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G12 Physics Part 1
8. Electromagnetic Waves · 8 / 8
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Chapter 8 of 8

8. Electromagnetic Waves

8 sections · 1,958 words

Introduction

55 words · 1 min read

Big idea: one linked electric–magnetic disturbance can carry energy across empty space.

Connect the charging-capacitor argument to the orientation, speed and spectrum of waves.

James Clerk Maxwell combined established electric and magnetic laws with a changing-electric-field contribution in the nineteenth century. Heinrich Hertz later produced and detected radio waves, supporting the electromagnetic-wave description.

8.1 INTRODUCTION

195 words · 1 min read

Light, radio and X-rays belong to one family even though their sources and interactions look different.

An electromagnetic wave carries coupled electric and magnetic fields through space. In vacuum it needs no material medium. This differs from a sound wave, which requires matter whose pressure and density can oscillate.

The central link is change: a changing magnetic field is associated with a circulating electric field, and a changing electric field contributes to a circulating magnetic field. Maxwell’s equations combine these statements with the laws governing charges and magnetic flux. Their wave solutions identify light as an electromagnetic phenomenon.

A spacecraft sends a radio message across an almost empty region. Does lack of air prevent the signal?

Radio is an electromagnetic wave, so a material medium is not required for propagation.

Result: No. The wave can propagate through vacuum. Sound could not carry the same message through empty space.

Common mistake: A wave needing no material medium still carries energy and momentum.

  1. Which field pair occurs in an electromagnetic wave?
  2. What distinguishes vacuum light from ordinary sound?

Answers

  1. Electric and magnetic fields.
  2. Light can propagate without matter; sound requires a material medium.

8.2 DISPLACEMENT CURRENT

454 words · 2 min read

A charging capacitor exposes a missing term in the steady-current form of Ampère’s law.

Draw one loop around a capacitor lead. A spanning surface cutting the wire encloses conduction current. Another surface with the same boundary can bulge between the capacitor plates, where no conduction charge crosses an ideal insulating gap. The magnetic circulation cannot depend on the imaginary surface you choose.

Maxwell supplied a displacement-current contribution Id=ϵ0 dΦE/dtI_d=\epsilon_0\,d\Phi_E/dt in vacuum. The corrected law is ∮B⋅dl=μ0(Ic+ϵ0 dΦE/dt)\oint\mathbf B\cdot d\mathbf l=\mu_0(I_c+\epsilon_0\,d\Phi_E/dt). The electric flux is ΦE=∫E⋅dA\Phi_E=\int\mathbf E\cdot d\mathbf A. A changing field is the essential condition; a static electric field contributes no displacement current.

For an ideal parallel-plate capacitor, neglect fringing: E=Q/(ϵ0A)E=Q/(\epsilon_0A) and ΦE=EA=Q/ϵ0\Phi_E=EA=Q/\epsilon_0. Therefore Id=ϵ0 d(Q/ϵ0)/dt=dQ/dt=IcI_d=\epsilon_0\,d(Q/\epsilon_0)/dt=dQ/dt=I_c. Across the gap use the displacement term; through the lead use conduction current. Do not count the full lead current twice for one selected surface.

A uniform electric field between vacuum plates changes at 5.0×108 V m−1 s−15.0\times10^8\,\mathrm{V\,m^{-1}\,s^{-1}} over 0.020 m20.020\,\mathrm{m^2}. Use ϵ0=8.854×10−12 F m−1\epsilon_0=8.854\times10^{-12}\,\mathrm{F\,m^{-1}}. Find displacement current.

Id=ϵ0A dE/dtI_d=\epsilon_0A\,dE/dt.

Result: Id=0.000089 AI_d=0.000089\,\mathrm A.

Displacement current has amperes as its unit but is not necessarily transport of free charges across the gap. It restores a surface-independent magnetic law consistent with charge conservation. In material media a fuller description includes polarization; the vacuum expression here is sufficient for the ideal capacitor model.

For uniform changing field over circular plates of radius R, a concentric loop of radius r inside the plate area encloses displacement current Idr2/R2I_d r^2/R^2. Symmetry then gives B(2πr)=μ0Idr2/R2B(2\pi r)=\mu_0I_d r^2/R^2, or B=μ0Idr/(2πR2)B=\mu_0I_dr/(2\pi R^2) for r<R. Outside the plate area, ignoring fringing and return-current details, the loop encloses the full displacement current and B=μ0Id/(2πr)B=\mu_0I_d/(2\pi r). The interior field grows with r instead of diverging at the axis.

