Class 12Physics · EM WavesFull chapter

Electromagnetic Waves

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Displacement Current and the Ampere–Maxwell Law

Quick answer Ampere's original circuital law breaks down for circuits containing a charging or discharging capacitor; Maxwell fixed this by introducing displacement current, a term due to a changing electric flux that produces a magnetic field exactly like a real current does.

Ampere's circuital law connects the magnetic field circulating around a closed loop to the conduction current Ic passing through any surface bounded by that loop: ∮B·dl = μ0Ic. This works perfectly for steady, unbroken currents such as the current in a long straight wire. But consider a wire carrying a charging current into one plate of a capacitor. If we choose a flat surface that cuts straight through the wire, the enclosed current is I. If instead we choose a bulged surface that passes through the gap between the plates (where no charge actually crosses), the enclosed conduction current is zero. Ampere's law, applied to the same loop, now gives two different answers for the same magnetic field — a contradiction.

Maxwell resolved this by noticing that although no charge flows across the gap, the electric field E between the plates is changing with time as charge accumulates, so the electric flux ΦE through the bulged surface is also changing. He proposed that a time-varying electric flux is exactly as effective at producing a magnetic field as a conduction current. He defined a new quantity, the displacement current, Id = ε0(dΦE/dt), and generalised Ampere's law to the Ampere–Maxwell law: ∮B·dl = μ0Ic + μ0ε0(dΦE/dt). With this correction, both surfaces give the same, consistent magnetic field, because on the bulged surface Ic = 0 but Id = I, while on the flat surface Ic = I and Id = 0.

Displacement current is not a flow of charge; it is a bookkeeping device that makes the total current (conduction + displacement) continuous around any closed circuit, even where an actual gap exists. It has the same units as current (ampere) and produces a genuine magnetic field, but it involves no moving charge and no heat dissipation (no I²R loss). This single correction was the missing piece that let Maxwell show that electric and magnetic fields could sustain each other in empty space, travelling as a wave — the electromagnetic wave.

Worked example. A parallel-plate capacitor with circular plates of radius R = 6.0 cm is being charged by a constant current I = 0.15 A. Find the rate of change of the electric field between the plates and confirm the displacement current equals the charging current.

  1. Plate area: A = πR² = π(0.06)² = 1.131 × 10⁻² m².
  2. The field between the plates is E = Q/(ε0A), so dE/dt = (dQ/dt)/(ε0A) = I/(ε0A) = 0.15/(8.85 × 10⁻¹² × 1.131 × 10⁻²) ≈ 1.5 × 10¹² V/(m·s).
  3. Displacement current: Id = ε0A(dE/dt) = ε0A × I/(ε0A) = I = 0.15 A.

So Id exactly equals the conduction current I flowing in the wires — this is true in general, not just for this geometry, and is what keeps the current continuous through the capacitor gap.

Ampere's circuital law (original) ∮ B · dl = μ₀ I_c T·m · Valid only when the current is steady/continuous through the chosen surface.
Electric flux Φ_E = E · A V·m · For a uniform field E over area A, e.g. between capacitor plates.
Displacement current I_d = ε₀ (dΦ_E / dt) A · Arises purely from a changing electric flux; has the same magnetic effect as a real current.
Ampere–Maxwell law ∮ B · dl = μ₀ I_c + μ₀ε₀ (dΦ_E / dt) T·m · Generalised form of Ampere's law valid in all situations, including capacitor charging.
Remember
  • Ampere's original law ∮B·dl = μ0Ic gives inconsistent results for a circuit with a capacitor unless a correction term is added.
  • A changing electric flux Φ_E between the capacitor plates acts as a source of magnetic field, called the displacement current: I_d = ε0(dΦ_E/dt).
  • The Ampere–Maxwell law ∮B·dl = μ0Ic + μ0ε0(dΦ_E/dt) is valid for both steady and time-varying fields.
  • Displacement current involves no actual charge motion and no resistive heating, but produces a real magnetic field.
  • In any single circuit branch, the displacement current between capacitor plates equals the conduction current in the connecting wires at every instant.
  • The need for displacement current directly led Maxwell to predict self-sustaining electromagnetic waves in vacuum.

