Class 12 Physics Notes – Chapter 8: Electromagnetic Waves

1. Displacement Current

According to Ampere’s circuital law, the line integral of the magnetic field is equal to μ₀ times the total current. However, Maxwell found an inconsistency in this law when applied to a charging capacitor. He suggested that a changing electric field produces a current in the empty space between the capacitor plates, which he called Displacement Current (i_d).

Formula: i_d = ε₀ · (dΦ_E / dt)

Where Φ_E is the electric flux. The total current is the sum of conduction current (i_c) and displacement current (i_d).

2. Maxwell’s Equations

Maxwell formulated a set of four equations that completely describe both electric and magnetic fields:

1. Gauss’s Law for Electricity: ∮ E · dA = Q / ε₀
2. Gauss’s Law for Magnetism: ∮ B · dA = 0 (This implies isolated magnetic monopoles do not exist)
3. Faraday’s Law: ∮ E · dl = – dΦ_B / dt
4. Ampere-Maxwell Law: ∮ B · dl = μ₀i_c + μ₀ε₀ (dΦ_E / dt)

3. Electromagnetic Waves

Accelerated charges radiate electromagnetic waves. In an electromagnetic wave, the electric field (E) and magnetic field (B) vary sinusoidally and are perpendicular to each other as well as perpendicular to the direction of propagation of the wave.

Properties of Electromagnetic Waves:

• They do not require any material medium for their propagation.
• They are transverse in nature.
• The speed of EM waves in a vacuum is equal to the speed of light: c = 1 / √(μ₀ε₀) = 3 × 10⁸ m/s.
• The ratio of the amplitudes of electric and magnetic fields is equal to the speed of the wave: E₀ / B₀ = c.
• They carry energy and momentum. The energy is shared equally between the electric and magnetic fields.

4. Electromagnetic Spectrum

The classification of EM waves according to frequency or wavelength is called the electromagnetic spectrum. Different types of waves in increasing order of frequency (or decreasing wavelength) are:

• Radio Waves: Wavelength > 0.1 m. Produced by accelerated motion of charges in conducting wires. Used in radio and TV communication.
• Microwaves: Wavelength 0.1 m to 1 mm. Produced by special vacuum tubes like klystrons and magnetrons. Used in radar systems and microwave ovens.
• Infrared Waves: Wavelength 1 mm to 700 nm. Produced by hot bodies and molecules. Used in physical therapy, remote controls, and night vision.
• Visible Light: Wavelength 700 nm to 400 nm. Emitted by excited atoms. Provides us information about the world via our eyes.
• Ultraviolet (UV) Rays: Wavelength 400 nm to 1 nm. Produced by very hot bodies like the sun. Used in water purifiers and LASIK eye surgery. (Absorbed by the ozone layer).
• X-Rays: Wavelength 1 nm to 10^-3 nm. Produced by bombarding a metal target with high-energy electrons. Used as a diagnostic tool in medicine (X-ray images).
• Gamma Rays: Wavelength < 10^-3 nm. Produced in nuclear reactions and radioactive decay. Used in medicine to destroy cancer cells.

Multiple Choice Questions (MCQs)

1. The concept of displacement current was introduced by:
a) Faraday
b) Ampere
c) Maxwell
d) Hertz
Answer: (c) Maxwell

2. Electromagnetic waves are produced by:
a) Stationary charges
b) Charges moving with uniform velocity
c) Accelerated or oscillating charges
d) Neutral particles
Answer: (c) Accelerated or oscillating charges

3. The speed of electromagnetic waves in a vacuum is given by:
a) μ₀ε₀
b) 1 / (μ₀ε₀)
c) 1 / √(μ₀ε₀)
d) √(μ₀ε₀)
Answer: (c) 1 / √(μ₀ε₀)

4. Which of the following EM waves has the highest frequency?
a) Radio waves
b) X-rays
c) Gamma rays
d) Microwaves
Answer: (c) Gamma rays

5. The direction of propagation of an electromagnetic wave is given by the direction of:
a) Vector E
b) Vector B
c) E × B
d) B × E
Answer: (c) E × B

6. Which waves are primarily used in RADAR systems for aircraft navigation?
a) Infrared waves
b) Microwaves
c) UV rays
d) X-rays
Answer: (b) Microwaves

7. The ratio of the amplitude of the electric field to the magnetic field (E₀ / B₀) for an EM wave in a vacuum is equal to:
a) 1
b) Speed of sound
c) Speed of light (c)
d) ε₀
Answer: (c) Speed of light (c)

8. The ozone layer in the atmosphere absorbs which of the following radiations from the sun?
a) Infrared
b) Microwaves
c) Ultraviolet (UV)
d) Visible light
Answer: (c) Ultraviolet (UV)

9. Gauss’s Law for magnetism states that ∮ B · dA = 0. This implies that:
a) Magnetic field is always zero
b) Magnetic monopoles do not exist
c) Magnetic field lines are straight
d) Magnetic force is conservative
Answer: (b) Magnetic monopoles do not exist

10. Which of the following is NOT an electromagnetic wave?
a) X-rays
b) Gamma rays
c) Sound waves
d) Infrared waves
Answer: (c) Sound waves

Textbook Exercise Solutions

Q 8.1: Figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 mm. The capacitor is being charged by an external source. 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.
(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
Solution:
(a) Given: r = 0.12 m, d = 5.0 mm = 5.0 × 10^-3 m, I = 0.15 A.
Area A = πr² = 3.14 × (0.12)² = 0.0452 m².
Capacitance C = ε₀A / d = (8.854 × 10^-12 × 0.0452) / (5.0 × 10^-3) ≈ 80.1 pF (80.1 × 10^-12 F).
Rate of change of potential difference dV/dt = I / C = 0.15 / (80.1 × 10^-12) = 1.87 × 10⁹ V s^-1.
(b) The displacement current is exactly equal to the conduction current during charging. So, i_d = i_c = 0.15 A.
(c) Yes, Kirchhoff’s first rule is valid at each plate if we consider both conduction current and displacement current. The current entering the plate (conduction) equals the current leaving it (displacement).