LawIntegral statementPhysical reading
Gauss electric∮E⋅dA=Q/ϵ0\oint\mathbf E\cdot d\mathbf A=Q/\epsilon_0Charge is the source of electric flux.
Gauss magnetic∮B⋅dA=0\oint\mathbf B\cdot d\mathbf A=0There is no net magnetic flux through a closed surface.
Faraday∮E⋅dl=−dΦB/dt\oint\mathbf E\cdot d\mathbf l=-d\Phi_B/dtChanging magnetic flux produces electric circulation.
Ampère–Maxwell∮B⋅dl=μ0(Ic+Id)\oint\mathbf B\cdot d\mathbf l=\mu_0(I_c+I_d)Conduction and changing electric flux contribute to magnetic circulation.
Common mistake: Displacement current does not mean electrons jump through the dielectric of an ideal capacitor.

  1. When is displacement current zero?
  2. What equals the conduction current while an ideal capacitor charges?

Answers

  1. When electric flux through the selected surface is constant.
  2. The displacement current through a suitable surface spanning the gap, neglecting fringing.

8.3 ELECTROMAGNETIC WAVES

503 words · 2 min read

A plane wave makes the geometry of electromagnetic propagation especially clear.

Accelerated charges can emit electromagnetic radiation. A charge oscillating in an antenna is one familiar source. Uniform motion in an inertial frame is not the same as acceleration; a static charge has an electric field but does not emit a continuing radiation wave in that frame.

Maxwell’s vacuum equations predict c=1/μ0ϵ0c=1/\sqrt{\mu_0\epsilon_0}. For school calculations use c≈3.00×108 m s−1c\approx3.00\times10^8\,\mathrm{m\,s^{-1}}. Frequency f, wavelength λ and speed obey c=fλc=f\lambda in vacuum. In matter the phase speed may differ; frequency stays fixed when a stationary boundary is crossed, while wavelength changes.

For a vacuum plane wave travelling along +x, one possible choice is E=E0sin⁡(kx−ωt) y^\mathbf E=E_0\sin(kx-\omega t)\,\hat{\mathbf y} and B=B0sin⁡(kx−ωt) z^\mathbf B=B_0\sin(kx-\omega t)\,\hat{\mathbf z}. Both fields are perpendicular to propagation and to each other. They are in phase, and E0=cB0E_0=cB_0. Their cross product specifies the energy-flow direction.

A side view shows E as a sine curve; annotate B with dots or crosses rather than drawing it parallel to E.
Draw it in an exam: Draw a propagation axis. Put E perpendicular to it, and B perpendicular to both. Use the right hand so E cross B points along propagation. Draw both fields in phase; do not infer equal units or equal amplitudes from equal sketch heights.
Normalized field amplitudes show orientation only. E and B have different units. At a fixed time their crests are in phase.

A vacuum radio wave has frequency 75 MHz75\,\mathrm{MHz}. Find its wavelength.

Convert MHz to hertz, then use λ=c/f\lambda=c/f.

Result: λ=4 m\lambda=4\,\mathrm m.

The electric-field amplitude is 90 V m−190\,\mathrm{V\,m^{-1}}. Find magnetic-field amplitude.

B0=E0/cB_0=E_0/c.

Result: B0=3e−7 TB_0=3e-7\,\mathrm T.

Instantaneous vacuum energy densities are uE=ϵ0E2/2u_E=\epsilon_0E^2/2 and uB=B2/(2μ0)u_B=B^2/(2\mu_0). The relation E=cB makes these equal in a plane wave. For a sinusoidal wave, the average total density is ⟨u⟩=ϵ0E02/2\langle u\rangle=\epsilon_0E_0^2/2. Average intensity is c⟨u⟩c\langle u\rangle.

For the field amplitude above, ⟨u⟩=3.5859e−8 J m−3\langle u\rangle=3.5859e-8\,\mathrm{J\,m^{-3}}. A wave also transports momentum. Normal absorption gives radiation pressure I/cI/c; ideal normal reflection gives 2I/c2I/c. These expressions assume an ideal plane beam and the stated boundary behaviour.

Common mistake: E and B have different units; equal sketch heights do not mean E equals B numerically.

  1. If E points +y and B points +z, where does energy flow?
  2. What is the phase difference between E and B in a vacuum travelling plane wave?
  3. Can a static charge be called a radiation source merely because it has a field?

Answers

  1. Along +x because y cross z is x.
  2. Zero.
  3. No. A static field and outgoing radiation are different situations.

8.4 ELECTROMAGNETIC SPECTRUM

387 words · 2 min read

The spectrum is a continuous range of frequencies; category names summarize common sources and interactions.

From long wavelength to short wavelength, the usual order is radio, microwave, infrared, visible, ultraviolet, X-ray and gamma. Frequency and photon energy increase along that order while vacuum wavelength decreases. The boundaries are conventional and sometimes overlap; X-ray and gamma names can also reflect how photons are produced.

