📌 Snapshot
- Displacement current, Maxwell's equations, the speed of light and the EM spectrum explained with numericals.
- Why a current that crosses no wire still makes a magnetic field, and how that idea predicted radio waves.
- How E and B sit in a wave, how $c = 1/\sqrt{\mu_0\varepsilon_0}$ arises, and how to read a wave equation in seconds.
- The full spectrum from radio to gamma rays in one table, with sources, detectors, uses and match the column traps.
📖 Detailed Notes
Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 8.
A magnetic field in a gap with nothing in it
Charge a capacitor through a wire. Current flows in the wire and, as you already know, it makes a magnetic field around the wire. But between the plates there is only air or vacuum. No charge crosses the gap, so there is no current there at all. Does the magnetic field stop at the plates?
In the 1860s James Clerk Maxwell showed that it does not, and chasing that answer led him to something much bigger: light itself is an electromagnetic wave. This chapter follows that chain of reasoning, from a puzzle in Ampere's law to the full electromagnetic spectrum.
The puzzle in Ampere's circuital law
Ampere's law says that the line integral of B around a closed loop equals μ₀ times the current passing through any surface bounded by the loop:
$$\oint B \cdot dl = \mu_0 i$$
Take a circular loop around the wire that charges a parallel plate capacitor. Choose a flat disc as the surface, and the wire's current passes through it, so B ≠ 0 on the loop. Now choose a different surface with the same rim: a bowl-shaped surface that bulges out and passes between the plates. No wire current pierces this bowl, so the law says $B = 0$ on the same loop. The same loop cannot have two different values of B. Something is missing.
The missing piece is the changing electric field between the plates. With plate area A and charge Q, the field is $E = Q/(\varepsilon_0 A)$, so the electric flux is $\Phi_E = Q/\varepsilon_0$. As Q grows with time:
$$\varepsilon_0 \dfrac{d\Phi_E}{dt} = \dfrac{dQ}{dt} = i$$
This quantity has the units of current and exactly equals the wire's current. Maxwell called it the displacement current:
$$i_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}$$
It is not charge moving. It is a changing electric field, and it creates a magnetic field in exactly the way a conduction current does. Between the plates the conduction current is zero and the displacement current equals i. In the wire the conduction current is i and the displacement current is zero. Whichever surface you choose, the total current i_c + i_d through it is the same, and the contradiction vanishes. The corrected law is the Ampere–Maxwell law:
$$\oint B \cdot dl = \mu_0 i_c + \mu_0\varepsilon_0 \dfrac{d\Phi_E}{dt}$$
Worked example: the field inside a charging capacitor. A parallel plate capacitor has circular plates of radius 10 cm and is charged by a steady current of 2.0 A. Find (a) the displacement current between the plates and the rate at which E changes, and (b) the magnetic field at 5 cm from the axis, between the plates.
| Step | What we do | Result |
|---|---|---|
| 1 | Displacement current must equal the charging current | $i_d = 2.0$ A |
| 2 | Plate area | $A = \pi(0.10)^2 = 3.14 \times 10^{-2}$ m² |
| 3 | $i_d = \varepsilon_0 A\,dE/dt$, so $dE/dt = i/(\varepsilon_0 A)$ | $2.0/(8.85 \times 10^{-12} \times 3.14 \times 10^{-2}) = 7.2 \times 10^{12}$ V m⁻¹ s⁻¹ |
| 4 | Apply Ampere–Maxwell to a loop of radius $r = 5$ cm. The displacement current is uniform over the plate, so the loop encloses the fraction $r^2/R^2$ of it | $B(2\pi r) = \mu_0 i (r^2/R^2)$ |
| 5 | Solve for B | $B = \mu_0 i r/(2\pi R^2) = (2 \times 10^{-7})(2.0)(0.05)/(0.10)^2 = 2.0 \times 10^{-6}$ T |
Common wrong step. Putting $B = 0$ between the plates because "no current flows there". The displacement current is as good a source of magnetic field as a wire.
Pause and predict: a capacitor is charged by a steady current. Is the displacement current between its plates larger, smaller or equal to the current in the connecting wire?
Check your answer: equal. The displacement current continues the wire's current across the gap, so the total current stays continuous around the circuit. This continuity is the whole reason Maxwell introduced the idea.
