By the end of this chapter you'll be able to…

  • 1Explain the inconsistency in Ampere's circuital law and how displacement current resolves it
  • 2Compute displacement current from the rate of change of electric flux
  • 3State the Ampere-Maxwell law and name Maxwell's four equations
  • 4Describe the transverse nature of electromagnetic waves and the orientation of E, B and the direction of travel
  • 5Relate the field amplitudes through E_0 = c B_0 and obtain c from the free-space constants
  • 6Apply c = nu lambda, omega = 2 pi nu and k = omega/c to a given wave
  • 7Order the electromagnetic spectrum and state how each band is produced and used
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Why this chapter matters
This short chapter completes classical electromagnetism. Displacement current is the correction that made Maxwell's equations consistent and predicted light itself, and the spectrum section is among the most reliably scoring material in the paper because it is largely factual.

Electromagnetic Waves

1. Check this before you revise anything

The "Additional Exercises" section has been removed from this chapter, as from all 14 chapters of the current Class 12 Physics book. The questions run contiguously from 8.1 to 8.10 with no gaps. At 14 pages this is one of the shortest chapters in the book.

Energy density is required by an exercise but never defined in the chapter. Exercise 8.10(c) asks you to "show that the average energy density of the field equals the average energy density of the field."

Searching the whole chapter for "energy density" returns exactly two hits — both inside that exercise. Searching for "intensity", "momentum", "radiation pressure" and "Poynting" returns zero hits each. Older editions carried a passage on electromagnetic waves transporting energy and momentum, including radiation pressure; only the qualitative sentence that these waves "carry energy" survives.

You can still answer it, using results from two earlier chapters:

  • comes from the energy stored in a capacitor, Chapter 2.
  • comes from Example 6.9, where the magnetic energy of a solenoid is rewritten in terms of , and .

With and , the two averages come out equal. Our solution shows the substitution in full.

Textbook sectionTopic
8.1Introduction
8.2Displacement current, and the Ampere-Maxwell law
8.3Electromagnetic waves: their sources and their nature
8.4The electromagnetic spectrum: radio, microwave, infrared, visible, ultraviolet, X-rays, gamma rays

A symbol trap if you work from an extracted PDF. As in Chapters 1 to 7, the micro symbol extracts as a plain "m", which affects and any microfarad or micrometre value.


2. Displacement Current (Textbook 8.2)

Maxwell noticed that Ampere's circuital law, as it stood, contradicts itself — and fixing the contradiction predicted electromagnetic waves.

The inconsistency. Consider a capacitor being charged. Take an Amperian loop around the connecting wire, and apply Ampere's law using two different surfaces bounded by that same loop:

  • A flat surface cutting the wire encloses the conduction current , giving .
  • A bulging surface passing between the capacitor plates encloses no conduction current, giving zero.

Same loop, same field, two different answers. Ampere's law in its original form is therefore incomplete.

Maxwell's resolution. Between the plates there is no charge flow, but there is a changing electric field. Maxwell proposed that a changing electric flux is itself a source of magnetic field, equivalent to a current:

The Ampere-Maxwell law then reads:

The displacement current is exactly equal to the conduction current at every instant during charging, which is what makes the two surfaces agree. Exercise 8.2(b) asks you to confirm this, and Exercise 8.1 works it out numerically for a 12 cm capacitor.

The word "current" here is a name, not a claim: nothing flows between the plates. What produces the magnetic field is the changing electric field.

Maxwell's four equations are Gauss's law for electricity, Gauss's law for magnetism, Faraday's law of induction, and this corrected Ampere-Maxwell law. Together they predicted waves travelling at m s — the measured speed of light — which is how light was identified as an electromagnetic wave. Hertz produced and detected such waves experimentally in 1887.


3. The Nature of Electromagnetic Waves (Textbook 8.3)

Sources (8.3.1). An electromagnetic wave is radiated by an accelerating charge. A stationary charge has a static field and radiates nothing; a charge in uniform motion produces a steady current and a static magnetic field, and still radiates nothing.

A charge oscillating at frequency radiates a wave of the same frequency . That single sentence answers Exercise 8.6 completely.

Nature (8.3.2). Electromagnetic waves are transverse. The oscillating electric and magnetic fields are:

  • perpendicular to each other,
  • both perpendicular to the direction of propagation,
  • and in phase, reaching their maxima and zeros together.

The direction of travel is along . For a wave travelling along , both and lie in the - plane, which is what Exercise 8.4 asks you to state.

