📌 Snapshot
- Maxwell removed a contradiction in Ampere's circuital law by introducing the displacement current id = ε₀ (dΦE/dt), giving the Ampere–Maxwell law.
- Time-varying electric and magnetic fields regenerate each other, leading to the prediction of electromagnetic waves that propagate at c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s — the speed of light.
- EM waves are transverse: E ⊥ B ⊥ direction of propagation, with E₀/B₀ = c; they need no material medium.
- The electromagnetic spectrum runs from γ-rays (λ ~ 10⁻¹² m) up to long radio waves (λ ~ 10⁶ m), classified by production/detection mechanisms.
- CUET regularly tests: displacement current definition/formula, characteristics of EM waves, c = 1/√(μ₀ε₀), E₀/B₀ = c, and matching spectrum bands to wavelengths/uses/sources.
📖 Detailed Notes
2.1 Core concepts
Two earlier results set the stage: an electric current produces a magnetic field (Ampere, Oersted) and a changing magnetic field produces an electric field (Faraday). James Clerk Maxwell asked the symmetric converse — does a changing electric field produce a magnetic field? — and answered yes (NCERT §8.1, p. 201). His route to the answer was through a paradox he discovered while applying Ampere's circuital law ∮B·dl = μ₀ i(t) to a charging parallel-plate capacitor (NCERT §8.2, p. 202, Eq. 8.1).
For a flat circular Amperian loop of radius r outside the capacitor and perpendicular to the wire, the conduction current i(t) pierces the disk, giving B(2πr) = μ₀ i(t) (NCERT §8.2, p. 202, Eq. 8.2). But the same closed loop also bounds a pot-shaped surface whose bottom slips between the capacitor plates, and a tiffin-box surface with a flat circular bottom S between the plates (NCERT Fig. 8.1, p. 203). Neither pot nor tiffin-box is pierced by any conduction current — yet Ampere's law should give the same B at point P regardless of which surface we choose. The two calculations clash: a non-zero B on one surface, zero B on the other (NCERT §8.2, p. 203).
Maxwell recognised that the missing physical input is the electric flux between the plates. With plate area A and instantaneous charge Q, the field is E = (Q/A)/ε₀, so ΦE = |E|A = Q/ε₀ (NCERT §8.2, p. 203, Eq. 8.3). When Q changes with time, ε₀ (dΦE/dt) = dQ/dt = i — exactly the conduction current in the external wire (NCERT §8.2, p. 203, Eq. 8.4). Identifying this quantity as a new "current" — the displacement current id = ε₀ (dΦE/dt) — restores consistency: id flows where ic cannot, and the total current i = ic + id is the same through any surface bounded by the loop (NCERT §8.2, p. 204, Eq. 8.5). Outside the capacitor only conduction current exists (id = 0); between the plates only displacement current exists (ic = 0, id = i).
The generalised statement is the Ampere–Maxwell law: ∮B·dl = μ₀ ic + μ₀ ε₀ (dΦE/dt) (NCERT §8.2, p. 204, Eq. 8.6). Maxwell wrote this together with three other equations — Gauss's law for electricity ∮E·dA = Q/ε₀, Gauss's law for magnetism ∮B·dA = 0 (no magnetic monopoles), and Faraday's law ∮E·dl = −dΦB/dt — to give the four Maxwell's equations in vacuum which, with the Lorentz force F = q(E + v × B), encode all of classical electromagnetism (NCERT §8.2 box, p. 205).
The symmetry is striking. Faraday says a changing B produces E; Ampere–Maxwell says a changing E produces B. If a region has time-varying fields, each field acts as the source of the other and the disturbance need not stay put — it can detach from any source and propagate. This is the qualitative mechanism by which Maxwell predicted electromagnetic waves (NCERT §8.2, p. 204–205).
Sources of EM waves (§8.3.1, p. 205). Stationary charges produce only electrostatic fields; uniformly moving charges (steady currents) produce only static magnetic fields. Neither radiates. Only accelerated charges radiate EM waves, and an oscillating charge of frequency ν radiates waves of exactly the same frequency ν. An electric dipole is the simplest source. The wave carries off energy at the expense of the source's kinetic energy. Experimental confirmation came in 1887 from Heinrich Hertz, who produced and detected radio-region waves; seven years later Jagadish Chandra Bose at Calcutta generated much shorter wavelengths of 25 mm to 5 mm; and Guglielmo Marconi transmitted EM waves over many kilometres, opening up wireless communication (NCERT §8.3.1, p. 205–206).
