ = c/ν

Substituting:

λ = (3 × 10⁸)/(30 × 10⁶)

λ = 10 m

Final Answer:

Electric and magnetic fields are perpendicular to each other and to the z-direction.

Wavelength = 10 m

QUESTION 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?

Solution:

Given:

Minimum frequency:

ν₁ = 7.5 MHz = 7.5 × 10⁶ Hz

Maximum frequency:

ν₂ = 12 MHz = 12 × 10⁶ Hz

Speed of electromagnetic waves:

c = 3 × 10⁸ m/s

Formula:

λ = c/ν

For 7.5 MHz:

λ₁ = (3 × 10⁸)/(7.5 × 10⁶)

λ₁ = 40 m

For 12 MHz:

λ₂ = (3 × 10⁸)/(12 × 10⁶)

λ₂ = 25 m

Therefore, the wavelength range is:

25 m to 40 m

Final Answer:

Corresponding wavelength band = 25 m to 40 m.

QUESTION 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?

Solution:

Given:

Frequency of oscillating charged particle:

ν = 10⁹ Hz

An oscillating charged particle is an accelerated charge.

An accelerated charge produces electromagnetic waves.

The frequency of the electromagnetic waves produced is equal to the frequency of oscillation of the charged particle.

Therefore:

Frequency of electromagnetic waves = 10⁹ Hz

Final Answer:

10⁹ Hz

QUESTION 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?

Solution:

Given:

Magnetic field amplitude:

B₀ = 510 nT

B₀ = 510 × 10⁻⁹ T

Speed of light:

c = 3 × 10⁸ m/s

We know:

E₀/B₀ = c

Therefore:

E₀ = cB₀

Substituting:

E₀ = 3 × 10⁸ × 510 × 10⁻⁹

E₀ = 153 N/C

Final Answer:

Amplitude of electric field = 153 N/C

QUESTION 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 E⃗ and B⃗.

Solution:

Given:

Electric field amplitude:

E₀ = 120 N/C

Frequency:

ν = 50.0 MHz

ν = 50 × 10⁶ Hz

Speed of light:

c = 3 × 10⁸ m/s

(a) Calculate B₀, ω, k and λ.

Step 1: Calculate magnetic field amplitude.

We know:

E₀/B₀ = c

Therefore:

B₀ = E₀/c

B₀ = 120/(3 × 10⁸)

B₀ = 4 × 10⁻⁷ T

Step 2: Calculate angular frequency.

ω = 2πν

ω = 2 × 3.14 × 50 × 10⁶

ω = 3.14 × 10⁸ rad/s

Step 3: Calculate wavelength.

λ = c/ν

λ = (3 × 10⁸)/(50 × 10⁶)

λ = 6 m

Step 4: Calculate wave number.

k = 2π/λ

k = 2 × 3.14/6

k = 1.05 rad/m

(b) Find expressions for E⃗ and B⃗.

The question does not specify the directions of propagation and fields. We can choose a convenient set of mutually perpendicular directions.

Let:

Wave propagate along the positive x-direction.
Electric field oscillate along the y-direction.
Magnetic field oscillate along the z-direction.

The electric and magnetic fields are in phase.

The general wave expressions are:

E⃗ = E₀ sin(kx − ωt) ĵ

B⃗ = B₀ sin(kx − ωt) k̂

Substituting the values:

Electric field:

E⃗ = 120 sin[(1.05)x − (3.14 × 10⁸)t] ĵ N/C

Magnetic field:

B⃗ = 4 × 10⁻⁷ sin[(1.05)x − (3.14 × 10⁸)t] k̂ T

Final Answer:

(a)

B₀ = 4 × 10⁻⁷ T

ω = 3.14 × 10⁸ rad/s

k = 1.05 rad/m

λ = 6 m

(b) The electric and magnetic field expressions are given above for the chosen directions of propagation and oscillation.

QUESTION 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?