Generation, Nature and Properties of Electromagnetic Waves

Quick answer Maxwell's four equations show that an oscillating electric field generates an oscillating magnetic field and vice versa, so the two fields sustain each other as a transverse wave that travels through vacuum at the speed of light.

Maxwell combined Gauss's law for electricity, Gauss's law for magnetism, Faraday's law of electromagnetic induction, and his own Ampere–Maxwell law into a single consistent set of equations. Faraday's law says a changing B produces an E; the Ampere–Maxwell law says a changing E produces a B. Together, these mean that in a region with no charges or conduction currents, an oscillating electric field creates an oscillating magnetic field, which in turn recreates an oscillating electric field a little further away, and so on — the disturbance propagates outward on its own, without needing a medium. This self-sustaining disturbance is an electromagnetic wave, and Maxwell calculated its speed in vacuum from purely electric and magnetic constants: c = 1/√(μ0ε0) ≈ 3 × 10⁸ m/s — a value that matched the already-measured speed of light so closely that Maxwell concluded light itself is an electromagnetic wave. Electromagnetic waves are produced whenever a charge is accelerated, most commonly by charges oscillating in a circuit or antenna; the German physicist Heinrich Hertz first generated and detected such waves (radio waves) in the laboratory in 1887, experimentally confirming Maxwell's theory.

A plane electromagnetic wave has three key features. First, it is transverse: the electric field E and magnetic field B oscillate perpendicular to the direction in which the wave travels (unlike sound, which is longitudinal). Second, E and B are also perpendicular to each other, and the direction of propagation is along E × B. Third, E and B oscillate in phase — they reach their maximum and zero values at the same instant and the same point in space — with their amplitudes always related by E0 = cB0. Electromagnetic waves carry energy, stored partly in the electric field and partly in the magnetic field; at every instant the two contributions are equal (uE = uB), and the total energy density is u = ½ε0E² + B²/(2μ0). The time-averaged rate at which this energy crosses unit area (the intensity) is I = ½ε0cE0². Electromagnetic waves also carry momentum, so they exert a (very small) radiation pressure on any surface they strike.

Worked example 1. The electric field amplitude of a plane electromagnetic wave in vacuum is E0 = 60 V/m. Find the amplitude of the magnetic field.

B0 = E0/c = 60/(3 × 10⁸) = 2.0 × 10⁻⁷ T = 200 nT. Note how small B0 is compared with E0 numerically — this is simply because c is a large number; the energies stored in the E and B fields are nevertheless equal.

Worked example 2. An FM radio wave has frequency 91 MHz. Find its wavelength using c = νλ.

λ = c/ν = (3 × 10⁸)/(91 × 10⁶) ≈ 3.3 m. This wavelength (a few metres) is why FM radio antennas are of that order in length.

Speed of EM waves in vacuum c = 1 / √(μ₀ε₀) m/s · ≈ 3 × 10⁸ m/s; matching the measured speed of light showed light is an EM wave.
Wave relation c = ν λ m/s · Connects frequency and wavelength for any EM wave in vacuum.
Amplitude relation E₀ = c B₀ — · Ratio of peak electric field to peak magnetic field always equals c.
Energy density u = ½ε₀E² + B²/(2μ₀) J/m³ · Instantaneous energy density; the electric and magnetic terms are equal at every instant (uE = uB).
Average intensity I = ½ ε₀ c E₀² W/m² · Time-averaged power crossing unit area, perpendicular to the direction of propagation.
Remember
  • An oscillating charge produces oscillating E and B fields that regenerate each other and propagate through vacuum as an electromagnetic wave.
  • Hertz's 1887 experiments produced and detected radio waves, experimentally confirming Maxwell's prediction.
  • EM waves are transverse: E, B and the direction of propagation are mutually perpendicular, with direction of travel along E × B.
  • E and B oscillate in phase, with amplitude ratio E0 = cB0.
  • Speed in vacuum, c = 1/√(μ0ε0) ≈ 3 × 10⁸ m/s, is the same for every electromagnetic wave regardless of frequency.
  • Energy is shared equally between the electric and magnetic fields at every instant; EM waves also carry momentum and exert radiation pressure.

