1. Chapter at a Glance
Maxwell showed that a changing electric field produces a magnetic field, just as a changing magnetic field produces an electric field. Together these changing fields travel through space as electromagnetic (EM) waves at the speed of light. This short chapter covers displacement current, the nature and properties of EM waves, and the electromagnetic spectrum with its uses. NEET asks one fairly direct question almost every year, often on the order of the spectrum or on E₀ = cB₀.
- Displacement current and the modified Ampere's law.
- Relation E₀ = cB₀ and c = 1/√(μ₀ε₀).
- Transverse nature; E, B and direction of travel mutually perpendicular.
- Energy density and intensity.
- Order of the EM spectrum, their sources and uses.
2. Displacement Current
- While a capacitor charges, no conduction current flows across the gap between its plates, yet a magnetic field exists there. Maxwell explained this with the displacement current, produced by the changing electric field.
- In the gap, the displacement current equals the conduction current in the wires: Id = C dV/dt.
- Outside the capacitor there is only conduction current; inside the gap there is only displacement current.
3. Nature of EM Waves
- EM waves are produced by accelerated charges. A charge moving at constant velocity does not radiate.
- They are transverse: E and B are perpendicular to each other and to the direction of propagation (the direction of E × B).
- E and B oscillate in phase and have the same frequency.
- They need no medium; they travel through vacuum at c. In a medium the speed is v = 1/√(με) = c/n.
- The frequency does not change when an EM wave passes from one medium to another; the wavelength and speed change.
4. Energy Carried by EM Waves
- The electric and magnetic fields carry equal shares of the energy.
- For a point source of power P, the intensity at distance r is I = P/(4πr²).
- EM waves also carry momentum, so they exert a small pressure on surfaces they strike.
5. The Electromagnetic Spectrum
| Wave (increasing frequency ↓) | Approx. wavelength | Produced by | Uses |
|---|---|---|---|
| Radio waves | > 0.1 m | Oscillating charges in antennas | Radio, TV, mobile communication |
| Microwaves | 1 mm – 0.1 m | Klystron, magnetron, Gunn diode | Radar, microwave ovens, satellite links |
| Infrared | 700 nm – 1 mm | Hot bodies, molecular vibrations | Remote controls, night vision, physiotherapy, greenhouse effect |
| Visible light | 400 – 700 nm | Electron transitions in atoms | Vision, photography |
| Ultraviolet | 1 – 400 nm | Sun, arc lamps | Sterilising water, LASIK eye surgery; absorbed by ozone |
| X-rays | 0.01 – 10 nm | High-speed electrons hitting a metal target | Medical imaging, cancer treatment, crystal study |
| Gamma rays | < 0.01 nm | Radioactive nuclei | Cancer therapy, food sterilisation |
Increasing frequency: Radio – Micro – Infrared – Visible – Ultraviolet – X-ray – Gamma. Remember "Rabbits Mate In Very Unusual eXotic Gardens". Wavelength decreases along the same order.
6. Formula Master Table
| Quantity | Formula |
|---|---|
| Displacement current | Id = ε₀ dΦE/dt = C dV/dt |
| Speed in vacuum | c = 1/√(μ₀ε₀) |
| Speed in a medium | v = 1/√(με) = c/√(μrεr) |
| Field amplitudes | E₀ = cB₀ |
| Wave relation | c = fλ ; k = 2π/λ ; ω = ck |
| Average energy density | u = ½ε₀E₀² |
| Intensity | I = ½ε₀cE₀² |
| Point source | I = P/4πr² |
7. Master Summary Table
| Property | Fact | NEET Note |
|---|---|---|
| Source | Accelerated charges | Uniform motion does not radiate |
| Nature | Transverse | Can be polarised |
| Phase of E and B | In phase | Both peak together |
| Medium | Not needed | Travel in vacuum at c |
| Energy share | Equal in E and B | uE = uB |
| Frequency in a new medium | Unchanged | λ and v change |
| Ozone layer | Absorbs UV | Protects life |
8. Mind Map
- ε₀ dΦ/dt
- Capacitor gap
- Transverse
- E₀ = cB₀
- No medium
- u = ½ε₀E₀²
- I = uc
- Radio → gamma
- Sources
- Uses
9. Tips and Tricks
- From an equation like E = E₀ sin(kx − ωt), speed = ω/k; wavelength = 2π/k; frequency = ω/2π.
- If B is along y and the wave travels along x, then E is along z (use E × B along travel).
- For frequency questions, f (Hz) = 3 × 10⁸/λ (m).
- "Heat waves" = infrared; "sterilising" = UV or gamma; "bones" = X-rays.
