Electromagnetic Waves · TOPIC LESSON
Connect Maxwell’s correction, transverse fields and energy transport in a clear vacuum-wave model.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
Electrostatics and magnetostatics describe restricted situations. Maxwell’s equations connect electric charge, current, electric field and magnetic field in both static and time-dependent cases. Electric charges source electric flux, magnetic field lines have no isolated beginnings or ends in the usual theory, changing magnetic flux produces circulating electric field, and currents together with changing electric flux produce circulating magnetic field.
The time-dependent coupling permits electromagnetic disturbances to propagate through space even where there is no material medium. A wave in vacuum does not require air, a string or an imagined mechanical ether. Its electric and magnetic fields are parts of one self-consistent traveling solution rather than two independent signals taking turns to create one another.
Accelerated charges can emit electromagnetic radiation, and an oscillating current distribution in an antenna is a familiar source. Not every time-varying field near a source is a simple outgoing plane wave. Near-field components can have different spatial and phase relations, so the far-field plane-wave formulas in this lesson must be applied within their stated regime.
The four field laws are most useful here as consistency relations. Gauss’s electric law relates closed-surface electric flux to enclosed charge; Gauss’s magnetic law gives zero net magnetic flux through a closed surface. Faraday and Ampère–Maxwell laws concern circulation around a closed boundary and time-changing flux through a spanning surface. Closed surfaces and closed loops are different geometrical objects.
02 / UNDERSTAND & APPLY
Consider a capacitor charging through a wire. A loop around the wire can bound a surface that cuts the wire, where conduction current flows, or a curved surface passing between the plates, where no conduction current crosses the vacuum gap. A magnetic-circulation law depending only on conduction current would give inconsistent answers for the same loop.
Maxwell’s correction in vacuum is I_d = ε₀ dΦ_E/dt, where Φ_E is electric flux through the chosen surface. The Ampère–Maxwell law reads ∮B·dl = μ₀(I_conduction + ε₀ dΦ_E/dt), with consistent boundary orientation. Displacement current contributes to magnetic circulation without requiring charge carriers to cross the gap.
For an ideal parallel-plate vacuum capacitor with negligible fringing, E = Q/(ε₀A) and Φ_E = EA = Q/ε₀. Thus I_d = dQ/dt, equal to the conduction current charging the plates. The equality concerns the appropriately chosen flux surface; it does not mean electrons physically jump across the insulating gap in this model.
A constant electric field has zero displacement current even if its magnitude is large. A changing field can produce displacement current where conduction current is absent. In material media, the macroscopic formulation involves electric displacement D and free current; do not replace every dielectric situation by ε₀E without accounting for polarization.
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ORIGINAL WORKED EXAMPLE
A vacuum parallel-plate capacitor has area 0.020 m² and a uniform electric field increasing at 3.0 × 10⁸ V m⁻¹ s⁻¹. Use ε₀ = 8.85 × 10⁻¹² F m⁻¹. Find displacement current through the full plate area.
Answer Approximately 53 μA.
03 / UNDERSTAND & APPLY
For a plane electromagnetic wave in vacuum, electric field E and magnetic field B are perpendicular to each other and to the propagation direction. Their instantaneous vectors satisfy B = (1/c)n̂ × E for propagation unit vector n̂. Therefore E × B points along propagation. Use this vector relation rather than guessing the third direction from a flat sketch.
A sinusoidal wave traveling in +x can be written E_y = E₀cos(kx − ωt + φ) and B_z = B₀cos(kx − ωt + φ), with E₀ = cB₀. The fields oscillate in phase in this simple traveling wave: their zeros and maxima occur together. Reversing both field signs half a cycle later leaves the energy-flow direction unchanged.
Wave speed in vacuum is c = 1/√(μ₀ε₀), approximately 3.00 × 10⁸ m s⁻¹. All electromagnetic frequencies travel at this same vacuum speed in the standard model. Wavelength and frequency satisfy c = fλ, while angular quantities satisfy ω = ck with k = 2π/λ and ω = 2πf.
In a medium, phase speed and field ratios depend on constitutive properties and frequency. The simple E/B = c relation applies to vacuum; in a simple lossless nondispersive medium the corresponding phase-speed relation can be used with care. Absorbing, dispersive or guided waves can require a more detailed description than the vacuum plane-wave formulas.