The Electromagnetic Spectrum: Radio Waves and Microwaves

Quick answer The electromagnetic spectrum arranges all EM waves by wavelength/frequency with no sharp boundaries; radio waves and microwaves are the longest-wavelength, lowest-frequency members, produced by oscillating currents in circuits and used for wireless transmission, radar and cooking.

Electromagnetic waves exist over an enormous continuous range of wavelengths and frequencies, collectively called the electromagnetic spectrum. Moving from long wavelength/low frequency to short wavelength/high frequency, the spectrum is broadly divided into radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. These bands overlap and have no rigid boundaries; a wave is classified mainly by how it is produced and detected. All of them travel at the same speed c in vacuum and differ only in wavelength/frequency (and hence photon energy).

Radio waves have wavelengths roughly from about 0.1 m to several hundred metres (frequencies from a few hertz up to about 10⁹ Hz). They are generated by accelerating charges in a conducting wire or antenna — an oscillating current in a circuit — and are used for AM/FM radio and television transmission. Longer-wavelength radio waves (AM band) travel further by reflecting off the ionosphere, while shorter waves (FM/TV band, VHF/UHF) travel more by line of sight.

Microwaves have shorter wavelengths, roughly from about 1 mm to 0.1 m (frequencies of order 10⁹–10¹¹ Hz). They are produced by special vacuum-tube devices such as klystrons, magnetrons and Gunn diodes rather than ordinary circuits, because such short wavelengths need very rapidly oscillating currents. Microwaves are used in radar systems (for detecting the speed and location of objects such as aircraft and vehicles), microwave ovens (their frequency is tuned to be absorbed efficiently by water molecules in food, heating it), and in analysing the fine structure of atoms and molecules.

Worked example. A microwave oven operates at a frequency of 2.45 GHz. Find its wavelength, and state which spectral band this falls in.

λ = c/ν = (3 × 10⁸)/(2.45 × 10⁹) ≈ 0.122 m ≈ 12.2 cm. Since this lies between about 1 mm and 0.1 m (up to a few tens of cm, by common convention), it is squarely in the microwave region — consistent with its use in microwave ovens.

Remember
  • The EM spectrum runs, in order of increasing frequency: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma ray, with overlapping, not sharply divided, bands.
  • All EM waves travel at the same speed c in vacuum; only wavelength/frequency differ across the spectrum.
  • Radio waves (longest wavelength) are produced by oscillating currents in antennas/circuits and used for radio and TV transmission.
  • Microwaves need special devices (klystron, magnetron, Gunn diode) to generate, because ordinary circuits cannot oscillate fast enough.
  • Microwaves are used in radar and microwave ovens, and for probing molecular/atomic structure.
  • A wave's position in the spectrum is identified by how it is produced/detected, not by a fixed wavelength cut-off.

The Electromagnetic Spectrum: Infrared, Visible Light and Ultraviolet

Quick answer Infrared is emitted by warm/hot bodies and molecular vibrations, visible light is the narrow band detected by the human eye, and ultraviolet is produced by very hot objects and electric discharges; all three have widespread everyday, medical and environmental roles.

Infrared (IR) waves have wavelengths roughly from 700 nm up to about 1 mm. They are produced by hot bodies and by vibrating/rotating molecules, and are often called heat waves because they are readily absorbed by most materials, increasing their thermal motion (heating them). The Sun is a strong source of infrared. Applications include remote controls, thermal/night-vision imaging, physiotherapy for muscular treatment, and infrared astronomy for studying cool objects hidden behind interstellar dust. Infrared radiation is also central to the greenhouse effect: incoming sunlight (mostly visible) warms the Earth's surface, which re-radiates the absorbed energy as infrared; greenhouse gases such as CO2 and water vapour absorb part of this outgoing infrared and re-emit it, trapping heat in the atmosphere.

Visible light occupies a very narrow band, roughly 400 nm (violet) to 700 nm (red), and is the only part of the spectrum the human eye can detect. It is produced mainly by electrons making transitions between energy levels in atoms and by hot, glowing bodies. Different wavelengths within this band are perceived as different colours, and a mixture of all of them appears white. Because it is such a small slice of the full spectrum, most of the universe's electromagnetic activity is invisible to us and can only be studied using instruments sensitive to other bands.