- Thinking E and B are 90° out of phase (they are in phase).
- Mixing up the order of UV and X-rays or of microwaves and infrared.
- Saying the magnetic field carries less energy because B₀ is numerically small.
- Thinking frequency changes on entering glass.
10. Important One-liners
- EM waves were predicted by Maxwell and first produced by Hertz.
- All EM waves travel at the same speed in vacuum.
- Gamma rays have the highest frequency and shortest wavelength.
- Radio waves have the longest wavelength.
- Infrared radiation keeps the Earth warm through the greenhouse effect.
- The ozone layer absorbs harmful ultraviolet radiation.
- Microwave ovens heat food by making water molecules oscillate.
- Displacement current exists even in vacuum.
11. Solved Numericals: Average to Advanced
Five average and five advanced numericals. Take c = 3 × 10⁸ m s⁻¹, ε₀ = 8.854 × 10⁻¹² F m⁻¹ and μ₀ = 4π × 10⁻⁷ T m A⁻¹. Click to reveal each answer and working.
Average level
Find the wavelength of a radio station broadcasting at 75 MHz.
Show answer & solution
Concept: λ = c/f.
- λ = 3 × 10⁸ / 75 × 10⁶.
- λ = 4 m.
Answer: 4 m
The electric field amplitude of an EM wave in vacuum is 45 V m⁻¹. Find the magnetic field amplitude.
Show answer & solution
Concept: B₀ = E₀/c.
- B₀ = 45/3 × 10⁸.
- B₀ = 1.5 × 10⁻⁷ T.
Answer: 1.5 × 10⁻⁷ T
Red light has a wavelength of 750 nm. Find its frequency. In which part of the spectrum is 750 nm found?
Show answer & solution
Concept: f = c/λ.
- f = 3 × 10⁸ / 750 × 10⁻⁹ = 4 × 10¹⁴ Hz.
- 750 nm is at the edge between visible red and near-infrared.
Answer: 4 × 10¹⁴ Hz; the red / near-infrared boundary
A 5 μF capacitor is being charged so that its voltage rises at 2 × 10³ V s⁻¹. Find the displacement current between its plates.
Show answer & solution
Concept: Id = C dV/dt.
- Id = 5 × 10⁻⁶ × 2 × 10³.
- Id = 1 × 10⁻² A = 10 mA.
Answer: 10 mA
Find the speed of an EM wave in a non-magnetic medium of relative permittivity 4.
Show answer & solution
Concept: v = c/√(μrεr).
- μr = 1, εr = 4 ⇒ √4 = 2.
- v = 3 × 10⁸/2 = 1.5 × 10⁸ m s⁻¹ (refractive index 2).
Answer: 1.5 × 10⁸ m s⁻¹
Advanced level
An EM wave has an electric field amplitude of 50 V m⁻¹. Find its average energy density and intensity.
Show answer & solution
Concept: u = ½ε₀E₀²; I = uc.
- u = ½ × 8.854 × 10⁻¹² × 2500 ≈ 1.11 × 10⁻⁸ J m⁻³.
- I = 1.11 × 10⁻⁸ × 3 × 10⁸ ≈ 3.32 W m⁻².
Answer: about 1.1 × 10⁻⁸ J m⁻³; about 3.3 W m⁻²
A 100 W bulb converts 5% of its power into visible light, radiated equally in all directions. Find the intensity of visible light 2 m away and the amplitude of its electric field.
Show answer & solution
Concept: I = P/(4πr²); I = ½ε₀cE₀².
- Visible power = 5 W; I = 5/(4π × 4) ≈ 0.0995 W m⁻².
- E₀ = √(2I/(ε₀c)) = √(0.199/(8.854 × 10⁻¹² × 3 × 10⁸)).
- = √(0.199/2.656 × 10⁻³) = √74.9 ≈ 8.66 V m⁻¹.
Answer: about 0.1 W m⁻²; about 8.7 V m⁻¹
The magnetic field of a wave in vacuum is By = 2 × 10⁻⁷ sin(500x + 1.5 × 10¹¹ t) T. Find its wavelength, frequency, the amplitude of E and its direction of travel.
Show answer & solution
Concept: compare with B₀ sin(kx + ωt).
- k = 500 m⁻¹ ⇒ λ = 2π/500 ≈ 1.26 × 10⁻² m.
- ω = 1.5 × 10¹¹ rad s⁻¹ ⇒ f = ω/2π ≈ 2.39 × 10¹⁰ Hz (check: ω/k = 3 × 10⁸ m s⁻¹ ✓).