Maxwell's four equations
Maxwell collected everything known about electricity and magnetism into four equations. Together with the Lorentz force $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$ they describe all of classical electromagnetism.
| Equation | Statement | In words |
|---|---|---|
| Gauss's law for E | $\oint E \cdot dA = Q/\varepsilon_0$ | Electric charge is the source of electric field lines |
| Gauss's law for B | $\oint B \cdot dA = 0$ | Magnetic field lines have no beginning or end; no isolated magnetic monopoles |
| Faraday's law | $\oint E \cdot dl = -d\Phi_B/dt$ | A changing magnetic field produces an electric field |
| Ampere–Maxwell law | $\oint B \cdot dl = \mu_0 i_c + \mu_0\varepsilon_0\, d\Phi_E/dt$ | Current and a changing electric field produce a magnetic field |
The last two form a pair. Faraday says a changing B makes an E. Maxwell's addition says a changing E makes a B. If you shake a charge, a changing E appears, which makes a changing B, which makes a changing E again, and so on. Each field regenerates the other, and the disturbance can detach from the charge and travel by itself. That is an electromagnetic wave.
Where waves come from
A stationary charge makes only an electrostatic field. A charge moving with constant velocity makes a steady magnetic field in addition. Neither radiates. Only an accelerated charge sends out electromagnetic waves, and a charge oscillating at frequency ν radiates waves of the same frequency ν. The energy carried away comes from the motion of the source.
Maxwell predicted these waves, and Heinrich Hertz produced and detected them in 1887. Jagadish Chandra Bose, working in Calcutta about seven years later, generated much shorter waves with wavelengths from 25 mm to 5 mm. Guglielmo Marconi then sent such waves over long distances and began wireless communication.
Pause and predict: an electron moves in a circle at a constant speed. Does it radiate?
Check your answer: yes. Its speed is constant, but its velocity keeps changing direction, so it is accelerating. Any acceleration of a charge, not only speeding up or slowing down, produces radiation.
What an electromagnetic wave looks like
Solving Maxwell's equations in free space gives a travelling wave with these properties. E and B are both perpendicular to the direction of travel and to each other, so the wave is transverse. They oscillate in phase, reaching their maximum and zero at the same instants. For a wave moving along +z with E along x and B along y:
$$E_x = E_0 \sin(kz - \omega t), \quad B_y = B_0 \sin(kz - \omega t)$$
where $k = 2\pi/\lambda$ and $\omega = 2\pi\nu$. The direction of travel is the direction of E × B. The wave speed comes out of the equations as
$$c = \omega/k = \nu\lambda = 1/\sqrt{\mu_0\varepsilon_0} \approx 3 \times 10^{8}\ \text{m/s}$$
Putting $\mu_0 = 4\pi \times 10^{-7}$ and $\varepsilon_0 = 8.85 \times 10^{-12}$ into the formula gives 2.998 × 10⁸ m/s, exactly the speed of light measured in experiments. This coincidence, a number from electric and magnetic constants matching the speed of light, was the clue that light is an electromagnetic wave.
The two amplitudes are linked by the speed:
$E_0/B_0 = c$, or $E_0 = cB_0$
So a tiny B goes with a large E in SI units: a magnetic field of 1 nT goes with an electric field of 0.3 V/m. The electric and magnetic energy densities are equal at every point of the wave, $1/2\,\varepsilon_0 E^2 = B^2/2\mu_0$.
EM waves need no medium. The fields support each other, so the waves cross a vacuum, which is how sunlight reaches Earth. In a material of permittivity ε and permeability μ, the speed is
$v = 1/\sqrt{\mu\varepsilon}$, $n = c/v$
and the frequency stays the same while the wavelength shrinks to $v/\nu$.