Ultraviolet (UV) radiation has wavelengths from about 1 nm to 400 nm, produced by very hot objects (like the Sun), specially designed lamps, and electric discharges through gases. UV carries more energy per photon than visible light, is capable of ionising atoms, and can damage living cells — prolonged exposure causes sunburn, premature skin ageing and can trigger skin cancer, and it can harm the eyes. Useful applications include sterilising water and surgical instruments (its germicidal effect kills bacteria), checking for forged banknotes/documents (fluorescence), LASIK eye surgery, and detecting invisible ink/fingerprints. Most of the harmful UV from the Sun is absorbed by the ozone (O3) layer in the upper atmosphere before it reaches the ground, which is why depletion of the ozone layer (caused by chlorofluorocarbons, CFCs) is an environmental concern.

Worked example. A hot filament radiates strongly at a wavelength of 550 nm. Find the frequency of this radiation and identify which part of the spectrum it belongs to.

ν = c/λ = (3 × 10⁸)/(550 × 10⁻⁹) ≈ 5.45 × 10¹⁴ Hz. Since 550 nm lies between 400 nm and 700 nm, this radiation is visible light — it would appear green to the eye.

Remember
  • Infrared (≈700 nm – 1 mm) is produced by hot bodies and molecular vibrations; used in remote controls, thermal imaging and physiotherapy.
  • Infrared re-radiated by Earth's surface, partly trapped by greenhouse gases, drives the natural greenhouse effect.
  • Visible light (≈400–700 nm) is the only band detected by the human eye, produced mainly by atomic electron transitions.
  • Ultraviolet (≈1–400 nm) is produced by very hot sources/electric discharges; it is energetic enough to ionise and damage tissue.
  • UV is used for sterilisation, LASIK surgery and detecting fluorescence; excess exposure causes sunburn and skin cancer risk.
  • The ozone layer absorbs most solar UV, shielding life on Earth; its depletion (by CFCs) increases harmful UV reaching the surface.

The Electromagnetic Spectrum: X-rays, Gamma Rays and Overall Summary

Quick answer X-rays are produced by decelerating high-energy electrons and gamma rays by nuclear processes; both are highly penetrating, high-frequency radiations with vital medical uses but require careful shielding because of their damaging effect on living tissue.

X-rays have wavelengths roughly from 10⁻⁸ m down to about 10⁻¹³ m (0.0001 nm to 10 nm), making them far more energetic than UV. They are produced when high-speed electrons, accelerated through a large potential difference in an X-ray tube, are suddenly decelerated on striking a heavy metal target. Because of their short wavelength, X-rays can penetrate soft tissue but are absorbed more by denser matter like bone, which is why they are used for medical diagnostic imaging (radiography) of bones and teeth, for airport baggage/security scanning, and in industry and crystallography to study the atomic arrangement inside crystals. X-rays are ionising radiation, so excessive or repeated exposure can damage or kill living cells; their medical use is always kept to the minimum necessary dose with proper shielding for staff.

Gamma rays lie at the extreme high-frequency, short-wavelength end of the spectrum, with wavelengths below about 10⁻¹⁰–10⁻¹⁴ m, overlapping the X-ray region but generally shorter. Unlike X-rays, which come from electron processes, gamma rays originate in the nucleus — emitted during radioactive decay of unstable nuclei and in nuclear reactions. They are the most penetrating and energetic of all electromagnetic radiations, requiring thick shielding (lead, concrete) to stop. Controlled use of gamma radiation includes cancer radiotherapy (destroying malignant cells), sterilising medical equipment and food, and gamma-ray astronomy, but uncontrolled exposure is highly hazardous to living organisms because of its strong ionising power.

Taken together, the full electromagnetic spectrum — radio, microwave, infrared, visible, ultraviolet, X-ray and gamma ray — is a single family of transverse waves differing only in wavelength/frequency, all travelling at speed c in vacuum, all obeying c = νλ, and all originating from accelerated charges (in circuits, atoms or nuclei, depending on the band). As frequency increases across this list, wavelength decreases and the radiation generally becomes more penetrating and more capable of ionising matter, which is why the higher bands (UV, X-ray, gamma) need careful handling while the lower bands (radio, microwave, infrared) are comparatively benign in normal use.