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ORIGINAL WORKED EXAMPLE
A vacuum plane wave propagates along +x. At an instant its electric field is +60 ĵ V m⁻¹. Find the magnetic field vector using c = 3.0 × 10⁸ m s⁻¹.
Answer +2.0 × 10⁻⁷ k̂ T.
04 / UNDERSTAND & APPLY
In cos(kx − ωt + φ), a constant phase requires k dx − ω dt = 0, so the pattern moves in +x with speed ω/k. The plus-time form cos(kx + ωt + φ) moves in −x. This sign test does not depend on the arbitrary phase constant or the sign of a particular instantaneous field value.
If the spatial coordinate is in metres and time in seconds, k has units rad m⁻¹ and ω rad s⁻¹. The cosine argument must be dimensionless. A coefficient of x is not automatically 1/λ, because k contains 2π. Likewise, the coefficient of t is not ordinary frequency f unless the expression already displays 2πft.
At a fixed location, two instants separated by one period have the same field phase. At a fixed time, points separated by one wavelength have the same phase. Points separated by half a wavelength have opposite E and B, but their E×B energy flow can still point in the same direction. Spatial and temporal snapshots should not be mixed in one reading.
ORIGINAL WORKED EXAMPLE
An electric field is E_y = 12 cos[(2π/0.50)x − (1.2π × 10⁹)t] V m⁻¹, with x in metres and t in seconds. Find wavelength, frequency, propagation direction and magnetic-field amplitude.
Answer λ = 0.50 m; f = 6.0 × 10⁸ Hz; direction +x; B₀ = 4.0 × 10⁻⁸ T.
05 / UNDERSTAND & APPLY
In vacuum, instantaneous electric energy density is u_E = ε₀E²/2 and magnetic energy density is u_B = B²/(2μ₀). For a plane wave with E = cB and c² = 1/(μ₀ε₀), these are equal at each event. Total u = ε₀E² = B²/μ₀, not merely one of the half-sized contributions.
The Poynting vector S = E × B/μ₀ gives electromagnetic energy flux, with units W m⁻². For the vacuum traveling plane wave, |S| = cu. The intensity I is its time average. For sinusoidal peak amplitudes, I = ε₀cE₀²/2 = cB₀²/(2μ₀). With rms field instead, I = ε₀ × c × Eᵣₘₛ² and Eᵣₘₛ = E₀/√2.
Intensity is proportional to field amplitude squared. Doubling E₀ quadruples intensity and average energy density. A total beam power requires integrating intensity over the illuminated cross section; for uniform intensity over area A normal to propagation, power is IA. An oblique surface intercepts the projected area, not automatically the same perpendicular beam area.
For an isotropic point-like radiator with no absorption, average intensity falls as P/(4πr²), so field amplitude scales as 1/r in the far zone. A collimated beam does not necessarily follow that same simple spherical-spreading law over every distance. Always state geometry before using an inverse-square intensity relation.
ORIGINAL WORKED EXAMPLE
A sinusoidal vacuum plane wave has peak electric field E₀ = 100 V m⁻¹. Use ε₀ = 8.85 × 10⁻¹² F m⁻¹ and c = 3.0 × 10⁸ m s⁻¹. Find average intensity and average total energy density.
Answer Intensity approximately 13 W m⁻²; average total energy density approximately 4.4 × 10⁻⁸ J m⁻³.
06 / UNDERSTAND & APPLY
A vacuum light pulse with energy E_pulse has momentum magnitude E_pulse/c in its propagation direction. When it is absorbed, momentum transfers to the target. For normal incidence on an ideal absorber, radiation pressure is I/c. For ideal specular reflection directly back, momentum reverses and pressure is 2I/c.
These simple pressure expressions assume the quoted intensity is incident intensity per area perpendicular to the beam and the surface is normal to it. At oblique incidence, both projected intercepted area and the normal momentum component matter. A diffuse or partly transmitting surface needs its actual angular and energy balance; do not simply assign every reflective object the normal-incidence factor two.
Radiation pressure can exist even though the wave has no rest mass. Energy and momentum are transported by the electromagnetic field. A small pressure does not mean no momentum; macroscopic forces can be small because c is very large. The target’s recoil energy is usually negligible compared with incident optical energy in elementary stationary-target calculations.