Worked example: reading a wave. A plane EM wave of frequency 25 MHz travels in vacuum along +x. At a point, the electric field is $\vec{E} = 6.3\,\hat{j}$ V/m. Find the wavelength, the wave number, the magnetic field amplitude and the direction of B.
| Step | What we do | Result |
|---|---|---|
| 1 | $\lambda = c/\nu$ | $3 \times 10^{8}/(25 \times 10^{6}) = 12$ m |
| 2 | $k = 2\pi/\lambda$ | 0.52 rad/m |
| 3 | $B_0 = E_0/c$ | $6.3/(3 \times 10^{8}) = 2.1 \times 10^{-8}$ T |
| 4 | E × B must point along +x, and $\hat{j} \times \hat{k} = \hat{i}$ | B is along +z |
Common wrong step. Writing $B_0 = cE_0$, which gives an absurd 1.9 × 10⁹ T. In SI units B is always the smaller number. The correct relation is $E_0 = cB_0$.
Worked example: decode the equation. The magnetic field in a wave is $B_y = (2 \times 10^{-7})\sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t)$ tesla. Find the wavelength, frequency, electric field amplitude and the direction of travel.
| Step | What we do | Result |
|---|---|---|
| 1 | Compare with $\sin(kx \pm \omega t)$: $k = 0.5 \times 10^{3} = 500$ rad/m | $\lambda = 2\pi/k = 1.26$ cm |
| 2 | $\omega = 1.5 \times 10^{11}$ rad/s | $\nu = \omega/2\pi = 2.39 \times 10^{10}$ Hz = 23.9 GHz |
| 3 | Check the speed | $\omega/k = 1.5 \times 10^{11}/500 = 3 \times 10^{8}$ m/s |
| 4 | $E_0 = cB_0$ | $3 \times 10^{8} \times 2 \times 10^{-7} = 60$ V/m |
| 5 | The sign inside the sine is plus, which means $kx + \omega t$ | The wave travels along −x |
Common wrong step. Reading $k = 500$ as the wavelength, or forgetting that $kx + \omega t$ moves the wave towards negative x.
Worked example: light enters glass. Light of wavelength 600 nm in vacuum enters glass of refractive index 1.5. Find its frequency, speed and wavelength in the glass.
| Step | What we do | Result |
|---|---|---|
| 1 | $\nu = c/\lambda$ in vacuum | $3 \times 10^{8}/(600 \times 10^{-9}) = 5 \times 10^{14}$ Hz |
| 2 | $v = c/n$ | 2 × 10⁸ m/s |
| 3 | Frequency is fixed by the source, so $\lambda' = v/\nu$ | 400 nm |
Common wrong step. Saying the frequency drops to 3.3 × 10¹⁴ Hz because the speed drops. The source decides frequency. Speed and wavelength change together.
The electromagnetic spectrum
All EM waves are the same kind of wave and differ only in wavelength and frequency. In vacuum every one of them travels at c. We split the spectrum into bands by how the waves are produced and detected. The boundaries between bands are conventional and overlap.
| Band | Wavelength range | Produced by | Detected by | Uses |
|---|---|---|---|---|
| Radio | greater than 0.1 m | Accelerated electrons in aerials | Receiver aerials | Radio and TV, cellular phones |
| Microwave | 0.1 m to 1 mm | Klystron, magnetron, Gunn diode | Point contact diodes | Radar, microwave ovens |
| Infrared | 1 mm to 700 nm | Vibrating atoms and molecules; hot bodies | Thermopile, bolometer | Physiotherapy, remote controls, satellite imaging |
| Visible | 700 nm to 400 nm | Electron transitions in atoms | Eye, photocells, film | Vision, photography |
| Ultraviolet | 400 nm to 0.6 nm | Inner shell electron transitions; very hot bodies | Photocells, film | Sterilising water, LASIK eye surgery |
| X-rays | 10 nm to 10⁻⁴ nm | X-ray tubes; inner shell transitions | Film, Geiger tube, ionisation chamber | Medical diagnosis, cancer treatment |
| Gamma rays | below 10⁻¹⁰ m | Radioactive decay of nuclei | Same as X-rays | Destroying cancer cells |
Going from radio to gamma, the wavelength decreases and the frequency and photon energy ($h\nu$) increase. Radio waves come from antennas metres long, visible light from electron jumps inside atoms, and gamma rays from the nucleus, which is about 10⁻¹⁴ m across. Smaller sources radiate shorter waves.