Worked example. A gamma ray emitted during a radioactive decay has wavelength 2 × 10⁻¹³ m. Compare its frequency with that of a typical X-ray of wavelength 2 × 10⁻¹⁰ m.

ν(gamma) = c/λ = (3 × 10⁸)/(2 × 10⁻¹³) = 1.5 × 10²¹ Hz. ν(X-ray) = (3 × 10⁸)/(2 × 10⁻¹⁰) = 1.5 × 10¹⁸ Hz. The gamma ray's frequency is 1000 times higher, consistent with gamma rays generally being far more energetic and penetrating than X-rays.

Remember
  • X-rays (≈10⁻⁴–10 nm) are produced by rapidly decelerating high-energy electrons in an X-ray tube; used in medical imaging, security scanning and crystallography.
  • Gamma rays (shortest wavelength, highest frequency) originate from nuclear processes such as radioactive decay, not from electron transitions.
  • Both X-rays and gamma rays are ionising and can damage living tissue, requiring shielding and controlled exposure.
  • Gamma rays are used in cancer radiotherapy and sterilisation; X-rays in diagnostic imaging and industrial testing.
  • Across the full spectrum, as frequency rises from radio to gamma, wavelength falls and the radiation becomes more penetrating/ionising.
  • Every band of the spectrum obeys the same relations c = νλ and travels at the same speed c in vacuum; only the source mechanism and typical use differ.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

∮ B · dl = μ₀ I_c
Ampere's circuital law (original)T·m
Φ_E = E · A
Electric fluxV·m
I_d = ε₀ (dΦ_E / dt)
Displacement currentA
∮ B · dl = μ₀ I_c + μ₀ε₀ (dΦ_E / dt)
Ampere–Maxwell lawT·m
c = 1 / √(μ₀ε₀)
Speed of EM waves in vacuumm/s
c = ν λ
Wave relationm/s
E₀ = c B₀
Amplitude relation
u = ½ε₀E² + B²/(2μ₀)
Energy densityJ/m³
I = ½ ε₀ c E₀²
Average intensityW/m²

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Displacement current easy

Displacement current between the plates of a charging capacitor arises because of:

Q2 Nature of EM waves easy

Which of the following is NOT a property of electromagnetic waves travelling in vacuum?

Q3 Electromagnetic spectrum easy

Arrange X-rays, microwaves, visible light and gamma rays in order of increasing wavelength.

Q4 Nature of EM waves medium

The electric field amplitude of a plane EM wave in vacuum is 300 V/m. What is the amplitude of its magnetic field?

Q5 Energy in EM waves medium

In a propagating electromagnetic wave, the energy carried is:

Q6 Displacement current medium

A parallel-plate capacitor is being charged by a constant current of 2 A. What is the displacement current between its plates?

Q7 Electromagnetic spectrum medium

Which part of the electromagnetic spectrum is commonly used to sterilise surgical instruments and drinking water because of its germicidal effect?

Q8 Electromagnetic spectrum medium

An FM radio station broadcasts at 88 MHz. What is the corresponding wavelength?

Q9 Displacement current hard

The Ampere–Maxwell law modifies Ampere's original circuital law by adding a term proportional to the:

Q10 Energy in EM waves hard

The peak electric field of a laser beam in vacuum is 6.0 × 10³ V/m. What is the average intensity of the beam? (Use I = ½ε0cE0², ε0 = 8.85 × 10⁻¹² F/m)

Q11 Displacement current hard

A circular parallel-plate capacitor of plate radius 5 cm carries a peak conduction current of 2 mA while charging. Using B = μ0I0r/(2πR²), find the amplitude of the magnetic field at r = 2 cm from the axis, between the plates.

Q12 Generation of EM waves hard

Which of Maxwell's equations explains how a magnetic field can be produced by a changing electric field even where the conduction current is zero, making electromagnetic wave propagation through vacuum possible?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 A parallel-plate capacitor made of two circular plates, each of radius 12 cm, separated by 5.0 cm, is being charged by a constant current of 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 junction rule valid at each plate of the capacitor? Explain.Displacement current

(a) Capacitance: C = ε₀A/d, with A = πR² = π(0.12)² = 4.524 × 10⁻² m² and d = 0.05 m.

