Chapter 07 · Electromagnetic Waves · Full book

Electromagnetic Waves

Core intuition: Chapter Roadmap

This chapter closes the electromagnetism block by showing that light itself is an electromagnetic wave—that everything built so far (fields, flux, Faraday's law, Ampère's law) was already the theory of light in disguise. The thread runs as follows.

Why light must be electromagnetic (Sec. ): Maxwell's constants \(\varepsilon_0\) and \(\mu_0\) predict a wave speed that matches the measured speed of light exactly.
Displacement current (Sec. ): the one missing term that makes Ampère's law consistent and lets a changing electric field source a magnetic field.
Origin and self-sustaining nature (Sec. ): how \(\vec{E}\) and \(\vec{B}\) regenerate each other and propagate with no medium.
Transverse character and polarisation (Sec. ): why the fields must lie across the direction of travel.
The field–field–propagation triad (Sec. ): three mutually perpendicular directions, with \(E = cB\) fixing the amplitudes.
Speed and the electromagnetic spectrum (Sec. , Sec. ): one speed in vacuum, a continuum of frequencies from radio waves to gamma rays.
Energy, intensity, and radiation pressure (Sec. ): the wave carries energy and momentum, and pushes on whatever it strikes.

Keep one question in view throughout: if light is just fields, what is waving? The answer—nothing material—is the whole point.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Wave That Needs No Medium: Sound cannot travel through vacuum. Water waves need water. Even a vibrating string needs the string. Light travels effortlessly through the emptiest regions of outer space, across billions of light-years of nothing. What exactly is waving?
The Force That Reaches Across the Gap: A capacitor is charging. Conduction current flows into one plate and out of the other, but no charge ever crosses the gap between the plates. Ampère's law says a current must be surrounded by a magnetic field. Is there a magnetic field in the gap? If so, what produced it? If not, is Ampère's law inconsistent with itself?
The Unification That Was Not Planned: Electricity and magnetism were studied for centuries as completely separate phenomena. One equation—written purely from mathematical consistency, not from any direct experiment—merged them into a single theory. How does an equation “discover” a new phenomenon?
Light Without a Source: Electromagnetic waves carry energy. They can propagate indefinitely through vacuum without any charges present along their path. If no charges are present to create fields, how do the fields sustain themselves? What maintains the wave in empty space?
The Number That Should Not Match: Maxwell derived the speed of electromagnetic waves from two purely electrical and magnetic constants—\(\varepsilon_0\) and \(\mu_0\)—measured in tabletop experiments with capacitors and coils. The answer turned out to equal the speed of light, which had been measured independently by astronomers. Two completely separate branches of physics, two independent sets of measurements, one identical number. Could this really be coincidence?

Each question targets a conceptual trap or a deeper truth hidden inside this chapter.
By the end, every one of them will be completely transparent.

Why Light Must Be an Electromagnetic Wave

By the time Maxwell began his work in the 1860s, two bodies of physics had each been developed into mature, well-tested theories. Optics established that light is a wave: it diffracts around obstacles, interferes in thin films, refracts at boundaries according to Snell's law. Its speed in vacuum had been measured, first by Rømer from astronomical observations of Jupiter's moons and later by Fizeau in a direct terrestrial experiment, at approximately \(3\times10^8\,\mathrm{m/s}\). What kind of wave, and what was waving, remained entirely unclear.

Separately, Faraday, Ampère, Oersted, and others had constructed the experimental foundation of electromagnetism: charges create electric fields, currents create magnetic fields, changing magnetic fields drive electric ones. These were catalogued as relationships between physical quantities, without a unifying mathematical framework.

Maxwell's contribution was to take all these experimental laws, cast them as a self-consistent set of field equations, and find that the equations had a logical gap—a gap that required a new term. Inserting that term, and doing nothing else, predicted the existence of waves in the electromagnetic field. He then calculated their propagation speed.