Q 8.2: A parallel plate capacitor 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.
Solution:
(a) Given: C = 100 pF = 100 × 10^-12 F, V_rms = 230 V, ω = 300 rad s^-1.
RMS conduction current I_rms = V_rms / X_C = V_rms × ω × C
I_rms = 230 × 300 × 100 × 10^-12 = 6.9 μA (6.9 × 10^-6 A).
(b) Yes, the conduction current in the wires is always equal to the displacement current between the plates.
(c) Formula for B inside capacitor at distance r from axis: B = (μ₀ r / 2πR²) · i_d
Amplitude of current I₀ = √2 × I_rms = 1.414 × 6.9 μA = 9.76 μA.
B₀ = (4π × 10^-7 × 0.03 / 2π × (0.06)²) × 9.76 × 10^-6
B₀ = (2 × 10^-7 × 0.03 / 0.0036) × 9.76 × 10^-6 = 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?
Solution:
The physical quantity that is the same for all of them is their speed in a vacuum. All electromagnetic waves travel with the same speed in a vacuum, which is c = 3 × 10⁸ m/s.

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?
Solution:
The electric field (E) and magnetic field (B) vectors are perpendicular to each other and also perpendicular to the direction of propagation (z-axis). Therefore, E and B lie in the x-y plane.
Given f = 30 MHz = 30 × 10⁶ Hz. Wavelength λ = c / f = (3 × 10⁸) / (30 × 10⁶) = 10 m.

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?
Solution:
For f₁ = 7.5 MHz: λ₁ = c / f₁ = (3 × 10⁸) / (7.5 × 10⁶) = 40 m.
For f₂ = 12 MHz: λ₂ = c / f₂ = (3 × 10⁸) / (12 × 10⁶) = 25 m.
The corresponding wavelength band 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?
Solution:
The frequency of the electromagnetic waves produced is exactly the same as the frequency of the oscillating charged particle. Therefore, the frequency of the EM waves is 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?
Solution:
Given: B₀ = 510 nT = 510 × 10^-9 T. We know E₀ / B₀ = c.
E₀ = c × B₀ = (3 × 10⁸) × (510 × 10^-9) = 153 V/m (or 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 v = 50.0 MHz. (a) Determine, B₀, ω, k, and λ. (b) Find expressions for E and B.
Solution:
(a) Given: E₀ = 120 N/C, f = 50.0 × 10⁶ Hz.
B₀ = E₀ / c = 120 / (3 × 10⁸) = 4 × 10^-7 T = 400 nT.
Angular frequency ω = 2πf = 2 × 3.14 × 50 × 10⁶ = 3.14 × 10⁸ rad/s.
Wavelength λ = c / f = (3 × 10⁸) / (50 × 10⁶) = 6.00 m.
Wave number k = 2π / λ = (2 × 3.14) / 6.00 = 1.05 rad/m.
(b) Assuming the wave propagates along the x-axis, and E is along the y-axis, then B is along the z-axis:
E = 120 sin(1.05x – 3.14 × 10⁸ t) jˆ (V/m)
B = 4.0 × 10^-7 sin(1.05x – 3.14 × 10⁸ t) kˆ (T)

Q 8.9: The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv to obtain the photon energy in units of eV for different parts of the electromagnetic spectrum.
Solution:
For a wavelength λ = 1 m (Radio waves):
E = hc / λ = (6.63 × 10^-34 × 3 × 10⁸) / 1 = 19.89 × 10^-26 J.
In eV: E = (19.89 × 10^-26) / (1.6 × 10^-19) = 1.24 × 10^-6 eV.
(Similarly, energy can be calculated for other wavelengths by plugging them into the formula E = 1240 / λ (in nm) eV).

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.
(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.
Solution:
(a) Given f = 2.0 × 10¹⁰ Hz. λ = c / f = (3 × 10⁸) / (2.0 × 10¹⁰) = 1.5 × 10^-2 m (1.5 cm).
(b) B₀ = E₀ / c = 48 / (3 × 10⁸) = 1.6 × 10^-7 T.
(c) Average energy density of electric field u_E = 1/2 ε₀ E²_rms = 1/4 ε₀ E₀².
Average energy density of magnetic field u_B = 1/2 (B²_rms / μ₀) = 1/4 (B₀² / μ₀).
We know E₀ = c B₀ and c² = 1 / (μ₀ε₀).
u_E = 1/4 ε₀ (c B₀)² = 1/4 ε₀ c² B₀² = 1/4 ε₀ [1 / (μ₀ε₀)] B₀² = 1/4 (B₀² / μ₀) = u_B.
Thus, u_E = u_B.