The amplitude relation:

Because is so large, is always numerically tiny beside . Exercises 8.7 and 8.8 use this in both directions.

The speed. Maxwell's equations give the speed from two electrostatic and magnetostatic constants alone:

In a medium the speed falls to , which is less than .

No medium is required. Unlike sound, electromagnetic waves propagate through vacuum — the fields sustain each other, a changing generating and a changing generating .

All electromagnetic waves travel at the same speed in vacuum, whatever their frequency. That is the whole of Exercise 8.3: X-rays, red light and radio waves differ in wavelength and frequency, but share exactly.

The usual wave relations apply throughout:


4. The Electromagnetic Spectrum (Textbook 8.4)

The bands differ only in frequency. They are continuous and overlapping, with no sharp boundaries, and are named by how they are produced rather than by any physical difference.

BandTypical wavelengthProduced byUses
Radio (8.4.1) mAccelerated charges in aerialsRadio, television, mobile communication
Microwaves (8.4.2) m to mmKlystrons, magnetrons, Gunn diodesRadar, microwave ovens, satellite links
Infrared (8.4.3) mm to nmHot bodies and moleculesHeating, night vision, remote controls
Visible (8.4.4) to nmAtoms and molecules in excited statesVision; the only band the eye detects
Ultraviolet (8.4.5) to nmSpecial lamps, very hot bodies, the SunSterilisation, LASIK eye surgery
X-rays (8.4.6) nm to nmBombarding a metal target with high-energy electronsMedical imaging, cancer treatment
Gamma rays (8.4.7) nmRadioactive nuclei and nuclear reactionsDestroying cancer cells, sterilising equipment

Frequency, energy and penetration all rise together as wavelength falls, since . This is the pattern Exercise 8.9 asks you to bring out: photon energies run from around eV for radio waves to about eV for gamma rays, and the source of each band matches its energy scale — aerial electrons for the weakest, atomic transitions in the middle, nuclear processes at the top.

Two atmospheric effects the chapter highlights:

  • The greenhouse effect. The atmosphere passes visible sunlight but absorbs the infrared re-radiated by the warmed Earth, trapping heat.
  • The ozone layer absorbs most solar ultraviolet, which would otherwise damage living tissue.

Microwave ovens work because the microwave frequency is matched to the rotational frequency of water molecules, so energy is absorbed efficiently by the water in the food and heats it throughout rather than only at the surface.