A few facts are worth knowing precisely. Visible light spans frequencies of roughly 4 × 10¹⁴ Hz to 7 × 10¹⁴ Hz. The ozone layer, 40 to 50 km above the ground, absorbs the harmful ultraviolet from the Sun, so ozone depletion is a concern. Infrared is absorbed strongly by water, CO₂ and NH₃. Radar uses microwaves because their short wavelength gives sharp, directional beams.
Beyond the textbook. For the exam: a microwave oven works because the microwave frequency matches the resonant frequency of water molecules, so energy transfers efficiently to them. In reality: ovens run at 2.45 GHz, which is well below water's strongest absorption. Water molecules are electric dipoles that keep trying to turn with the rapidly reversing field, and their jostling with neighbours turns that motion into heat. This frequency lets the waves penetrate a few centimetres into food. It is also in a band set aside for such uses. Why it matters: the key idea, that polar molecules absorb the field's energy, is the same, but the word "resonance" is a simplification.
Beyond the textbook. For the exam: the greenhouse effect is the trapping of infrared radiation, re-emitted by the warm Earth, by gases like CO₂ and water vapour. In reality: the natural greenhouse effect is what keeps the planet habitable. Without it the average surface temperature would be well below freezing, about −18 °C instead of roughly +15 °C. Rising CO₂ from fossil fuels strengthens the effect and causes the warming we are now measuring. Why it matters: the same absorption property that makes infrared useful for heating in physiotherapy makes it the main player in climate.
Beyond the textbook. For the exam: X-rays come from bombarding a metal target with fast electrons, and gamma rays come from radioactive decay. In reality: the wavelength ranges overlap, and an X-ray and a gamma ray of the same wavelength are physically identical. The names tell you the origin, not a boundary. Why it matters: never expect a clean wavelength cutoff between neighbouring bands, only a typical range for each.
Pause and predict: which has more energy per photon: a radio wave photon or an X-ray photon?
Check your answer: the X-ray photon, by a factor of millions or more. Photon energy is $h\nu$, and X-rays have a far higher frequency. This is why X-rays ionise atoms and damage tissue, while radio waves pass through you harmlessly.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "Displacement current is charges flowing across the gap." | The word "current" suggests moving charge | No charge crosses. It is $\varepsilon_0\, d\Phi_E/dt$, a changing electric field |
| "$E_0 = B_0/c$" | The c has to go somewhere | $E_0 = cB_0$, so $B_0 = E_0/c$ |
| "$c = \sqrt{\mu_0\varepsilon_0}$" | The square root is memorised, the reciprocal forgotten | $c = 1/\sqrt{\mu_0\varepsilon_0}$ |
| "EM waves need a medium, like sound." | Every wave you met in Class XI did | They travel in vacuum. That is how sunlight reaches us |
| "EM waves are longitudinal." | Sound is | E and B are perpendicular to the direction of travel: transverse |
| "A charge at constant speed radiates." | It is moving | Only accelerated charges radiate. Circular motion at constant speed is accelerated |
| "Gamma rays are always shorter than X-rays." | Gamma looks like the top of the list | The ranges overlap. The names reflect origin: nuclei versus electrons |
| "Only infrared waves heat things." | They are called heat waves | Any absorbed EM wave heats. Infrared is called heat radiation because hot bodies emit it and many molecules absorb it strongly |
Formula sheet
| Quantity | Formula | SI unit | When to use |
|---|---|---|---|
| Displacement current | $i_d = \varepsilon_0\, d\Phi_E/dt$ | ampere | Between capacitor plates; $i_d = i_c$ in the circuit |
| Ampere–Maxwell law | $\oint B \cdot dl = \mu_0(i_c + i_d)$ | T m | Field of a changing E |
| Wave equations | $E_x = E_0\sin(kz - \omega t)$, $B_y = B_0\sin(kz - \omega t)$ | V/m, T | Reading amplitude and direction |
| Wave number | $k = 2\pi/\lambda$ | rad/m | Compare with the wave equation |
| Wave speed | $c = \omega/k = \nu\lambda$ | m/s | Frequency and wavelength conversions |
| Speed in vacuum | $c = 1/\sqrt{\mu_0\varepsilon_0}$ | m/s | Numerical value or dimensions |
| Amplitude ratio | $E_0 = cB_0$ | V/m | Find one field from the other |
| Speed in a medium | $v = 1/\sqrt{\mu\varepsilon}$, $n = c/v$ | m/s | Waves inside materials |