RegionTypical production or detectionUseful connection
RadioAccelerated currents in transmitting antennasBroadcasting and communication
MicrowaveElectronic oscillators and antennasRadar, satellite links and molecular heating
InfraredThermal emission and molecular vibrationsThermal sensing and remote controls
VisibleElectronic processes in lamps and atomsVision and imaging
UltravioletHot sources and electronic transitionsFluorescence; excessive exposure can damage tissue
X-rayFast-electron deceleration or inner-shell transitionsImaging and diffraction; ionizing radiation
GammaNuclear and high-energy processesNuclear spectroscopy and specialized treatment; ionizing radiation

Visible light occupies only a small part of the spectrum. Within it, red has longer wavelength than violet in vacuum. A hot object emits a broad spectrum rather than only one infrared wavelength. Infrared is not itself “heat”: absorption can transfer wave energy into thermal energy.

The atmosphere transmits some bands better than others. This affects communication and astronomical observation. A short wavelength does not automatically mean a wave passes through every material; absorption depends on the material and interaction mechanisms.

Compare a 0.060 m0.060\,\mathrm m microwave with a 600 nm600\,\mathrm{nm} visible wave. Use the same vacuum speed.

Frequency is inversely proportional to wavelength. The wavelength ratio is 0.060/(600×10−9)0.060/(600\times10^{-9}).

Result: The visible frequency is $100000 $ times the microwave frequency.

All these waves obey the same vacuum propagation speed. Their different effects follow from frequency, photon energy, intensity and interactions with matter. Neither speed alone nor the brightness seen by an eye can rank the risks of invisible radiation.

Common mistake: Microwaves, infrared and visible light are not different-speed waves in vacuum.

  1. Which has greater vacuum wavelength: infrared or ultraviolet?
  2. Why is there no single sharp physical boundary separating every spectrum category?
  3. What remains equal for radio and gamma waves in vacuum?

Answers

  1. Infrared.
  2. The spectrum is continuous and several labels are conventional or source-based.
  3. Their propagation speed c.

SUMMARY

59 words · 1 min read
  • Displacement current in vacuum: Id=ϵ0dΦE/dtI_d=\epsilon_0d\Phi_E/dt.
  • Plane vacuum waves: c=1/μ0ϵ0c=1/\sqrt{\mu_0\epsilon_0}, c=fλc=f\lambda, E0=cB0E_0=cB_0.
  • E, B and propagation are mutually perpendicular; E and B oscillate in phase.
  • Frequency rises from radio through microwave, infrared, visible, ultraviolet, X-ray to gamma.
Use the equations with their assumptions, sign conventions and units. Explain each symbol before substituting values.

POINTS TO PONDER

37 words · 1 min read

  • A field can exist without a propagating radiation wave.
  • Spectrum labels are useful categories, not separate kinds of vacuum propagation.
  • The displacement-current term preserves consistency; it is not conduction through an insulating gap.

EXERCISES

268 words · 1 min read

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

8.1 Explain the capacitor-surface problem addressed by Maxwell’s correction.

8.2 Vacuum electric flux grows at 2.0×109 V m s−12.0\times10^9\,\mathrm{V\,m\,s^{-1}}. Find displacement current using the stated epsilon value.

8.3 During steady ideal capacitor charging at 0.40 A, what is the displacement current in its gap?

8.4 Find wavelength at 125 MHz125\,\mathrm{MHz}.

8.5 Find vacuum frequency for 0.015 m0.015\,\mathrm m wavelength.

8.6 Find B amplitude if E amplitude is 150 V m−1150\,\mathrm{V\,m^{-1}}.

8.7 A wave travels +z and E points +x. Which direction does B point?

8.8 Does a plane wave carry more electric-field energy than magnetic-field energy in vacuum?

8.9 Arrange infrared, gamma and radio by increasing frequency.

8.10 What changes when monochromatic light enters a stationary transparent medium?

8.11 Why is an oscillating charge relevant to radiation?

8.12 A wave is completely reflected at normal incidence. Compare pressure with absorption at the same intensity.

Answers

8.1 Different spanning surfaces of the same loop must predict the same magnetic circulation; changing electric flux accounts for the gap surface.

8.2 0.017708 A0.017708\,\mathrm A.

8.3 0.40 A in the ideal parallel-plate approximation.

8.4 2.4 m2.4\,\mathrm m.

8.5 20000000000 Hz20000000000\,\mathrm{Hz}.

8.6 5e−7 T5e-7\,\mathrm T.

8.7 +y, since x cross y points +z.

8.8 No. Their instantaneous energy densities are equal for the travelling plane-wave model.

8.9 Radio, infrared, gamma.

8.10 Its speed and wavelength can change; its frequency remains fixed.

8.11 Oscillation involves acceleration, which can produce outgoing electromagnetic radiation.

8.12 Ideal reflection gives twice the pressure because the normal momentum reverses.

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