Nature of EM waves (§8.3.2, p. 206). Solving Maxwell's equations in free space yields plane-wave solutions in which E and B are mutually perpendicular and both perpendicular to the direction of propagation — i.e., EM waves are transverse. For a wave travelling along z with E along x and B along y, Ex = E₀ sin(kz − ωt) and By = B₀ sin(kz − ωt) (NCERT Eqs. 8.7a, 8.7b, p. 206). The angular frequency ω and wave vector k are related by ω = ck where the constant c = 1/√(μ₀ε₀); equivalently νλ = c (NCERT Eqs. 8.9a, 8.9b, p. 207). When μ₀ and ε₀ are inserted numerically, c works out to ≈ 3 × 10⁸ m/s — the same as the measured speed of light. This single observation forced the conclusion that light is an electromagnetic wave, unifying optics with electricity and magnetism.
The amplitudes of the two oscillating fields are not independent: B₀ = E₀/c, so the magnetic-field amplitude is smaller than the electric-field amplitude by a factor of c (NCERT Eq. 8.10, p. 207). EM waves are self-sustaining: unlike sound or water waves they require no material medium and travel through vacuum. In a medium of permittivity ε and permeability μ the wave speed is v = 1/√(με), which is less than c; this dependence is the microscopic origin of the refractive index (NCERT Eq. 8.11, p. 207). The vacuum speed c is independent of wavelength to within metres-per-second and is now used to define the metre.
Electromagnetic spectrum (§8.4, p. 208–211). Classification is by frequency/wavelength and reflects production and detection methods, not sharp physical boundaries.
- Radio waves — λ > 0.1 m. Produced by the accelerated motion of free electrons in conducting wires and aerials. AM band 530 kHz–1710 kHz; short-wave bands up to 54 MHz; TV 54–890 MHz; FM 88–108 MHz; cellular telephones in the UHF band (NCERT §8.4.1, p. 209).
- Microwaves — 0.1 m to 1 mm (GHz range). Produced by klystrons, magnetrons and Gunn diodes. Short wavelengths and good directionality make them ideal for radar (aircraft navigation, speed guns for cars and tennis serves). In a microwave oven the frequency is chosen to match the resonant frequency of water molecules, transferring energy efficiently as kinetic energy and raising the temperature of food (NCERT §8.4.2, p. 209).
- Infrared (heat) waves — 1 mm to 700 nm. Produced by the vibration of atoms and molecules in hot bodies. Strongly absorbed by water, CO₂ and NH₃, making them useful in physical therapy and Earth-observing satellites. Infrared LEDs power TV/video remotes. Infrared trapping by atmospheric CO₂ and water vapour is the greenhouse effect that keeps Earth warm (NCERT §8.4.3, p. 210).
- Visible light — 700 nm to 400 nm, frequency 4 × 10¹⁴ Hz to 7 × 10¹⁴ Hz. Produced when electrons in atoms drop between energy levels; detected by the eye, photocells and photographic film (NCERT §8.4.4, p. 210).
- Ultraviolet — 4 × 10⁻⁷ m down to 6 × 10⁻¹⁰ m. Produced by special lamps, very hot bodies and the Sun; absorbed by the ozone layer at 40–50 km altitude, so CFC-driven ozone depletion is a serious concern. Causes tanning by stimulating melanin production; used in LASIK eye surgery (laser-assisted in situ keratomileusis) and to sterilise water (NCERT §8.4.5, p. 210).
- X-rays — 10⁻⁸ m to 10⁻¹³ m. Produced by bombarding a metal target with high-energy electrons (X-ray tubes) or from inner-shell electron transitions; used for medical diagnosis and to treat certain cancers. Because they damage tissue, exposure must be minimised (NCERT §8.4.6, p. 211).
- Gamma rays — λ < 10⁻¹⁰ m down to below 10⁻¹⁴ m. Produced by nuclear reactions and radioactive decay; used in medicine to destroy cancer cells (NCERT §8.4.7, p. 211). A key conceptual takeaway (NCERT Points to Ponder, p. 213) is that the only basic difference between bands is wavelength/frequency — every band travels at c in vacuum. The wavelength typically correlates with the size of the radiating system: γ-rays from atomic nuclei (~10⁻¹⁴ m), X-rays from heavy atoms, visible light from atomic electron transitions, radio waves from antennas the size of metres.
