CLASS 12 PHYSICS · CHAPTER 8 · STUDY NOTES
Electromagnetic Waves
An exploration of Maxwell's unification of electricity, magnetism, and light, covering displacement current, the nature of electromagnetic waves, and the complete electromagnetic spectrum.
EXPLORE
The Need for Displacement Current
Compare two surfaces that share the same loop around a charging wire.Two surfaces, one magnetic circulation
Read the assumptions
Uniform fields between ideal circular plates, negligible fringing and slow charging. In general both conduction and displacement terms can occur in the same region. Displacement current describes changing electric flux; it does not mean charges cross an ideal insulator.
Original teaching drawing · NCERT Physics Part I, chapter 8, PDF pages 1–3 · simplified models
- Displacement current
- The electric-flux term I_d = ε₀ dΦ_E/dt in vacuum. It contributes to magnetic circulation without requiring charge to cross a gap.
| Contribution | What changes? | Ideal capacitor example |
|---|---|---|
| Conduction current | Actual charge flow | Wire connected to a plate |
| Displacement term | Electric flux | Uniform changing field in the gap |
One surface cuts the wire; another curves through the capacitor gap. Counting only conduction current would give different answers for the same magnetic circulation. Maxwell included the changing electric flux to make the answers agree.
For an ideal capacitor with uniform field, Φ_E = EA = Q/ε₀. Therefore ε₀ dΦ_E/dt = dQ/dt, equal to the external wire current. Electrons accumulate on a plate or leave it through a wire; they do not cross the insulating gap.
Let’s look a little closer.
Go deeper: What changes, and what stays still
Displacement current depends on the rate of change of electric flux, not the field magnitude alone. A charged capacitor held at constant voltage has an electric field but no continuing displacement current.
Conduction and displacement are two contributions to the same field equation. In general they need not be confined to different regions. The clean wire-versus-gap split is an ideal slow-charging model with fringing neglected.
Take a moment. What do you think?
PAUSE & TRY · think first, then check with a free account
Question
An ideal capacitor gains 6 µC of charge in 3 ms at a constant rate. Find the wire current and the displacement current through a surface spanning the gap.
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 1, 2, 3.
EXPLORE
The Ampere-Maxwell Law
Use circular symmetry to find the magnetic field between charging plates.A changing electric field has a magnetic field
Read the assumptions
Uniform fields between ideal circular plates, negligible fringing and slow charging. In general both conduction and displacement terms can occur in the same region. Displacement current describes changing electric flux; it does not mean charges cross an ideal insulator.
Original teaching drawing · NCERT Physics Part I, chapter 8, PDF pages 3–5 · simplified models
- Ampere–Maxwell law
- ∮ B · dl = μ₀(I_c + ε₀ dΦ_E/dt). Magnetic circulation depends on conduction current and changing electric flux through a chosen surface.
For circular plates of radius R with uniform changing electric field, an Amperian circle of radius r ≤ R encloses the fraction r²/R² of the total displacement current I. Hence B(2πr) = μ₀I r²/R², giving B = μ₀Ir/(2πR²).
For r ≥ R, the whole displacement current is enclosed, so B = μ₀I/(2πr). The two expressions meet at the plate edge. The field grows linearly inside and falls as 1/r outside in this ideal symmetric model.
Go deeper: Choosing a surface does not change the answer
The boundary loop and its orientation determine the circulation. The right-hand rule relates the direction of traversing the loop to the surface normal; reversing both preserves the equation.
Negligible fringing, uniform charging and slowly varying fields make this model tractable. The complete law includes both contributions wherever needed; it is not a rule that displacement current exists only inside capacitors.
PAUSE & TRY · think first, then check with a free account
Question
Circular plates have R = 0.10 m and charging current 0.20 A. Compare the magnetic fields at r = 0.05 m and r = 0.20 m.
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 3, 4, 5.
EXPLORE
Maxwell's Equations in Vacuum
Read a separate field picture for each law, rather than one repeated process arrow.Four laws, four different pictures
Read the assumptions
The arrows describe fields, not particles travelling in circles. Maxwell’s equations apply together. No isolated magnetic monopole has been observed; the usual classical model gives zero magnetic flux through every closed surface.
Original teaching drawing · NCERT Physics Part I, chapter 8, PDF pages 4–5 · simplified models
- Maxwell’s equations
- Four classical laws connecting electric and magnetic fields to charge, current and field changes.
Gauss’s electric law connects net flux through a closed surface to enclosed charge. Gauss’s magnetic law gives zero net magnetic flux: magnetic field lines do not start or end at isolated poles in the standard classical model.
Faraday’s law relates circulating electric field to changing magnetic flux. Ampere–Maxwell’s law relates magnetic circulation to conduction current and changing electric flux. Together, the vacuum equations admit travelling electromagnetic waves.
