Chapter 8: Electromagnetic Waves Application Questions | physics Class 12 CBSE

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Chapter 8: Electromagnetic Waves Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point, E = 6.3 ĵ V/m. Determine the magnitude and direction of B at that point.

Using B₀ = E₀/c, the magnitude of B is B = 6.3/(3 × 10⁸) = 2.1 × 10⁻⁸ T. The relation B₀ = E₀/c is derived from Maxwell's equations and holds for all EM waves in free space.
The wave propagates along the x-direction and E is along the y-direction (ĵ). Since E, B, and the direction of propagation must be mutually perpendicular, B must be along the z-direction.
Using vector algebra, E × B must point along the direction of propagation (+x̂). Since ĵ × k̂ = î, B is along the +z direction. Thus B = 2.1 × 10⁻⁸ k̂ T.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 2

Applied Concept

The magnetic field in a plane electromagnetic wave is By = (2 × 10⁻⁷) T sin(0.5 × 10³ x + 1.5 × 10¹¹ t). Find the wavelength and frequency of the wave, and write the corresponding expression for the electric field.

Comparing with the standard form B = B₀ sin(kx + ωt): k = 0.5 × 10³ rad/m, so λ = 2π/k = 2π/(0.5 × 10³) ≈ 1.26 cm. The angular frequency ω = 1.5 × 10¹¹ rad/s, giving ν = ω/2π ≈ 23.9 GHz.
The amplitude of E is E₀ = B₀c = (2 × 10⁻⁷)(3 × 10⁸) = 60 V/m. The wave propagates along the x-direction with B along y.
The electric field is perpendicular to both B and the direction of propagation, so it lies along the z-axis: Ez = 60 sin(0.5 × 10³ x + 1.5 × 10¹¹ t) V/m.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

With reference to a parallel plate capacitor connected to an AC source, explain why displacement current is needed to maintain the continuity of current in the circuit.

In an AC circuit with a capacitor, conduction current flows in the external wires as charge builds up on the plates. However, no physical charge crosses the gap between the plates — hence conduction current is zero inside the capacitor.
Without displacement current, Kirchhoff's junction rule (which requires current continuity) would be violated at the capacitor plates: current flows in but apparently does not flow out.
Maxwell's displacement current iₐ = ε₀(dΦ_E/dt) flows between the plates due to the changing electric field, effectively completing the circuit. The total current (conduction + displacement) is continuous throughout the circuit.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A radio station broadcasts at a frequency of 600 kHz. What is the wavelength of the transmitted wave? If another station broadcasts at 100 MHz (FM), what is its wavelength?

For the AM station at 600 kHz: λ = c/ν = (3 × 10⁸)/(6 × 10⁵) = 500 m. This is in the AM band (530 kHz to 1710 kHz) as described in the chapter.
For the FM station at 100 MHz: λ = c/ν = (3 × 10⁸)/(10⁸) = 3 m. This falls in the FM band (88–108 MHz) which is used for frequency modulated radio broadcasting.
The AM wavelength (500 m) is vastly larger than the FM wavelength (3 m), explaining why AM antennas and propagation characteristics differ significantly from FM systems.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A microwave oven heats food containing water very efficiently but does not heat dry glass or plastic. Explain this observation using properties of electromagnetic waves.

The frequency of microwaves in a microwave oven is specifically selected to match the resonant frequency of water molecules. When irradiated, water molecules absorb the microwave energy efficiently and convert it to kinetic energy (heat).
Dry glass and plastic contain no (or very little) water, and their molecular resonant frequencies do not match the microwave frequency used in the oven. As a result, they absorb very little microwave energy and remain cool.
This application of resonant absorption of electromagnetic energy by specific molecules is a direct consequence of the wave-matter interaction principle — the degree of absorption depends on how well the wave frequency matches the natural oscillation frequency of the molecules.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 6

Applied Concept

With reference to the chapter, explain why one cannot get tanned or sunburned while sitting behind a glass window, despite the sun's radiation reaching the glass.

Sunburn and tanning are caused by ultraviolet (UV) radiation from the sun, not by visible light. UV radiation has wavelengths from about 400 nm down to 0.6 nm and sufficient energy to damage skin cells and stimulate melanin production.
Ordinary glass absorbs ultraviolet radiation and does not transmit it. Visible light (400–700 nm) passes through glass relatively easily, which is why you can see through glass, but the UV component is blocked.
Since UV radiation is the causative agent of tanning and sunburn, and since it is absorbed by the glass before reaching the person inside, no tanning or sunburn occurs — even though the person may feel warm from the infrared radiation and see light through the window.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

Explain the role of the ozone layer in protecting life on Earth from UV radiation, and why its depletion by CFCs is a matter of international concern.