Solution:

We know:

Energy of a photon:

E = hν

Also:

ν = c/λ

Therefore:

E = hc/λ

Given:

Planck's constant:

h = 6.63 × 10⁻³⁴ J s

Speed of light:

c = 3 × 10⁸ m/s

Charge of electron:

e = 1.6 × 10⁻¹⁹ C

To convert energy from joules to electronvolts:

E(eV) = E(J)/(1.6 × 10⁻¹⁹)

Therefore:

E(eV) = hc/(eλ)

Substituting:

E(eV) = [(6.63 × 10⁻³⁴)(3 × 10⁸)]/[(1.6 × 10⁻¹⁹)λ]

E(eV) = 1.24 × 10⁻⁶/λ

Here, wavelength λ is measured in metres.

Photon energies in different regions of the electromagnetic spectrum:

Region	Approximate wavelength	Photon energy
Radio waves	Greater than 0.1 m	Below 1.2 × 10⁻⁵ eV
Microwaves	0.1 m to 1 mm	1.2 × 10⁻⁵ to 1.2 × 10⁻³ eV
Infrared rays	1 mm to 700 nm	1.2 × 10⁻³ to 1.8 eV
Visible light	700 nm to 400 nm	1.8 to 3.1 eV
Ultraviolet rays	400 nm to 1 nm	3.1 to 1.2 × 10³ eV
X-rays	1 nm to 10⁻³ nm	1.2 keV to 1.2 MeV
Gamma rays	Less than 10⁻³ nm	Above 1.2 MeV

Relation between photon energy and sources:

The energy of electromagnetic radiation is related to the energy changes occurring in its source.

Gamma rays are produced by nuclear transitions because nuclear energy differences are very large.
X-rays are produced by processes involving high-energy electrons, including inner-shell electronic transitions.
Visible and ultraviolet radiation are generally associated with electronic transitions in atoms and molecules.
Infrared radiation is commonly associated with molecular vibrations and rotations.
Radio waves are produced by accelerated charges in conducting aerials.

Final Answer:

Photon energy increases with frequency and decreases with wavelength. The different photon energy ranges are related to the energy changes and physical processes occurring in the sources of radiation.

QUESTION 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 = 3 × 10⁸ m s⁻¹.]

Solution:

Given:

Frequency:

ν = 2.0 × 10¹⁰ Hz

Electric field amplitude:

E₀ = 48 V/m

Speed of light:

c = 3 × 10⁸ m/s

(a) Calculate wavelength.

We know:

c = νλ

Therefore:

λ = c/ν

Substituting:

λ = (3 × 10⁸)/(2 × 10¹⁰)

λ = 1.5 × 10⁻² m

Wavelength = 1.5 cm

(b) Calculate amplitude of magnetic field.

We know:

E₀/B₀ = c

Therefore:

B₀ = E₀/c

Substituting:

B₀ = 48/(3 × 10⁸)

B₀ = 1.6 × 10⁻⁷ T

(c) Show that average energy density of electric field equals average energy density of magnetic field.

The instantaneous energy density of the electric field is:

uᴇ = ½ ε₀E²

The instantaneous energy density of the magnetic field is:

uᴮ = B²/(2μ₀)

For a sinusoidal electromagnetic wave:

E = E₀ sin(kx − ωt)

B = B₀ sin(kx − ωt)

The average value of sin² over one complete cycle is:

Average of sin² = 1/2

Therefore, average electric field energy density is:

ūᴇ = ¼ ε₀E₀²

Average magnetic field energy density is:

ūᴮ = B₀²/(4μ₀)

We know:

E₀ = cB₀

Therefore:

B₀ = E₀/c

Substituting in the magnetic energy density:

ūᴮ = E₀²/(4μ₀c²)

We know:

c² = 1/(μ₀ε₀)

Therefore:

1/(μ₀c²) = ε₀

Hence:

ūᴮ = ε₀E₀²/4

But:

ūᴇ = ε₀E₀²/4

Therefore:

ūᴇ = ūᴮ

Hence proved.

Calculate their common value:

ūᴇ = ¼ ε₀E₀²

ūᴇ = ¼ × 8.85 × 10⁻¹² × 48²

ūᴇ ≈ 5.1 × 10⁻⁹ J/m³

Final Answer:

(a) Wavelength = 1.5 × 10⁻² m

(b) Magnetic field amplitude = 1.6 × 10⁻⁷ T

Each average energy density = 5.1 × 10⁻⁹ J/m³

PART 3: IMPORTANT DERIVATIONS

1. Derive the expression for displacement current.

According to Maxwell, a changing electric field produces a displacement current.