C = (8.85 × 10⁻¹² × 4.524 × 10⁻²)/0.05 ≈ 8.0 × 10⁻¹² F = 8.0 pF.

Rate of change of potential difference: since I = C(dV/dt), dV/dt = I/C = 0.15/(8.0 × 10⁻¹²) ≈ 1.87 × 10¹⁰ V/s.

(b) Displacement current: By the Ampere–Maxwell law, the displacement current between the plates always equals the conduction current in the connecting wires, so I_d = I_c = 0.15 A.

(c) Kirchhoff's junction rule: If only conduction current is considered, the rule appears to fail at a capacitor plate, since current 0.15 A flows into the plate through the wire but no conduction current crosses the gap to the other plate. However, once the displacement current (0.15 A) between the plates is included as the 'continuation' of the current, the total current (conduction + displacement) is continuous and equal on both sides — so Kirchhoff's junction rule remains valid when displacement current is properly accounted for.

2 A parallel-plate capacitor with circular plates of radius R = 6.0 cm has a capacitance C = 100 pF. It is connected to a 230 V AC supply with an 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) Find the amplitude of the magnetic field B at a point 3.0 cm from the axis, between the plates.Displacement current

(a) rms conduction current: Capacitive reactance X_C = 1/(ωC) = 1/(300 × 100 × 10⁻¹²) = 1/(3 × 10⁻⁸) = 3.33 × 10⁷ Ω.

I_rms = V_rms/X_C = 230/(3.33 × 10⁷) ≈ 6.9 × 10⁻⁶ A = 6.9 μA.

(b) Comparison: Yes. By the Ampere–Maxwell law the displacement current between the plates equals the conduction current in the leads at every instant, so the displacement current also has rms value 6.9 μA.

(c) Magnetic field amplitude at r = 3 cm: Inside the capacitor gap (r < R), B(r) = μ₀I₀r/(2πR²), where I₀ is the peak current.

I₀ = √2 × I_rms = 1.414 × 6.9 × 10⁻⁶ ≈ 9.76 × 10⁻⁶ A.

B = (μ₀/2π) × I₀r/R² = (2 × 10⁻⁷) × (9.76 × 10⁻⁶ × 0.03)/(0.06)² = (2 × 10⁻⁷) × (2.93 × 10⁻⁷)/(3.6 × 10⁻³) ≈ 1.63 × 10⁻¹¹ T.

3 What physical quantity is the same for X-rays of wavelength 10⁻¹⁰ m, red light of wavelength 6800 Å, and radio waves of wavelength 500 m?Nature of EM waves

Their speed is the same. All three are electromagnetic waves, and every electromagnetic wave travels through vacuum at the same speed, c ≈ 3 × 10⁸ m/s, irrespective of its wavelength or frequency. (Their wavelengths and frequencies are all very different, since c = νλ must hold in each case.)

4 A plane electromagnetic wave travels in vacuum along the 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?Nature of EM waves

Since an electromagnetic wave is transverse, both E and B must be perpendicular to the direction of propagation (the z-axis) and perpendicular to each other; for example, E could be along the x-axis and B along the y-axis, chosen so that E × B points along +z (the direction of travel).

Wavelength: λ = c/ν = (3 × 10⁸)/(30 × 10⁶) = 10 m.

5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?Electromagnetic spectrum

Using λ = c/ν for the two ends of the frequency range:

At ν = 7.5 MHz: λ₁ = (3 × 10⁸)/(7.5 × 10⁶) = 40 m.

At ν = 12 MHz: λ₂ = (3 × 10⁸)/(12 × 10⁶) = 25 m.

So the wavelength band extends from 25 m to 40 m (as frequency increases, wavelength decreases, so the higher frequency 12 MHz corresponds to the shorter wavelength 25 m).

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 this oscillator?Generation of EM waves

The frequency of the electromagnetic wave radiated is exactly the same as the frequency of oscillation of the source charge, since an accelerating charge radiates energy at its own frequency of oscillation. Hence the electromagnetic wave produced also has frequency 10⁹ Hz.