Go deeper: Fields and sources
A field at one point can arise from charges elsewhere even if the local charge density is zero. Likewise, zero net magnetic flux through a closed surface does not mean zero magnetic field at each point.
Changing E and B are coupled parts of one solution. Saying that they continuously “create each other” is a useful shorthand, but not a literal sequence of separate waves taking turns. Maxwell predicted electromagnetic propagation; Hertz later demonstrated radio waves experimentally.
PAUSE & TRY · think first, then check with a free account
Question
A uniform magnetic field passes through an empty closed cube. Is the magnetic field zero because net flux is zero? Explain.
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 4, 5.
EXPLORE
Nature and Properties of EM Waves
Follow the field directions and the direction in which energy travels.Two perpendicular fields travel together
Read the assumptions
The orthogonal, in-phase field picture is a travelling plane wave in vacuum, away from a source. Near fields need not have this geometry. Arrow lengths for E and B have separate units and scales. Playback is slow teaching motion, not the actual speed of light.
Original teaching drawing · NCERT Physics Part I, chapter 8, PDF pages 5–8 · simplified models
- Vacuum plane electromagnetic wave
- A travelling wave whose E and B fields are transverse, mutually perpendicular and in phase, with E₀ = cB₀.
For a plane wave travelling along +x in vacuum, choose E along y and B along z. Their cross product E × B points along +x. Both fields reverse together after half a cycle, so the direction of energy flow stays the same.
The wave needs no material medium. In vacuum c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s, and the electric and magnetic contributions to energy density are equal. An accelerating charge can radiate; charges moving uniformly in vacuum do not radiate simply because they move.
Go deeper: Where this picture applies
The neat orthogonal, in-phase picture describes a travelling plane wave away from sources. Near a charge or antenna, the fields can include near-field components that do not have those simple relations.
For a simple homogeneous, lossless medium the phase speed is 1/√(με). Real materials can disperse, absorb and respond differently at different frequencies. E and B have different units; E₀ = cB₀ does not justify comparing their numerical values as if they measured the same quantity.
Electromagnetic radiation carries energy and momentum. Absorption or reflection can exert radiation pressure; the zero average of an oscillating field does not mean zero average energy transport.
PAUSE & TRY · think first, then check with a free account
Question
A vacuum plane wave travels along −x and has electric field along +y at one instant. Which way is its magnetic field pointing?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 5, 6, 7, 8.
EXPLORE
Radio Waves and Microwaves
Compare three uses that depend on how waves interact with matter.Antennas, echoes and turning molecules
Read the assumptions
Frequency, wavelength and energy vary continuously. The spectrum names have conventional boundaries, and microwave is also commonly treated as part of the radio range. Typical sources and uses are examples, not exclusive definitions.
Original teaching drawing · NCERT Physics Part I, chapter 8, PDF pages 9–11 · simplified models
- Electromagnetic spectrum
- The continuous range of electromagnetic frequencies. Named bands have conventional boundaries, and microwave is often included within the radio range.
Alternating currents in an antenna can emit radio waves. A receiving antenna responds to the arriving field with an induced alternating signal. Charges oscillate locally; they do not travel from the transmitter to the receiver.
Microwaves support directional communication and radar. Radar compares emission and echo times to estimate distance d = cΔt/2. In an oven, absorbed microwave energy heats food through electrical response of its molecules; it is not a special resonance frequency of water.
Go deeper: One wave family, different engineering
Wavelength affects antenna size, diffraction and beam formation. Communication systems choose bands for practical propagation and equipment properties, rather than because radio and microwave are fundamentally different kinds of waves.
The molecule illustration is qualitative: polar water molecules try to respond to the changing field, with lag and dissipation. Food composition and geometry affect absorption; water, fats and sugars can absorb microwave energy. Radar’s factor of two comes from the outward and return journey.
PAUSE & TRY · think first, then check with a free account
Question
A radar echo returns 4.0 µs after a pulse is sent. Estimate the target distance, ignoring delays in the equipment.
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 9, 11.
EXPLORE
Infrared, Visible, and Ultraviolet Rays
Connect wavelength to emission, vision and absorption.One spectrum — many interactions
Read the assumptions
Frequency, wavelength and energy vary continuously. The spectrum names have conventional boundaries, and microwave is also commonly treated as part of the radio range. Typical sources and uses are examples, not exclusive definitions.
