Displacement current and Maxwell's idea
While a capacitor is being charged, the wire carries a current, yet no charge crosses the gap between the plates. Link to earlier classes : Ampere's circuital law from Chapter 4 and capacitors from Chapter 2.
Maxwell noticed that Ampere's law fails for a charging capacitor. He added a term called the displacement current, produced by a changing electric field, and so completed the law. A changing electric field therefore produces a magnetic field, just as a changing magnetic field produces an electric field.
Rule 1: Displacement current — I_d = ε₀ dΦ_E ÷ dt (Φ_E = electric flux). The total current is I = I_c + I_d, and the Ampere–Maxwell law is ∮ B · dl = μ₀ (I_c + I_d).
- Easy : A capacitor is charged by a conduction current of 2 A. Between its plates there is no conduction current, but I_d = 2 A, so the current is continuous.
- Medium : For a capacitor of 10 pF whose voltage rises at 10⁹ V s⁻¹, I = C dV/dt = 10 × 10⁻¹² × 10⁹ = 0.010 A, which is also I_d between the plates.
- Exam : For I_d = 2 A, the flux must change at dΦ_E/dt = 2 ÷ 8.85 × 10⁻¹² = 2.26 × 10¹¹ V m s⁻¹.
Rule 2: Importance — The displacement current is the link that predicts electromagnetic waves: changing E makes B, changing B makes E, and the two sustain each other as they travel through empty space.
- Easy : Without I_d, the magnetic field around a charging capacitor could not be explained.
- Medium : Maxwell's equations together describe all of electricity, magnetism and light.
- Exam : Hertz confirmed the predicted waves experimentally in 1887.
I_d = ε₀ dΦ_E/dt; I_c + I_d is continuous; changing E gives B.
Thinking the displacement current is a flow of real charge. It is a changing electric field.
Memory clue : "Changing E makes B; changing B makes E: waves are born."
Spot the difference
| Current | Cause | Where |
|---|---|---|
| Conduction I_c | flow of charges | wires |
| Displacement I_d | changing electric field | between capacitor plates |
Practice questions
- 1Who introduced the displacement current?
- 2Write the formula for the displacement current.
- 3Where does a displacement current exist while a capacitor is charging?
- 4A conduction current of 2 A charges a capacitor. What is the displacement current between the plates?
- 5A 10 pF capacitor's voltage rises at 10⁹ V s⁻¹. Find the displacement current.
- 6Find dΦ_E/dt that gives a displacement current of 2 A.
- 7Write the Ampere–Maxwell law.
- 8Why was the displacement current needed?
Show answers (try them yourself first)
- 1James Clerk Maxwell.
- 2I_d = ε₀ dΦ_E ÷ dt.
- 3Between the plates.
- 42 A.
- 5I_d = C dV/dt = 0.010 A (10 mA).
- 62 ÷ 8.85 × 10⁻¹² ≈ 2.26 × 10¹¹ V m s⁻¹.
- 7∮ B · dl = μ₀ (I_c + I_d).
- 8Without it, Ampere's law gave contradictory answers for a charging capacitor.
Electromagnetic waves and their properties
Light, radio waves, the signal of a mobile phone and the heat of a heater are the same kind of wave with different frequencies. Link to earlier classes : waves from Class 11 and light from earlier classes.
An electromagnetic wave is made of electric and magnetic fields oscillating in phase at right angles to each other and to the direction of travel. They are transverse, need no medium and travel in vacuum at the speed of light, c = 3 × 10⁸ m s⁻¹. They are produced by accelerated charges.
Rule 1: Speed and field ratio — c = 1 ÷ √(μ₀ ε₀), c = f λ and E₀ ÷ B₀ = c (E₀, B₀ = amplitudes). In a medium of refractive index n, v = c ÷ n.
- Easy : E₀ = 90 V m⁻¹ ⇒ B₀ = 90 ÷ 3 × 10⁸ = 3 × 10⁻⁷ T.
- Medium : f = 6 × 10¹⁴ Hz ⇒ λ = 3 × 10⁸ ÷ 6 × 10¹⁴ = 500 nm.
- Exam : In glass of n = 1.5 the speed is 3 × 10⁸ ÷ 1.5 = 2 × 10⁸ m s⁻¹, and the frequency stays the same.
Rule 2: Energy and intensity — The average energy density is u = ½ ε₀ E₀² (electric and magnetic parts are equal), and the intensity is I = u c = ½ ε₀ c E₀². The waves also carry momentum, p = U ÷ c.
- Easy : E₀ = 90 V m⁻¹ ⇒ u = ½ × 8.85 × 10⁻¹² × 8100 = 3.6 × 10⁻⁸ J m⁻³.
- Medium : The intensity is 3.6 × 10⁻⁸ × 3 × 10⁸ ≈ 10.8 W m⁻².
- Exam : Radiation pressure on a surface comes from this momentum, which is why a comet's tail points away from the Sun.