ORIGINAL WORKED EXAMPLE
A 6.0 W beam strikes a perfectly reflecting surface normally and is reflected straight back. Find force magnitude using c = 3.0 × 10⁸ m s⁻¹.
Answer 4.0 × 10⁻⁸ N in the incident propagation direction.
07 / UNDERSTAND & APPLY
Ordered from lower frequency and longer wavelength toward higher frequency and shorter wavelength, the familiar bands are radio, microwave, infrared, visible, ultraviolet, X-ray and gamma ray. Their boundaries are conventional and can overlap in usage, especially where production mechanism matters. All are electromagnetic waves and share vacuum speed c.
Radio and microwave systems commonly use oscillating currents and antennas; infrared is important in thermal emission and molecular transitions; visible light spans the eye’s response range; ultraviolet, X-rays and gamma rays can involve higher-energy electronic or nuclear processes. These are characteristic associations, not exclusive production rules assigning every band to only one source.
Photon energy hf increases with frequency. Increasing intensity at fixed frequency increases energy flux, not the energy of each photon. Frequency and intensity therefore influence interactions in different ways. “Higher frequency” and “more powerful beam” are not synonymous descriptions.
Polarization describes the transverse electric-field direction pattern. A linearly polarized plane wave has E along one fixed transverse line, while circular and elliptical polarization involve rotation of the transverse vector. The magnetic field follows from the propagation direction and E; it is not an independently chosen polarization axis.
At a stationary interface, frequency is preserved while speed and wavelength can change. Refraction, reflection, absorption and dispersion depend on the material and frequency. The spectrum label alone does not determine how every material responds; for example, a material transparent in one band may absorb strongly in another.
08 / UNDERSTAND & APPLY
First decide whether the question gives peak or rms fields and whether the wave is in vacuum. Draw the E, B and propagation directions as an ordered right-handed relation, then check dimensions.
ORIGINAL PRACTICE · 1
No.
Maxwell’s field equations permit propagation through vacuum.
ORIGINAL PRACTICE · 2
Zero.
I_d = ε₀ dΦ_E/dt, so a nonzero but constant flux gives zero derivative.
ORIGINAL PRACTICE · 3
They are equal.
Φ_E = Q/ε₀ gives ε₀dΦ_E/dt = dQ/dt.
ORIGINAL PRACTICE · 4
+y.
B is along n̂×E, so k̂×î=ĵ; E×B then points +z.
ORIGINAL PRACTICE · 5
1.0 × 10⁻⁶ T.
B₀ = E₀/c.
ORIGINAL PRACTICE · 6
0.30 m.
λ = c/f.
ORIGINAL PRACTICE · 7
−x.
k dx + ω dt=0 gives dx/dt=−ω/k.
ORIGINAL PRACTICE · 8
They are equal.
E=cB and c²=1/(μ₀ε₀) make ε₀E²/2=B²/(2μ₀).
ORIGINAL PRACTICE · 9
It increases ninefold.
Intensity is proportional to amplitude squared.
ORIGINAL PRACTICE · 10
No.
The averaging factor is already included in Eᵣₘₛ²=E₀²/2.
ORIGINAL PRACTICE · 11
Intensity is one quarter; amplitude one half.
Spherical spreading gives I∝1/r² and amplitude∝√I.
ORIGINAL PRACTICE · 12
Reflection gives twice the pressure.
Momentum reversal doubles the normal transfer relative to absorption.
ORIGINAL PRACTICE · 13
Ultraviolet.
It has higher frequency and photon energy hf.
ORIGINAL PRACTICE · 14
No.
Frequency is fixed across the stationary interface; wavelength changes with phase speed.
09 / UNDERSTAND & APPLY
Name the propagation direction and use n̂×E to obtain B. Read wavelength and frequency from the full phase coefficients, including 2π. Distinguish a snapshot field from a time average and peak amplitude from rms amplitude. If a formula gives a vacuum result, do not silently apply it to an absorbing medium or a near field.
For energy and force, identify whether the supplied quantity is intensity, intercepted total power or pulse energy. Apply the relevant geometry before converting energy flux into momentum transfer. For capacitor displacement current, use changing flux rather than the field value alone. These checks connect the equations without treating every electromagnetic situation as the same plane-wave picture.
KEEP THE BIG IDEAS
READ WITH CONTEXT
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
Analytics use a separately reviewed, incomplete historical corpus. Opening a topic does not change its counts or denominator.