| Photon energy | $E = h\nu$ | joule | Comparing bands |
60-second revision
| Topic | Lock this in |
|---|---|
| Displacement current | $i_d = \varepsilon_0\, d\Phi_E/dt$. A changing E acts like a current. Zero in the wire, equal to i between the plates |
| Maxwell's equations | Gauss (E), Gauss (B), Faraday, Ampere–Maxwell |
| Sources | Only accelerated charges radiate, at the same frequency as their oscillation |
| Wave nature | Transverse, E ⊥ B ⊥ direction of travel, in phase, no medium needed |
| Speed | $c = 1/\sqrt{\mu_0\varepsilon_0} = 3 \times 10^{8}$ m/s. $E_0 = cB_0$. Energy shared equally between E and B |
| Spectrum order, long to short wavelength | Radio, microwave, infrared, visible, ultraviolet, X-ray, gamma |
| Origins | Radio: aerials. Microwave: magnetron. IR: hot bodies. UV, X-ray: electrons. Gamma: nucleus |
| Hertz and Bose | Hertz detected radio waves in 1887. Bose made 25 mm to 5 mm waves in Calcutta |
| Environment | Ozone absorbs UV. Greenhouse gases trap infrared |
Where this chapter connects
- Previous: Alternating Current covers the LC oscillations that are the circuit behind every radio transmitter.
- Next: Ray Optics and Optical Instruments treats visible light as straight line rays, once you know it is an EM wave of tiny wavelength.
- Foundations: Electromagnetic Induction is Faraday's law, one half of the pair that makes waves possible, and Electric Charges and Fields supplies Gauss's law.
- Later link: Dual Nature of Radiation and Matter shows why photon energy $h\nu$ decides which band can eject electrons.
🎯 Practice MCQs
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Q1. A parallel plate capacitor is being charged by a steady current of 2 A flowing in the connecting wires. The displacement current between the plates is
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Answer: C
Between the plates the changing electric flux gives $i_d = \varepsilon_0\, d\Phi_E/dt = dQ/dt$, which equals the conduction current in the wire, 2 A. Option (A) assumes no current flows in the gap. Options (B) and (D) have no basis, because the displacement current must be exactly equal to the wire current to keep the total current continuous.
Q2. A plane electromagnetic wave travels in a non magnetic medium ($\mu = \mu_0$) whose permittivity is four times $\varepsilon_0$. The speed of the wave in this medium is
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Answer: C
$v = 1/\sqrt{\mu\varepsilon} = 1/\sqrt{\mu_0 \times 4\varepsilon_0} = c/2 = 1.5 \times 10^{8}$ m/s. Option (A) wrongly multiplies c by 2. Option (B) is the vacuum speed. Option (D) divides by 4 instead of by the square root of 4.
Q3. Which of the following statements about an electromagnetic wave in vacuum are correct? I. The electric and magnetic fields are perpendicular to each other. II. Both fields are perpendicular to the direction of propagation. III. The electric and magnetic fields oscillate in phase. IV. The wave needs a material medium to travel.
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Answer: A
The wave is transverse with E and B mutually perpendicular and both perpendicular to the direction of travel, and E and B reach their maxima together. Statement IV is false because electromagnetic waves travel through vacuum. Options (C) and (D) include IV, and option (B) leaves out the true statement III.
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Q4. In a plane electromagnetic wave in vacuum, the amplitude of the magnetic field is 510 nT. The amplitude of the electric field is approximately
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Answer: C
$E_0 = cB_0 = (3 \times 10^{8})(510 \times 10^{-9}) = 153$ V/m. Option (A) divides by c instead of multiplying. Option (B) is off by a factor of ten in the arithmetic. Option (D) equates the numerical values of E and B, ignoring c.
Q5. A plane electromagnetic wave travels along the negative x direction. At an instant its electric field points along the positive y direction. The magnetic field at that instant points along
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Answer: B
The direction of travel is the direction of $E \times B$. We need $\hat{y} \times \hat{B} = -\hat{x}$. Since $\hat{y} \times \hat{z} = \hat{x}$, the required $\hat{B}$ is $-\hat{z}$. Option (A) would give a wave moving along +x. Options (C) and (D) put B parallel to E, which can never happen in a transverse wave.