Original teaching drawing · NCERT Physics Part I, chapter 8, PDF pages 9–11 · simplified models
- Visible light
- The approximate wavelength range to which the human eye responds, commonly taken as about 400–700 nm. Infrared lies toward longer wavelengths and ultraviolet toward shorter ones.
| Band | Representative wavelength | Example source |
|---|---|---|
| Infrared | 10 µm | Warm matter |
| Visible | 550 nm | Electronic transitions |
| Ultraviolet | 100 nm | Hot stars and excited atoms |
Warm objects emit thermal radiation, often strongly in the infrared at everyday temperatures. Infrared is useful for thermal imaging, but any absorbed electromagnetic radiation can transfer energy and heat matter.
Visible wavelengths let us see, and a prism separates them because refractive index varies with wavelength. Ultraviolet has higher photon energy than visible light; its interaction with the atmosphere and materials depends strongly on wavelength. Different parts of the UV range are not absorbed equally.
Go deeper: Colour is a response, not a sharp physical boundary
Spectrum names do not mark sudden changes in the electromagnetic equations. The limits of visible vision are approximate and vary with conditions. A visible colour gradient is meaningful, but extending those colours into infrared or UV would be a false colour representation.
Atmospheric gases absorb substantial portions of ultraviolet, with ozone particularly important. Do not infer that every material called “glass” blocks all ultraviolet. Absorption and transmission require a specified wavelength and material.
PAUSE & TRY · think first, then check with a free account
Question
A photon has wavelength 600 nm and another 300 nm in vacuum. Which has the higher frequency and photon energy, and by what factor?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 9, 10, 11.
EXPLORE
X-Rays and Gamma Rays
Distinguish a source mechanism from an absolute energy boundary.Different sources can produce overlapping energies
Read the assumptions
Frequency, wavelength and energy vary continuously. The spectrum names have conventional boundaries, and microwave is also commonly treated as part of the radio range. Typical sources and uses are examples, not exclusive definitions.
Original teaching drawing · NCERT Physics Part I, chapter 8, PDF pages 11 · simplified models
- Photon energy
- E_photon = hf. Higher frequency gives higher photon energy, independent of whether the radiation is labelled X-ray or gamma.
A common X-ray source accelerates electrons onto a metal target. Their slowing produces a continuous spectrum; transitions involving inner atomic shells also produce characteristic X-ray lines. Different absorption by tissue and bone helps form an image.
An excited nucleus can emit a gamma photon as it changes to a lower energy state. Gamma radiation can also arise through other high-energy processes. X-ray and gamma energy ranges overlap: the labels often refer to source mechanisms, not a universal frequency dividing line.
Go deeper: Radiation is not a charged particle
Both are electromagnetic photons and are not deflected by electric or magnetic fields as an electron would be. Their interactions with matter depend on photon energy and the material.
The nuclear energy-level sketch describes one transition, not a picture of an electron orbiting a nucleus. X-ray and gamma illustrations are source models, not procedures for producing radiation or clinical guidance.
PAUSE & TRY · think first, then check with a free account
Question
A 50 keV photon was emitted in a nuclear transition. Another 50 keV photon came from an X-ray tube. Must their speeds in vacuum differ?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF page 11.
EXPLORE
Worked Examples: Wave Parameters
Change the horizontal spacing and vertical amplitude independently.Read wavelength and field amplitude separately
Read the assumptions
We use c ≈ 3 × 10⁸ m/s. The space graph is a snapshot and the time graph is measured at one location. Frequency is converted from MHz to Hz before calculating wavelength. The fields in a vacuum plane wave satisfy E₀ = cB₀.
Original teaching drawing · NCERT Physics Part I, chapter 8, PDF pages 8, 14 · simplified models
- Wave parameters
- f counts cycles per second, λ is the distance between repeating spatial phases, and ω = 2πf with k = 2π/λ.
Use c = fλ and E₀ = cB₀ for a vacuum plane wave. Convert MHz to Hz, nm to m and nT to T before substituting. Frequency determines wavelength; amplitude determines the field strength.
For E = E₀ sin(kx−ωt), a fixed phase moves along +x. With a plus sign, kx+ωt, it moves along −x. Choose the magnetic direction so that E × B matches that propagation direction.
Frequency fixes spatial and temporal repetition; amplitude fixes field height. Check units and propagation direction separately.
Go deeper: An original complete example
Let f = 15 MHz and E₀ = 90 V/m. Then λ = 20 m, B₀ = 300 nT, k = π/10 rad/m and ω = 30π × 10⁶ rad/s. A space snapshot and a time trace describe different measurements of the same wave.
If frequency doubles while E₀ is held fixed, λ halves but B₀ stays 300 nT. This separates frequency from amplitude rather than assuming that a shorter wavelength automatically means a stronger wave.
PAUSE & TRY · think first, then check with a free account
Question
A vacuum plane wave has f = 25 MHz and B₀ = 200 nT. Find λ and E₀. If only frequency halves, what changes?
Sign in to see the answer
Write your own answer and compare it with ours. It’s free.
Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 8, 14.