Displacement current is defined as:

Iᵈ = ε₀ (dΦᴇ/dt)

For a parallel-plate capacitor:

Electric flux:

Φᴇ = EA

Where:

E = Electric field
A = Area of plate

Therefore:

Iᵈ = ε₀ d(EA)/dt

For constant plate area:

Iᵈ = ε₀A(dE/dt)

For a parallel-plate capacitor:

E = Q/(ε₀A)

Differentiating:

dE/dt = 1/(ε₀A) × dQ/dt

Substituting:

Iᵈ = ε₀A × [1/(ε₀A)] × dQ/dt

Therefore:

Iᵈ = dQ/dt

But:

I꜀ = dQ/dt

Hence:

Iᵈ = I꜀

Thus, displacement current is equal to conduction current in a charging capacitor.

2. Derive the expression for the speed of electromagnetic waves in vacuum.

According to Maxwell's electromagnetic theory, the speed of electromagnetic waves in vacuum is determined by the permeability and permittivity of free space.

The expression is:

c = 1/√(μ₀ε₀)

Where:

μ₀ = Permeability of free space
ε₀ = Permittivity of free space

Using the values:

μ₀ = 4π × 10⁻⁷ N/A²

ε₀ = 8.85 × 10⁻¹² F/m

Substituting:

c = 1/√[(4π × 10⁻⁷)(8.85 × 10⁻¹²)]

Therefore:

c ≈ 3 × 10⁸ m/s

Hence, electromagnetic waves travel through vacuum with the speed of light.

3. Derive the relation between electric and magnetic field amplitudes.

For an electromagnetic wave travelling in vacuum, the electric and magnetic fields are related by:

E₀/B₀ = c

Therefore:

E₀ = cB₀

Where:

E₀ = Maximum electric field
B₀ = Maximum magnetic field
c = Speed of light in vacuum

This shows that the ratio of electric field amplitude to magnetic field amplitude is equal to the speed of light.

4. Show that electromagnetic waves are transverse in nature.

An electromagnetic wave consists of oscillating electric and magnetic fields.

The electric field is perpendicular to the magnetic field.

Both fields are perpendicular to the direction of propagation.

If the wave travels along the x-direction, the electric field may oscillate along the y-direction and the magnetic field along the z-direction.

Therefore:

E⃗ ⟂ B⃗

E⃗ ⟂ Direction of propagation

B⃗ ⟂ Direction of propagation

Hence, electromagnetic waves are transverse waves.

PART 4: IMPORTANT NUMERICAL QUESTIONS

1. Calculate the frequency of an electromagnetic wave having a wavelength of 600 nm in vacuum.

Given:

λ = 600 nm = 600 × 10⁻⁹ m

c = 3 × 10⁸ m/s

Formula:

c = νλ

ν = c/λ

ν = (3 × 10⁸)/(600 × 10⁻⁹)

ν = 5 × 10¹⁴ Hz

2. Calculate the wavelength of an electromagnetic wave having a frequency of 100 MHz.

Given:

ν = 100 MHz = 100 × 10⁶ Hz

c = 3 × 10⁸ m/s

Formula:

λ = c/ν

λ = (3 × 10⁸)/(100 × 10⁶)

λ = 3 m

3. The amplitude of the magnetic field of an electromagnetic wave is 2 × 10⁻⁷ T. Calculate the amplitude of its electric field.

Given:

B₀ = 2 × 10⁻⁷ T

c = 3 × 10⁸ m/s

Formula:

E₀ = cB₀

E₀ = 3 × 10⁸ × 2 × 10⁻⁷

E₀ = 60 N/C

4. Calculate the energy of a photon having frequency 6 × 10¹⁴ Hz.

Given:

ν = 6 × 10¹⁴ Hz

h = 6.63 × 10⁻³⁴ J s

Formula:

E = hν

E = 6.63 × 10⁻³⁴ × 6 × 10¹⁴

E = 3.978 × 10⁻¹⁹ J

5. Calculate the displacement current if the rate of change of electric flux is 5 × 10⁶ V m/s multiplied by the appropriate area factor such that dΦᴇ/dt = 5 × 10⁶ V m/s.