Q6. The magnetic field of an electromagnetic wave in vacuum is $B_y = (2 \times 10^{-7})\sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t)$ tesla, with x in metres and t in seconds. The amplitude of the electric field and the direction of propagation are
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Answer: B
$E_0 = cB_0 = 3 \times 10^{8} \times 2 \times 10^{-7} = 60$ V/m. The argument $kx + \omega t$ means the wave moves towards negative x. The ratio $\omega/k = 3 \times 10^{8}$ m/s confirms the speed. Option (A) misreads the sign. Option (C) divides by c. Option (D) confuses the direction of propagation with the direction of the field.
Q7. Match List I with List II. List I: (i) Microwaves (ii) Infrared (iii) Radio waves (iv) Ultraviolet. List II: (p) 0.1 m to 1 mm (q) 1 mm to 700 nm (r) greater than 0.1 m (s) 400 nm to 0.6 nm
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Answer: A
Microwaves lie between 0.1 m and 1 mm. Infrared lies between 1 mm and 700 nm, just beyond the red end of visible light. Radio waves are longer than 0.1 m. Ultraviolet runs from 400 nm down to about 0.6 nm. The other options misplace the neighbouring bands.
Q8. Match each type of electromagnetic wave with its usual method of production. List I: (i) Microwaves (ii) X-rays (iii) Gamma rays (iv) Infrared. List II: (p) Radioactive decay of nuclei (q) Magnetron or klystron valve (r) Bombarding a metal target with fast electrons (s) Vibration of atoms and molecules in hot bodies
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Answer: A
Microwaves are made in magnetrons and klystrons, X-rays in tubes where fast electrons strike a metal target, gamma rays in nuclear decay, and infrared by vibrating atoms and molecules in warm bodies. Option (B) swaps X-rays with gamma rays, and option (C) swaps microwaves with infrared.
Q9. Assertion (A): Sunlight reaches the Earth even though there is almost no matter between the Sun and the Earth. Reason (R): Electromagnetic waves are transverse in nature.
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Answer: B
Both statements are true, but the transverse nature of the wave is not why it crosses empty space. It travels in vacuum because a changing electric field produces a magnetic field and vice versa, so the wave sustains itself without a medium. Waves on a string are also transverse yet need a medium, so transverse nature alone explains nothing.
Q10. Which of the following charge configurations radiates electromagnetic waves?
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Answer: C
A radiating charge must be accelerated. Motion in a circle at constant speed involves a continuous change in the direction of velocity, so the charge has centripetal acceleration and radiates. Options (A) and (B) have zero acceleration, so only static fields exist. Option (D) is therefore wrong.
Q11. Which of the following statements are correct? I. The ozone layer absorbs most of the harmful ultraviolet radiation from the Sun. II. The greenhouse effect is caused by the trapping of infrared radiation by gases like carbon dioxide and water vapour. III. X-rays are produced by the radioactive decay of nuclei. IV. Microwaves are used in radar.
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Answer: A
Statements I, II and IV are standard results. Statement III is false because radioactive decay produces gamma rays, while X-rays come from fast electrons striking a metal target or from inner shell electron transitions. Every other option includes III.
Q12. A radio station broadcasts at a frequency of 90 MHz. The wavelength of its waves in air is approximately
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Answer: A
$\lambda = c/\nu = 3 \times 10^{8}/(90 \times 10^{6}) = 3.33$ m. Option (B) is the wavelength at 1 GHz. Options (C) and (D) are out by factors of ten and a thousand.
Q13. In an electromagnetic wave in vacuum, the ratio of the average electric energy density to the average magnetic energy density is
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Answer: A
The electric energy density is $1/2\,\varepsilon_0 E^2$ and the magnetic is $B^2/2\mu_0$. Substituting $E = cB$ and $c^2 = 1/\mu_0\varepsilon_0$ gives $1/2\,\varepsilon_0 c^2 B^2 = B^2/2\mu_0$, so they are equal. The wave shares its energy equally between the two fields, so options (B), (C) and (D) are wrong.
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