The ozone layer, located in the atmosphere at an altitude of about 40–50 km, absorbs most of the ultraviolet radiation emitted by the sun before it reaches the earth's surface.
UV radiation in large quantities has harmful biological effects: it damages DNA and cellular structures, causes skin cancer, tanning, and can harm ecosystems. Without ozone absorption, life on earth would be exposed to lethal UV doses.
Chlorofluorocarbons (CFCs) such as freon, when released into the atmosphere, chemically destroy ozone molecules, thinning the ozone layer. This allows more UV radiation to reach the earth's surface, increasing the risk of skin cancer and ecological damage, making it a global environmental emergency.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

An electromagnetic wave has an electric field amplitude E₀ = 120 N/C and frequency ν = 50 MHz. Find B₀, ω, k, and λ.

The magnetic field amplitude: B₀ = E₀/c = 120/(3 × 10⁸) = 4 × 10⁻⁷ T = 400 nT.
Angular frequency: ω = 2πν = 2π × 50 × 10⁶ ≈ 3.14 × 10⁸ rad/s. Wavelength: λ = c/ν = (3 × 10⁸)/(50 × 10⁶) = 6 m.
Wave number: k = 2π/λ = 2π/6 ≈ 1.05 rad/m. Alternatively k = ω/c = (3.14 × 10⁸)/(3 × 10⁸) ≈ 1.05 rad/m. This confirms the relation ω = ck.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

Earth satellites use infrared detectors to monitor crop growth and military activities. Explain why infrared waves are preferred over visible light for such satellite imaging.

Infrared waves are emitted by all warm bodies and can be detected even at night or through cloud cover (which blocks visible light), making IR satellite imaging effective around the clock and in adverse weather.
Different crops, healthy versus diseased vegetation, and warm military equipment emit infrared radiation at different intensities, allowing infrared satellite imaging to distinguish between them based on thermal signatures.
Infrared detectors (thermopiles, bolometers, IR photographic film) are sensitive to this range, and the wavelength difference between IR emission from crops (~microns) and background terrain allows detailed mapping of agricultural and military activities invisible to visible-light cameras.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 10

Applied Concept

A radio can tune to any station in the band 7.5 MHz to 12 MHz. What is the corresponding wavelength band? Which part of the EM spectrum does this correspond to?

For ν₁ = 7.5 MHz: λ₁ = c/ν₁ = (3 × 10⁸)/(7.5 × 10⁶) = 40 m. For ν₂ = 12 MHz: λ₂ = c/ν₂ = (3 × 10⁸)/(12 × 10⁶) = 25 m.
The corresponding wavelength band is from 25 m to 40 m.
Both wavelengths are much greater than 0.1 m, placing them firmly in the radio wave region of the electromagnetic spectrum. This range falls within the short wave bands (frequencies up to 54 MHz are used for short wave radio).

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Chapter 8: Electromagnetic Waves | Application Questions

A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point, E = 6.3 ĵ V/m. Determine the magnitude and direction of B at that point.

The magnetic field in a plane electromagnetic wave is By = (2 × 10⁻⁷) T sin(0.5 × 10³ x + 1.5 × 10¹¹ t). Find the wavelength and frequency of the wave, and write the corresponding expression for the electric field.

With reference to a parallel plate capacitor connected to an AC source, explain why displacement current is needed to maintain the continuity of current in the circuit.

A radio station broadcasts at a frequency of 600 kHz. What is the wavelength of the transmitted wave? If another station broadcasts at 100 MHz (FM), what is its wavelength?

A microwave oven heats food containing water very efficiently but does not heat dry glass or plastic. Explain this observation using properties of electromagnetic waves.

With reference to the chapter, explain why one cannot get tanned or sunburned while sitting behind a glass window, despite the sun's radiation reaching the glass.

Explain the role of the ozone layer in protecting life on Earth from UV radiation, and why its depletion by CFCs is a matter of international concern.

An electromagnetic wave has an electric field amplitude E₀ = 120 N/C and frequency ν = 50 MHz. Find B₀, ω, k, and λ.

Earth satellites use infrared detectors to monitor crop growth and military activities. Explain why infrared waves are preferred over visible light for such satellite imaging.

A radio can tune to any station in the band 7.5 MHz to 12 MHz. What is the corresponding wavelength band? Which part of the EM spectrum does this correspond to?

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