In the figure : An electromagnetic wave travelling along x with E along y and B along z, in phase.
c = 1/√(μ₀ε₀) = fλ; E₀/B₀ = c; transverse; no medium needed.
Thinking that the frequency changes when the wave enters a medium. Speed and wavelength change; frequency does not.
Memory clue : "E over B is c: three hundred million."
Spot the difference
| Wave | Needs a medium? | Type |
|---|---|---|
| Sound | yes | longitudinal |
| Electromagnetic | no | transverse |
Practice questions
- 1Are electromagnetic waves longitudinal or transverse?
- 2Do they need a medium to travel?
- 3What is their speed in vacuum?
- 4The electric field amplitude is 90 V m⁻¹. Find the magnetic field amplitude.
- 5Find the wavelength of light of frequency 6 × 10¹⁴ Hz.
- 6Find the speed of light in glass of refractive index 1.5.
- 7Find the intensity of a wave with E₀ = 90 V m⁻¹.
- 8Find the wavelength of an FM radio wave of 100 MHz.
Show answers (try them yourself first)
- 1Transverse.
- 2No.
- 33 × 10⁸ m s⁻¹.
- 4B₀ = E₀ ÷ c = 3 × 10⁻⁷ T.
- 5λ = c ÷ f = 500 nm.
- 6v = c ÷ n = 2 × 10⁸ m s⁻¹.
- 7I = ½ε₀cE₀² ≈ 10.8 W m⁻².
- 8λ = 3 × 10⁸ ÷ 10⁸ = 3 m.
The electromagnetic spectrum
The remote of a TV (infrared), the oven in a kitchen (microwaves), the X-ray in a hospital and the colours of a rainbow are all parts of one spectrum. Link to earlier classes : the visible spectrum from Class 10 and photons in the next chapter.
The electromagnetic spectrum arranges all the waves in order of wavelength, from the long radio waves to the short gamma rays. All travel at c in vacuum; they differ in frequency, wavelength and the way they are produced and detected.
Rule 1: The order and uses — In increasing frequency : radio waves (longer than about 0.1 m; communication, broadcasting), microwaves (about 1 mm to 0.1 m; radar, ovens), infrared (700 nm to 1 mm; remotes, thermal imaging, warmth), visible light (400 nm to 700 nm; sight), ultraviolet (1 nm to 400 nm; sterilising, vitamin D, absorbed by the ozone layer), X-rays (about 1 pm to 10 nm; medical imaging, crystal studies) and gamma rays (shorter than about 10 pm; cancer treatment, from nuclei).
- Easy : A 100 MHz FM station has λ = 3 m, a radio wave.
- Medium : A microwave oven at 2.45 GHz has λ = 3 × 10⁸ ÷ 2.45 × 10⁹ = 12.2 cm.
- Exam : Green light of 500 nm lies in the visible band, between ultraviolet and infrared.
Rule 2: Photon energy — E = h f = h c ÷ λ, or E (eV) = 1240 ÷ λ (nm). Higher frequency means higher energy, and more penetrating and more damaging radiation.
- Easy : A photon of 500 nm has E = 1240 ÷ 500 = 2.48 eV.
- Medium : An X-ray photon of 0.1 nm has E = 1240 ÷ 0.1 = 12 400 eV = 12.4 keV.
- Exam : The greenhouse effect: the Earth's surface radiates infrared, which carbon dioxide and water vapour absorb and re-radiate towards the ground.
In the figure : The spectrum on a wavelength scale from radio waves to gamma rays, with the visible band marked.
Radio < micro < infrared < visible < UV < X < gamma (in frequency); E = hf.
Thinking that X-rays and gamma rays are different from light. They are the same kind of wave, only with shorter wavelength.
Memory clue : "Rabbits Mate In Very Unusual Xylophone Games: Radio, Micro, Infrared, Visible, UV, X, Gamma."
Spot the difference
| Wave | Typical use |
|---|---|
| Microwaves | radar, mobile signals, ovens |
| Infrared | remote controls, night vision |
| Ultraviolet | sterilising water |
| X-rays | bone imaging |
Practice questions
- 1Which part of the spectrum is used in a TV remote?
- 2Which waves have the highest frequency?
- 3Which waves are absorbed by the ozone layer?
- 4Find the wavelength of a 100 MHz FM radio wave.
- 5Find the wavelength of a 2.45 GHz microwave oven.
- 6Find the energy of a 500 nm photon in eV.
- 7Find the energy of an X-ray photon of wavelength 0.1 nm.
- 8Explain in a sentence how the greenhouse effect warms the Earth.
Show answers (try them yourself first)
- 1Infrared.
- 2Gamma rays.
- 3Ultraviolet rays (harmful UV).
- 4λ = 3 m.
- 5λ ≈ 12.2 cm.
- 6E = 1240 ÷ 500 = 2.48 eV.
- 7E = 1240 ÷ 0.1 = 12 400 eV ≈ 12.4 keV.
- 8Gases like carbon dioxide absorb the infrared radiated by the ground and send part of it back, so the surface stays warmer.