Summary

  • Ampere's original law gives two different answers for two surfaces sharing one loop when a capacitor charges — that inconsistency is what displacement current fixes.
  • : a changing electric field produces a magnetic field just as a current does.
  • Ampere-Maxwell law: , with at every instant during charging.
  • Nothing physically flows between the plates; "current" here is a name for a changing flux.
  • Maxwell's four equations predicted waves at m s, identifying light as electromagnetic; Hertz confirmed it in 1887.
  • Only an accelerating charge radiates; a stationary or uniformly moving charge does not.
  • A charge oscillating at frequency radiates at the same frequency .
  • Electromagnetic waves are transverse, with , both perpendicular to the propagation direction, and in phase; travel is along .
  • , so the magnetic amplitude is always numerically far smaller.
  • ; in a medium .
  • No medium is needed — the two fields sustain each other.
  • All electromagnetic waves share the same speed in vacuum, whatever their frequency.
  • , , .
  • The spectrum runs radio, microwave, infrared, visible, ultraviolet, X-ray, gamma — continuous, overlapping, named by how each is produced.
  • Frequency, photon energy and penetrating power all rise as wavelength falls.
  • The greenhouse effect traps re-radiated infrared; the ozone layer absorbs solar ultraviolet.
  • Microwave ovens match the rotational frequency of water molecules.
  • Energy density, intensity, momentum and radiation pressure have been removed from this chapter, though Exercise 8.10(c) still requires energy density — take from Chapter 2 and from Example 6.9.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Displacement current
i_d = epsilon_0 x (rate of change of electric flux)
A changing electric field acts as a source of magnetic field; nothing physically flows between the plates
Ampere-Maxwell law
The line integral of B around a closed loop equals mu_0 times the sum of the conduction and displacement currents
This is Ampere's law corrected, and it removes the two-surface contradiction
Equality of the two currents
During charging, i_d equals i_c at every instant
This is what makes the flat and bulging surfaces give the same answer
Speed of light from free-space constants
c = 1 divided by the square root of mu_0 epsilon_0, giving 3 x 10^8 m/s
Maxwell obtained this from purely electrical and magnetic constants, which is how light was identified as an electromagnetic wave
Speed in a medium
v = 1 divided by the square root of mu epsilon
Always less than c, since both constants exceed their free-space values
Amplitude relation
E_0 = c B_0
The magnetic amplitude is numerically tiny beside the electric one because c is so large
Wave relations
c = nu lambda, omega = 2 pi nu, and k = omega/c = 2 pi/lambda
The standard wave relations apply unchanged across the whole spectrum
Source of electromagnetic waves
Only an ACCELERATING charge radiates
A stationary charge or one in uniform motion produces static fields and radiates nothing
Frequency of the radiated wave
A charge oscillating at frequency nu radiates a wave of the same frequency nu
This single statement answers Exercise 8.6 in full
Orientation of the fields
E and B are perpendicular to each other and both perpendicular to the direction of travel, and they are in phase
Propagation is along the direction of E cross B, so the waves are transverse
Photon energy
E = h nu, with h = 6.63 x 10^-34 J s
Energy rises with frequency, so gamma rays carry far more per photon than radio waves
Electric energy density
u_E = (1/2) epsilon_0 E squared
Not defined in this chapter; carried over from the capacitor energy of Chapter 2 and needed for Exercise 8.10(c)
Magnetic energy density
u_B = B squared divided by 2 mu_0
Not defined in this chapter either; it follows from Example 6.9 on the magnetic energy of a solenoid
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Believing that charge actually flows between the capacitor plates
Nothing flows there. Displacement current is a name for epsilon_0 times the rate of change of electric flux, and it is the CHANGING ELECTRIC FIELD that produces the magnetic field.
WATCH OUT
Assuming a uniformly moving charge radiates electromagnetic waves
Only accelerating charges radiate. A charge in uniform motion is simply a steady current, producing a static magnetic field and no wave at all.
WATCH OUT
Expecting different parts of the spectrum to travel at different speeds in vacuum
All electromagnetic waves travel at exactly c in vacuum, whatever their frequency. That is the entire point of Exercise 8.3.
WATCH OUT
Taking E and B to be out of phase with each other
In a plane electromagnetic wave in vacuum they are in phase, reaching their maxima and their zeros together. They are perpendicular in direction but synchronised in time.
WATCH OUT
Concluding that the magnetic field is unimportant because B_0 is numerically small
The small number is an artefact of the units: B_0 = E_0/c, and c is very large. The two fields carry equal average energy density, which is exactly what Exercise 8.10(c) asks you to show.
WATCH OUT
Looking for the energy density formula inside Chapter 8
It is not there. Both occurrences of the phrase in this chapter are inside Exercise 8.10 itself. Take u_E = half epsilon_0 E squared from Chapter 2 and u_B = B squared over 2 mu_0 from Example 6.9.
WATCH OUT
Preparing intensity, momentum or radiation pressure of electromagnetic waves
All have been removed. Searching the chapter returns zero hits for intensity, momentum, radiation pressure and Poynting. Only the qualitative statement that these waves carry energy survives.
WATCH OUT
Treating the spectrum boundaries as sharp
The bands are continuous and overlapping, and are distinguished by how they are produced rather than by any physical difference. Only the order and rough magnitudes are examinable.
WATCH OUT
Mixing up which atmospheric layer does what
The ozone layer absorbs solar ultraviolet. The greenhouse effect is separate: the atmosphere passes visible sunlight but absorbs the infrared re-radiated by the warmed Earth.
WATCH OUT
Giving the frequency of a radiated wave as a multiple of the source oscillation
It is equal to it, not a multiple. A charge oscillating at 10^9 Hz radiates at 10^9 Hz.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Electromagnetic Waves?