Previous-year board questions 4

Q1 Identify the electromagnetic waves used for: (i) satellite communication, (ii) treatment of muscular strain, (iii) detecting fracture of bones. Write these three types of waves in the ascending order of their frequency. 2022 3 marks

(i) Satellite communication: Microwaves — because of their short wavelength they can be transmitted as narrow beams to and from satellites without much diffraction/spreading.

(ii) Treatment of muscular strain: Infrared waves — they are absorbed by tissue and produce a heating effect that relieves muscular pain (used in physiotherapy).

(iii) Detecting fracture of bones: X-rays — they pass more easily through soft tissue than through denser bone, producing a shadow image that reveals fractures.

Ascending order of frequency: microwaves (lowest frequency), then infrared waves, then X-rays (highest frequency).

Q2 The electric field of a plane electromagnetic wave propagating along the x-axis in vacuum is given by E_y = 50 sin[(1.05 × 10⁷)x − (3.14 × 10¹⁵)t] V/m. Calculate: (a) the wavelength and frequency of the wave, (b) the speed of the wave (and verify that it equals c), (c) the amplitude of the magnetic field B_z. 2023 5 marks

Comparing with the standard form E_y = E₀ sin(kx − ωt), we read off k = 1.05 × 10⁷ rad/m and ω = 3.14 × 10¹⁵ rad/s.

(a) Wavelength and frequency: λ = 2π/k = 6.283/(1.05 × 10⁷) ≈ 5.98 × 10⁻⁷ m ≈ 600 nm.

ν = ω/2π = (3.14 × 10¹⁵)/6.283 ≈ 5.0 × 10¹⁴ Hz.

(b) Speed: v = ω/k = νλ = (5.0 × 10¹⁴)(5.98 × 10⁻⁷) ≈ 2.99 × 10⁸ m/s ≈ 3 × 10⁸ m/s = c, confirming this is a genuine vacuum electromagnetic wave.

(c) Magnetic field amplitude: B₀ = E₀/c = 50/(3 × 10⁸) ≈ 1.67 × 10⁻⁷ T, directed along the z-axis so that E × B points along +x (the direction of propagation).

Q3 At a certain point in the path of a plane electromagnetic wave travelling in vacuum, the energy density of the electric field is 3.0 × 10⁻⁸ J/m³ at some instant. Find (a) the total energy density (electric + magnetic) at that instant, and (b) the corresponding magnitude of the magnetic field. 2022 4 marks

(a) Total energy density: In an electromagnetic wave, the electric and magnetic energy densities are equal at every instant and every point (since E = cB pointwise, ½ε₀E² = ½ε₀(cB)² = B²/2μ₀ using c² = 1/μ₀ε₀). So u_B = u_E = 3.0 × 10⁻⁸ J/m³, and the total is:

u = u_E + u_B = 2 × 3.0 × 10⁻⁸ = 6.0 × 10⁻⁸ J/m³.

(b) Magnetic field magnitude: From u_B = B²/(2μ₀): B² = 2μ₀u_B = 2 × (4π × 10⁻⁷) × (3.0 × 10⁻⁸) = 2 × 1.2566 × 10⁻⁶ × 3.0 × 10⁻⁸ ≈ 7.54 × 10⁻¹⁴ T².

B = √(7.54 × 10⁻¹⁴) ≈ 2.75 × 10⁻⁷ T.

(Check: from u_E, E = √(2u_E/ε₀) = √(6×10⁻⁸/8.85×10⁻¹²) ≈ 82.3 V/m, and B = E/c = 82.3/(3×10⁸) ≈ 2.74 × 10⁻⁷ T — consistent.)

Q4 Why is the ozone layer crucial for the survival of life on Earth? Name the part of the electromagnetic spectrum that is absorbed by it, and mention one harmful effect that results if this layer becomes depleted. 2021 2 marks

The ozone (O₃) layer in the upper atmosphere absorbs most of the ultraviolet (UV) radiation coming from the Sun before it can reach the Earth's surface. This is crucial because UV radiation is energetic enough to damage living cells and DNA.

Part of spectrum absorbed: Ultraviolet radiation.

Harmful effect of depletion: If the ozone layer is depleted (for example, by chlorofluorocarbons, CFCs), more UV radiation reaches the ground, increasing the incidence of skin cancer, cataracts and other eye damage, weakening the human immune system, and harming crops and aquatic ecosystems.

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