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Displacement current (id) | The current id = ε₀ (dΦE/dt) due to a time-varying electric flux; acts as a source of magnetic field exactly like conduction current. | 203–204 |
| Conduction current (ic) | Current carried by conductors due to the flow of charges. | 203 |
| Ampere–Maxwell law | ∮B·dl = μ₀ ic + μ₀ ε₀ (dΦE/dt) — generalised Ampere's law including displacement current. | 204 |
| Gauss's law for electricity | ∮E·dA = Q/ε₀ — relates electric flux to enclosed charge. | 205 |
| Gauss's law for magnetism | ∮B·dA = 0 — no isolated magnetic monopoles. | 205 |
| Faraday's law | ∮E·dl = −dΦB/dt — emf equals rate of change of magnetic flux. | 205 |
| Maxwell's equations | The four equations of electromagnetism in vacuum. | 205 |
| Electromagnetic wave | Coupled time-varying electric and magnetic fields propagating through space; transverse and self-sustaining. | 201, 207 |
| Speed of EM waves in vacuum | c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s; same for all wavelengths. | 207 |
| E₀/B₀ ratio | E₀/B₀ = c (i.e., B₀ = E₀/c). | 207 |
| Wave number (k) | k = 2π/λ — magnitude of propagation vector. | 207 |
| Angular frequency (ω) | ω = 2πν; satisfies ω = ck for vacuum EM waves. | 207 |
| Wave speed in medium | v = 1/√(με), where ε and μ are the medium's permittivity and permeability. | 207 |
| Electromagnetic spectrum | Classification of EM waves by frequency/wavelength, from γ-rays (~10⁻¹² m) to long radio waves (~10⁶ m). | 208 |
| Radio waves | EM waves with λ > 0.1 m, produced by accelerated electrons in aerials. | 209 |
| Microwaves | EM waves of 0.1 m–1 mm produced by klystrons/magnetrons. | 209 |
| Infrared (heat waves) | EM waves of 1 mm–700 nm produced by molecular vibration. | 210 |
| Visible light | EM waves of 700–400 nm, detected by the eye. | 210 |
| Ultraviolet | EM waves of 400 nm to 0.6 nm, mostly absorbed by ozone. | 210 |
| X-rays | EM waves of 10 nm–10⁻⁴ nm produced by bombarding metal targets. | 211 |
| Gamma rays | EM waves of λ < 10⁻¹⁰ m from nuclear/radioactive sources. | 211 |
| Greenhouse effect | Trapping of re-radiated infrared from earth's surface by gases like CO₂ and water vapour. | 210 |
2.3 Diagrams / processes to remember
- Fig. 8.1 (p. 203): Parallel-plate capacitor with three Amperian surfaces — flat disk (a), pot-shaped surface (b) whose bottom lies between the plates, and a tiffin-box surface (c). The three surfaces share the same circular rim, so Ampere's law must give the same B at point P regardless of which surface we choose — the contradiction that motivates id.
- Fig. 8.2 (p. 204): The electric and magnetic fields E and B between the capacitor plates at point M, and a cross-sectional view showing concentric circular B-lines around the displacement-current axis — the perpendicularity of E and B emerging directly from displacement current.
- Fig. 8.3 (p. 206): A linearly polarised plane EM wave travelling along +z with Ex along the x-axis and By along the y-axis. The two sinusoids are in phase and of perpendicular polarisation — a visual proof that E ⊥ B ⊥ direction of propagation.
- Fig. 8.4 (p. 209): The full electromagnetic spectrum on a logarithmic axis from 10 Hz (long radio) up to 10²³ Hz (γ-rays), with corresponding wavelengths from 10⁷ m down to 10⁻¹⁴ m and a side-bar showing the 400–700 nm visible band with violet → red colours.
- Table 8.1 (p. 211): Wavelength range, production mechanism, and detector for each band of the spectrum — high-yield for CUET matching items. Worth memorising: aerials/receiver aerials for radio; klystron–point-contact diode for microwaves; vibrating molecules–thermopile/bolometer for IR; atomic transitions–eye for visible; inner-shell electrons–photocell for UV; X-ray tubes–Geiger tube/ionisation chamber for X-rays; radioactive decay–Geiger tube for γ-rays.
- Worked Example 8.1 (p. 208): Given E = 6.3 ĵ V/m for a wave moving along x at 25 MHz, B is found from B₀ = E₀/c = 2.1 × 10⁻⁸ T with direction along z from E × B pointing along the propagation direction.
- Worked Example 8.2 (p. 208): From By = (2 × 10⁻⁷) sin(0.5 × 10³ x + 1.5 × 10¹¹ t), comparison gives λ = 1.26 cm, ν = 23.9 GHz, and E₀ = B₀c = 60 V/m along the z-axis.