Given:

dΦᴇ/dt = 5 × 10⁶ V m/s

ε₀ = 8.85 × 10⁻¹² F/m

Formula:

Iᵈ = ε₀(dΦᴇ/dt)

Iᵈ = 8.85 × 10⁻¹² × 5 × 10⁶

Iᵈ = 4.425 × 10⁻⁵ A

PART 5: IMPORTANT FORMULA SHEET

Displacement current:
Iᵈ = ε₀(dΦᴇ/dt)
Electric flux:
Φᴇ = EA
Displacement current for uniform electric field:
Iᵈ = ε₀A(dE/dt)
Ampere-Maxwell law:
∮ B⃗ · dl⃗ = μ₀(I꜀ + Iᵈ)
Speed of electromagnetic waves:
c = 1/√(μ₀ε₀)
Speed, frequency and wavelength relation:
c = νλ
Frequency:
ν = c/λ
Wavelength:
λ = c/ν
Angular frequency:
ω = 2πν
Wave number:
k = 2π/λ
Electric and magnetic field relation:
E₀ = cB₀
Magnetic field amplitude:
B₀ = E₀/c
Energy of a photon:
E = hν
Energy of a photon in terms of wavelength:
E = hc/λ
Electric field energy density:
uᴇ = ½ ε₀E²
Magnetic field energy density:
uᴮ = B²/(2μ₀)
Average electric field energy density:
ūᴇ = ¼ ε₀E₀²
Average magnetic field energy density:
ūᴮ = B₀²/(4μ₀)
Electromagnetic wave speed:
c = 3 × 10⁸ m/s
Planck's constant:
h = 6.63 × 10⁻³⁴ J s

PART 6: ELECTROMAGNETIC SPECTRUM – QUICK TABLE

Electromagnetic Wave	Important Uses
Radio waves	Radio and television broadcasting
Microwaves	Radar and satellite communication
Infrared rays	Remote controls and thermal imaging
Visible light	Vision and photography
Ultraviolet rays	Sterilisation and fluorescence
X-rays	Medical imaging and crystal studies
Gamma rays	Cancer radiotherapy and nuclear research

Order of electromagnetic waves

Increasing frequency:

Radio waves → Microwaves → Infrared → Visible light → Ultraviolet → X-rays → Gamma rays

Increasing wavelength:

Gamma rays → X-rays → Ultraviolet → Visible light → Infrared → Microwaves → Radio waves

Increasing photon energy:

Radio waves → Microwaves → Infrared → Visible light → Ultraviolet → X-rays → Gamma rays

PART 7: IMPORTANT SI UNITS

Physical Quantity	SI Unit
Electric field	N/C or V/m
Magnetic field	Tesla (T)
Electric flux	N m²/C
Displacement current	Ampere (A)
Conduction current	Ampere (A)
Permittivity	F/m
Permeability	H/m
Frequency	Hertz (Hz)
Wavelength	Metre (m)
Angular frequency	rad/s
Wave number	rad/m
Photon energy	Joule (J) or electronvolt (eV)
Speed of electromagnetic waves	m/s
Energy density	J/m³

PART 8: QUICK REVISION POINTS

Electromagnetic waves were theoretically predicted by James Clerk Maxwell.
Electromagnetic waves are produced by accelerating electric charges.
Electromagnetic waves do not require a material medium.
Electromagnetic waves are transverse in nature.
Electric and magnetic fields are perpendicular to each other.
Both electric and magnetic fields are perpendicular to the direction of propagation.
The speed of electromagnetic waves in vacuum is 3 × 10⁸ m/s.
Displacement current is associated with a changing electric field.
Maxwell introduced displacement current to modify Ampere's circuital law.
In a charging capacitor, displacement current equals conduction current.
The relation between electric and magnetic field amplitudes is E₀ = cB₀.
All electromagnetic waves travel with the same speed in vacuum.
Radio waves have the longest wavelengths and lowest frequencies.
Gamma rays have the shortest wavelengths and highest frequencies.
Photon energy is directly proportional to frequency.
The energy density of the electric field equals the energy density of the magnetic field when averaged over a complete cycle.
Microwaves are used in radar and satellite communication.
Infrared rays are associated with thermal radiation.
X-rays are used in medical imaging.
Gamma rays are used in radiotherapy and nuclear research.