- E₀ = cB₀ = 3 × 10⁸ × 2 × 10⁻⁷ = 60 V m⁻¹.
- (kx + ωt) means the wave travels in the −x direction.
Answer: about 1.26 cm; about 2.4 × 10¹⁰ Hz (microwave); 60 V m⁻¹; along −x
A parallel-plate capacitor with circular plates of radius 5 cm is charged by a steady current of 0.25 A. Find the rate of change of the electric field between the plates.
Show answer & solution
Concept: Id = ε₀A dE/dt equals the charging current.
- A = π(0.05)² ≈ 7.85 × 10⁻³ m².
- dE/dt = I/(ε₀A) = 0.25/(8.854 × 10⁻¹² × 7.85 × 10⁻³).
- dE/dt ≈ 3.6 × 10¹² V m⁻¹ s⁻¹.
Answer: about 3.6 × 10¹² V m⁻¹ s⁻¹
The magnetic field amplitude of a light wave is 4 × 10⁻⁸ T. Find E₀ and the average energy density, and show that the electric and magnetic parts carry equal energy.
Show answer & solution
Concept: E₀ = cB₀; u = B₀²/(2μ₀) = ½ε₀E₀².
- E₀ = 3 × 10⁸ × 4 × 10⁻⁸ = 12 V m⁻¹.
- Using B: u = (4 × 10⁻⁸)²/(2 × 4π × 10⁻⁷) ≈ 6.37 × 10⁻¹⁰ J m⁻³.
- Using E: u = ½ × 8.854 × 10⁻¹² × 144 ≈ 6.37 × 10⁻¹⁰ J m⁻³ — the same, so each field contributes half of the total when averaged.
Answer: 12 V m⁻¹; about 6.4 × 10⁻¹⁰ J m⁻³
12. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
Which of these has the shortest wavelength?
- Microwaves
- Ultraviolet rays
- X-rays
- Infrared rays
Show solution
In order of decreasing wavelength: microwaves > infrared > UV > X-rays. So X-rays have the shortest wavelength of the four.
Answer: (C)
EM waves are produced by:
- Stationary charges
- Charges moving with constant velocity
- Accelerated charges
- Neutral particles
Show solution
Only accelerated charges produce changing fields that radiate. A stationary charge gives only a static E field, and a charge in uniform motion gives steady E and B fields but no radiation.
Answer: (C)
In an EM wave, the phase difference between the electric and magnetic fields is:
- 0
- π/2
- π
- π/4
Show solution
E and B oscillate in phase. They are perpendicular in direction, which students often confuse with a 90° phase difference.
Answer: (A)
The radiation used to kill germs in water purifiers is:
- Infrared
- Microwaves
- Ultraviolet
- Radio waves
Show solution
UV radiation damages the DNA of microbes, so it is used to sterilise water. Infrared only heats, and microwaves and radio waves carry too little energy per photon to do this.
Answer: (C)
Displacement current exists in the region between the plates of a capacitor only when:
- The capacitor is fully charged
- The electric field between the plates is changing
- A dielectric is present
- The plates are touching
Show solution
Id = ε₀ dΦE/dt is non-zero only while E is changing, that is, while the capacitor charges or discharges. A dielectric is not needed.
Answer: (B)
An EM wave travels along +z with its electric field along +x. Its magnetic field is along:
- +x
- +y
- −y
- +z
Show solution
E × B must point along the direction of travel (+z). Since x̂ × ŷ = ẑ, B is along +y. B cannot be along x or z, because it must be perpendicular to both E and the direction of travel.
Answer: (B)
Which of the following does NOT change when light passes from air into glass?
- Speed
- Wavelength
- Frequency
- Amplitude of E in general
Show solution
The frequency is fixed by the source, so it stays the same. The speed drops to c/n and the wavelength drops to λ/n. The amplitude also usually changes because of partial reflection.
Answer: (C)
The ratio of the energy densities of the electric field and the magnetic field in an EM wave is:
- 1 : 1
- c : 1
- 1 : c
- c² : 1
Show solution
uE = ½ε₀E² and uB = B²/(2μ₀). With E = cB and c² = 1/(μ₀ε₀), the two are equal. The ratio c : 1 is the ratio of the field amplitudes E/B, not of the energies.
Answer: (A)
13. Quick Revision Summary
- Id = ε₀ dΦE/dt completes Ampere's law.
- c = 1/√(μ₀ε₀); E₀ = cB₀; E, B in phase; transverse.
- u = ½ε₀E₀²; I = uc; equal energy in E and B.
- Spectrum by rising frequency: radio, micro, IR, visible, UV, X, γ.
- Frequency stays fixed across media; λ and v change.