2.4 Common confusions / NTA trap points
- Treating displacement current as a real flow of charge — it is not. It is due to a time-varying electric field but produces magnetic field exactly like conduction current (NCERT p. 203–204).
- Confusing the wavelength order: increasing wavelength is γ-rays < X-rays < UV < visible < IR < microwaves < radio. Many students reverse this (NCERT Summary point 6, p. 212).
- Forgetting that all EM waves travel at the same speed c in vacuum — they differ only in wavelength/frequency (NCERT Points to Ponder 1, p. 213).
- Mixing up the amplitude relation: it is B₀ = E₀/c (not E₀ = B₀/c). NTA distractors swap c into the wrong position (NCERT Eq. 8.10, p. 207).
- Writing c = √(μ₀ε₀) instead of c = 1/√(μ₀ε₀) — a common slip when nervous (NCERT Eq. 8.9a, p. 207).
- Assigning the wrong source: microwave ovens use microwaves (resonant with water molecules), radar uses microwaves, radioactive decay produces γ-rays, inner-shell transitions/bombarding metal produces X-rays — NTA loves matching these (Table 8.1, p. 211).
- Believing EM waves need a medium; in fact they are self-sustaining in vacuum (p. 207).
- Calling EM waves longitudinal — they are transverse, with E and B oscillating perpendicular to the direction of propagation (NCERT §8.3.2, p. 206).
- Forgetting that a uniformly moving charge does not radiate — only accelerated (oscillating) charges do (NCERT §8.3.1, p. 205).
- Confusing UV absorption: the ozone layer at 40–50 km absorbs solar UV; ordinary glass also absorbs UV, which is why one cannot tan through a window (NCERT §8.4.5, p. 210).
- Mis-identifying who first generated short-wavelength microwaves: it was J.C. Bose at Calcutta (25 mm to 5 mm), not Hertz or Marconi (NCERT §8.3.1, p. 206).
- Forgetting the symmetric companion of Faraday: a changing E induces B (Ampere–Maxwell) just as a changing B induces E (Faraday) — this symmetry is the reason EM waves can exist (NCERT §8.2, p. 204).
2.5 Key formulas table
| Symbol | Formula | Meaning | NCERT page |
|---|---|---|---|
| id | id = ε₀ (dΦE/dt) | Displacement current from changing electric flux | 203, Eq. 8.4 |
| ΦE | ΦE = Q/ε₀ (between capacitor plates) | Electric flux between plates of charge Q | 203, Eq. 8.3 |
| i (total) | i = ic + id = ic + ε₀ (dΦE/dt) | Total current = conduction + displacement | 204, Eq. 8.5 |
| Ampere–Maxwell | ∮B·dl = μ₀ ic + μ₀ ε₀ (dΦE/dt) | Generalised Ampere's law | 204, Eq. 8.6 |
| Gauss (E) | ∮E·dA = Q/ε₀ | Maxwell equation 1 | 205 |
| Gauss (B) | ∮B·dA = 0 | Maxwell equation 2 (no monopoles) | 205 |
| Faraday | ∮E·dl = −dΦB/dt | Maxwell equation 3 | 205 |
| Plane EM wave (E) | Ex = E₀ sin(kz − ωt) | Electric component travelling along +z | 206, Eq. 8.7a |
| Plane EM wave (B) | By = B₀ sin(kz − ωt) | Magnetic component, in phase with E | 206, Eq. 8.7b |
| Wave number | k = 2π/λ | Magnitude of propagation vector | 206, Eq. 8.8 |
| Dispersion | ω = ck | Frequency–wavevector relation | 207, Eq. 8.9a |
| Frequency–wavelength | νλ = c | EM wave dispersion in vacuum | 207, Eq. 8.9b |
| Amplitude ratio | B₀ = E₀/c | Magnetic to electric amplitude ratio | 207, Eq. 8.10 |
| Vacuum speed | c = 1/√(μ₀ε₀) ≈ 3 × 10⁸ m/s | Speed of light in vacuum | 207, Eq. 8.9a |
| Medium speed | v = 1/√(με) | EM wave speed in a medium | 207, Eq. 8.11 |
| Photon energy | E = hν | Energy of a photon of frequency ν | 214, Ex. 8.9 |
| EM wave power | (cylindrical capacitor) i = ε₀ A (dE/dt) | Displacement current through plate area A | 203 |
| Refractive index | n = c/v = √(με/μ₀ε₀) | Ratio of vacuum to medium speed | 207 |
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. The displacement current id introduced by Maxwell to remove the inconsistency in Ampere's circuital law is given by:
▸ Show answer & explanation
Answer: B
Displacement current equals ε₀ times the rate of change of electric flux. (A) describes a magnetic-flux quantity and (C) is Faraday's induced emf, not displacement current.