9 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

9 questions~6 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Ampere's original law gives two answers for two surfaces on one loop when a capacitor charges
  • Displacement current i_d = epsilon_0 times the rate of change of electric flux fixes it
  • Ampere-Maxwell law adds i_d to the conduction current; i_d equals i_c during charging
  • Nothing physically flows between the plates; the changing electric field is the source
  • Maxwell's four equations predicted waves at 3 x 10^8 m/s, identifying light as electromagnetic
  • Hertz produced and detected electromagnetic waves experimentally in 1887
  • Only an accelerating charge radiates; stationary or uniformly moving charges do not
  • A charge oscillating at frequency nu radiates at the same frequency nu
  • Electromagnetic waves are transverse, with E and B perpendicular to each other and to the direction of travel
  • E and B oscillate in phase; propagation is along E cross B
  • E_0 = c B_0, so B_0 is numerically far smaller
  • c = 1 over the square root of mu_0 epsilon_0; in a medium v = 1 over the square root of mu epsilon
  • No medium is required, since the two fields sustain each other
  • All electromagnetic waves share the same speed in vacuum whatever their frequency
  • c = nu lambda, omega = 2 pi nu, k = omega/c = 2 pi/lambda
  • Spectrum order: radio, microwave, infrared, visible, ultraviolet, X-rays, gamma rays
  • Photon energy E = h nu rises as wavelength falls, and so does penetrating power
  • The greenhouse effect traps re-radiated infrared; the ozone layer absorbs solar ultraviolet
  • Microwave ovens match the rotational frequency of water molecules
  • Energy density, intensity, momentum and radiation pressure have been removed, though Exercise 8.10(c) still needs energy density
  • The Additional Exercises block has been removed, leaving Exercises 8.1 to 8.10

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit V: Electromagnetic Waves, no chapter-wise split published by CBSE

Question typeMarks eachTypical countWhat it tests
Displacement Current and the Ampere-Maxwell Law2-31The two-surface inconsistency, the definition of displacement current, and its equality with the conduction current
Nature of Electromagnetic Waves, Wave Relations, Amplitude Relation and Energy Density of an Electromagnetic Wave3-41Transverse character, field orientation, E_0 = c B_0, the wave relations, and equal energy densities
The Electromagnetic Spectrum and Photon Energy Across the Spectrum2-31Order of the bands, how each is produced and used, and E = h nu
Prep strategy
  • Learn the spectrum order and one production method and one use per band, since this is the most reliably scoring part of the chapter
  • State the two-surface argument explicitly whenever displacement current is asked about
  • Convert every frequency and wavelength into SI units before substituting
  • Remember that E and B are perpendicular in space but in phase in time
  • Import the energy density formulas from Chapters 2 and 6, since this chapter does not state them

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Mobile and radio communication

Aerials radiate and receive radio waves, with the band chosen to match the range and data rate required.

Microwave ovens

The microwave frequency matches the rotational frequency of water molecules, so the water in food absorbs the energy and heats the food throughout.

Medical imaging and therapy

X-rays image bone and tissue because of their penetrating power, while gamma rays are used to destroy cancer cells and sterilise equipment.

Infrared sensing

Warm objects emit infrared, which night vision equipment and thermal cameras detect; remote controls use short infrared pulses to send commands.

Ultraviolet sterilisation

Ultraviolet lamps destroy micro-organisms in water and on surfaces, using the same photon energies the ozone layer screens out of sunlight.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

1
Memorise the spectrum order together with one production method and one use per band
2
Write the two-surface argument in words before any algebra when displacement current is asked about
3
Convert megahertz and nanometres to SI units in the first line of working
4
State that E and B are perpendicular in direction but in phase in time, since both halves earn marks
5
Quote c = 3 x 10^8 m/s and use it consistently rather than switching to 2.998
6
For energy density questions, name the chapter each formula comes from, since Chapter 8 does not state them
7
Give both the magnitude and the direction whenever a field orientation is asked for

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH
The Poynting vector, which gives the direction and rate of energy flow in an electromagnetic wave
STRETCH
Radiation pressure and momentum transfer, and the solar sail as a practical application
STRETCH
Derivation of the wave equation directly from Maxwell's four equations in differential form
STRETCH
Polarisation of electromagnetic waves, and how the transverse nature makes it possible at all
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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainDisplacement current in a charging capacitorAmpere-Maxwell law applied numerically

A capacitor of two circular plates of radius cm, separated by cm, is charged by a steady current of A. Find the capacitance, the rate of change of potential difference, and the displacement current between the plates.

Stuck? Show the approach

Compute the capacitance from the geometry, then use I = C dV/dt, and finally apply the equality of conduction and displacement currents.

Show the full solution

The plate area is m, so pF. From , the rate of change of potential difference is V s. The displacement current equals the conduction current at every instant, so A. Kirchhoff's first rule therefore still holds across the capacitor gap, provided displacement current is counted alongside conduction current.

Answer: C = 8.0 pF; dV/dt = 1.9 x 10^10 V/s; displacement current = 0.15 A
The trap

Assuming no current exists between the plates because no charge crosses the gap. The changing electric field does the same job, and its equivalent current has exactly the same value.

JEE MainField amplitudes and wave parametersApplying E_0 = c B_0 and the wave relations

An electromagnetic wave in vacuum has electric field amplitude N C and frequency MHz. Find , , and , and describe the orientation of the two fields.