Q2. The speed of electromagnetic waves in free space is given by:
▸ Show answer & explanation
Answer: C
Maxwell's equations give the vacuum speed as the reciprocal of the square root of the product of vacuum permeability and permittivity, numerically about 3 × 10⁸ m/s.
Q3. Which of the following statements about an electromagnetic wave propagating in vacuum is correct?
▸ Show answer & explanation
Answer: C
EM waves are transverse with E ⊥ B ⊥ direction of propagation. (D) is wrong because EM waves are transverse, not longitudinal.
🔒 10 more practice MCQs
Create a free account to unlock every MCQ in this chapter — answers and explanations included. No payment needed.
Already registered? Just log in and they'll all appear here.
Q4. Inside the region between the plates of a charging parallel-plate capacitor:
▸ Show answer & explanation
Answer: C
No charges flow between the plates, so ic = 0, but the changing E-field gives a displacement current equal to the conduction current in the external wire.
Q5. Match the bands of the electromagnetic spectrum (Column I) with their typical wavelength ranges (Column II): Column I — Column II P. Microwaves 1. 1 mm to 700 nm Q. Infrared 2. > 0.1 m R. Radio 3. 400 nm to 1 nm S. Ultraviolet 4. 0.1 m to 1 mm
▸ Show answer & explanation
Answer: A
Microwaves 0.1 m–1 mm; infrared 1 mm–700 nm; radio > 0.1 m; ultraviolet 400 nm–1 nm. Only option (A) matches all four.
Q6. In a plane electromagnetic wave in vacuum, the amplitude of the magnetic field is B₀ = 510 nT. The amplitude of the electric field is closest to:
▸ Show answer & explanation
Answer: B
E₀ = B₀ × c = (510 × 10⁻⁹ T)(3 × 10⁸ m/s) ≈ 153 V/m. Option (A) wrongly divides by c instead of multiplying.
Q7. Which of the following is the correct ordering of electromagnetic waves in order of **increasing wavelength**?
▸ Show answer & explanation
Answer: B
Wavelength increases from γ-rays (shortest) to radio waves (longest). Option (A) is decreasing wavelength; (D) incorrectly places γ-rays after X-rays.
Q8. Assertion (A): Microwave ovens use microwaves of a specific frequency to heat food efficiently. Reason (R): The frequency of microwaves used in ovens is chosen to match the resonant frequency of water molecules, so that energy is efficiently transferred to the kinetic energy of these molecules.
▸ Show answer & explanation
Answer: A
Both statements are taken directly, and R explains exactly why microwave ovens heat food containing water.
Q9. Which of the following types of charge configurations radiate electromagnetic waves?
▸ Show answer & explanation
Answer: C
Only accelerated (e.g., oscillating) charges radiate; the frequency of the EM wave equals the oscillation frequency of the charge.
Q10. The dimensional formula of (μ₀ε₀)⁻¹ is the same as that of:
▸ Show answer & explanation
Answer: B
Since c = 1/√(μ₀ε₀), squaring gives c² = 1/(μ₀ε₀), so (μ₀ε₀)⁻¹ has the dimensions of velocity squared.
Q11. X-rays are commonly produced by:
▸ Show answer & explanation
Answer: B
Hot bodies give IR; radioactive decay gives γ-rays; klystrons give microwaves; X-ray tubes use high-energy electrons hitting metal.
Q12. In a plane electromagnetic wave travelling along the +x direction, the electric field oscillates sinusoidally along the +y direction. The magnetic field must oscillate along:
▸ Show answer & explanation
Answer: C
Since E × B must point along the direction of propagation (+x), and E is along +y (ĵ), we need ĵ × B̂ = î, so B̂ = k̂, i.e., +z direction.
Q13. The greenhouse effect in Earth's atmosphere is mainly due to:
▸ Show answer & explanation
Answer: B
Visible sunlight reaches the surface, is absorbed and re-emitted as infrared at longer wavelengths; CO₂ and H₂O vapour absorb this IR, warming the atmosphere.
📊 Previous-Year Questions
Practise with real CUET Physics previous-year papers — every question solved, with the correct answer and a step-by-step explanation.
View solved CUET PYQ papers →Ready to drill Physics?
Unlock all MCQs, chapter tests, mocks & PYQs for ₹199/year.
Get UniDrill Pro