Stuck? Show the approach

Use E_0 = c B_0 for the magnetic amplitude, then the standard wave relations for the rest.

Show the full solution

T, that is nT. Then rad s, rad m, and m. The two fields are perpendicular to each other and both perpendicular to the direction of propagation, and they oscillate in phase, since the wave travels along .

Answer: B_0 = 400 nT; omega = 3.14 x 10^8 rad/s; k = 1.05 rad/m; lambda = 6.0 m; E and B perpendicular, both transverse, in phase
The trap

Reporting that B is negligible because 400 nT looks small. That is a units artefact, and the two fields carry equal energy density.

JEE AdvancedEqual energy densitiesA result the chapter never defines

A plane electromagnetic wave has frequency Hz and electric field amplitude V m. Find the wavelength and , and show that the average energy densities of the electric and magnetic fields are equal.

Stuck? Show the approach

Find the wavelength and magnetic amplitude first, then write both energy densities and substitute B_0 = E_0/c together with c squared = 1 over mu_0 epsilon_0.

Show the full solution

m, that is cm, and T. Averaging the square of a sinusoid gives half its peak, so and . Substituting gives , and since this becomes , identical to . Numerically both are about J m.

Answer: lambda = 1.5 cm; B_0 = 1.6 x 10^-7 T; both average energy densities equal about 5.1 x 10^-9 J per cubic metre
The trap

Searching Chapter 8 for the energy density formula. It is not there: take u_E from the capacitor energy of Chapter 2 and u_B from Example 6.9 on the solenoid.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 BoardHigh
JEE MainMedium
NEETMedium
JEE AdvancedLow

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

It gave two different answers for the same loop. Take a loop around the wire feeding a charging capacitor. A flat surface bounded by that loop is pierced by the conduction current, so the law gives mu_0 times I. A bulging surface bounded by the same loop passes between the plates, where no charge crosses, so the law gives zero. Since both surfaces share the same boundary, the law as written was self-contradictory. Maxwell resolved it by adding a term for the changing electric flux between the plates.

No. Displacement current is a name, not a description of charge motion. What exists between the plates is an electric field that changes as the capacitor charges, and Maxwell's insight was that a changing electric field produces a magnetic field just as a real current does. Defining i_d as epsilon nought times the rate of change of electric flux makes the two surfaces agree, because i_d turns out to equal the conduction current exactly at every instant during charging.

Because the speed follows from Maxwell's equations as one over the square root of mu nought epsilon nought, and neither constant depends on frequency. X-rays, visible light and radio waves therefore travel at exactly the same speed in vacuum, differing only in wavelength and frequency, which trade off through c = nu lambda. In a material medium the speed drops to one over the square root of mu epsilon, and there it can depend on frequency, which is what produces dispersion.

It is small only as a number, because B_0 = E_0/c and c is about 3 x 10^8. The physical significance of the two fields is identical: the average energy density of the electric field equals that of the magnetic field, which is what Exercise 8.10(c) asks you to prove. Substituting B_0 = E_0/c into B_0 squared over 4 mu_0, and using c squared = 1 over mu_0 epsilon_0, returns exactly one quarter epsilon_0 E_0 squared.

You import both formulas from earlier chapters. The electric energy density u_E = half epsilon nought E squared comes from the energy stored in a capacitor in Chapter 2. The magnetic energy density u_B = B squared over 2 mu nought follows from Example 6.9 in Chapter 6, where the magnetic energy of a solenoid is rewritten in terms of B, A and l. Average each over a cycle by halving, then substitute B_0 = E_0/c and c squared = 1 over mu_0 epsilon_0, and the two expressions become identical.

Only accelerated motion. A stationary charge has a purely static electric field and radiates nothing. A charge moving with constant velocity constitutes a steady current, producing a static magnetic field alongside its electric field, and still radiates nothing. It is acceleration that produces the coupled changing fields which detach and propagate. An oscillating charge is the standard case, and it radiates at exactly its own oscillation frequency, which is what Exercise 8.6 tests.

Not from this chapter. Searching the current text returns zero hits for intensity, momentum, radiation pressure and Poynting. Older editions carried a passage on electromagnetic waves transporting both energy and momentum, and on the pressure they exert when absorbed, but only the qualitative statement that these waves carry energy survives. Concentrate instead on displacement current, the nature of the waves, and the spectrum, which is where the marks are.
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Last reviewed on 18 August 2026. Written and reviewed by subject-matter experts — read about our process.
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