The Predicted Speed

The two constants that characterise the behaviour of electric and magnetic fields in vacuum are: \[ \varepsilon_0 = 8.85\times10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}} \quad\text{(electric permittivity of free space)} \] \[ \mu_0 = 4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \quad\text{(magnetic permeability of free space).} \]

Both constants were measured by Faraday-era experimenters using benchtop apparatus—capacitors and coils, no optics whatsoever. Maxwell's wave equations predict that electromagnetic disturbances travel through free space at a speed: \[ c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}. \] Substituting the measured values: \[ c = \frac{1}{\sqrt{(8.85\times10^{-12})(4\pi\times10^{-7})}} \approx 3.0 \times 10^8\,\mathrm{m/s}. \]

This is, within experimental precision, exactly the speed of light.

The agreement is not close in the way that two independent estimates of an uncertain quantity sometimes agree roughly. It is exact to every significant figure available from nineteenth-century measurement. Two entirely separate branches of physics—one studying charged pith balls and solenoids, the other studying prisms and interference fringes—had converged on a single number.

Core intuition: Why This Coincidence Cannot Be a Coincidence

If the measured values of \(\varepsilon_0\) and \(\mu_0\) had given any speed other than the known speed of light, there would be two possibilities:

One of the measurements was wrong.
Light is not an electromagnetic wave.

The measurements were reproducible across many independent laboratories. The only remaining conclusion is that light is an electromagnetic wave—not similar to one, not analogous to one, but physically identical to an oscillating electric and magnetic field propagating through space.

What Unified Electricity and Magnetism

Before Maxwell, the connection between electricity and magnetism came entirely from experiment: Oersted found that a current deflects a compass needle; Faraday found that a changing magnetic field drives a current. These were surprising, non-obvious links, but they were cast as separate empirical facts with no overarching explanation.

Maxwell's equations did something deeper: they showed that electric and magnetic fields are not independent quantities that happen to influence each other under special circumstances. They are two aspects of a single electromagnetic field—related by the geometry of space and the speed \(c\). An observer in one frame of reference may measure a purely electric field; a second observer moving relative to the first will measure both electric and magnetic components. The decomposition into “\(E\)-field” and “\(B\)-field” is reference-frame dependent. The electromagnetic field itself is the invariant object.

This is the unification that Maxwell achieved. Not a merger of two separate disciplines, but the discovery that they had been describing the same thing from two different vantage points.

What This Means for Light

A beam of light is, at every point along its path, an oscillating electric field \(\vec{E}\) and an oscillating magnetic field \(\vec{B}\), mutually perpendicular, both perpendicular to the direction of propagation, travelling at speed \(c\). There is no medium required: the fields sustain themselves through mutual induction. A changing \(\vec{E}\) produces a changing \(\vec{B}\); that changing \(\vec{B}\) sustains the \(\vec{E}\); the disturbance propagates forward.

We will build this picture carefully in Sections and . For now the essential fact is:

Core intuition: Light Is Electromagnetic Radiation

Light is not produced by electromagnetic phenomena, nor merely analogous to an electromagnetic wave. Light is an oscillating electromagnetic field. The wave equations that govern electric and magnetic fields in vacuum are the same wave equations that describe light—because they are the same phenomenon. This means every property of light—its speed, its transverse polarisation, its ability to propagate through vacuum, the relationship between its electric and magnetic amplitudes—follows from Maxwell's equations without \begin{ideabox}[title={Hertz: From Prediction to Spark}] Maxwell predicted electromagnetic waves in 1865 but did not live to see them produced deliberately. In 1887 Heinrich Hertz generated and detected them in the laboratory. A spark across a gap in a tuned circuit launched an oscillating disturbance; a second loop several metres away, with its own tiny gap and no wire connecting the two, sparked in sympathy each time the first did. Hertz measured the wavelength (from the nodes of a standing-wave pattern) and the frequency (set by his circuit), multiplied them, and recovered a speed equal to \(c\). He further showed the waves reflect, refract, and polarise exactly as light does. A prediction made from pure mathematical consistency had become a spark one could watch. The waves Hertz made are what we now call radio waves.

any additional assumptions. \end{ideabox}

Misconception alert: “Light Travels Through the Electromagnetic Field”

The misconception: Light is a wave that travels through the electromagnetic field the way sound travels through air—the field is the medium. Why it is wrong: Light does not travel through the electromagnetic field. Light is the electromagnetic field oscillating. There is no distinction between the wave and the medium: the oscillating \(\vec{E}\) and \(\vec{B}\) are the wave, not a displacement of some pre-existing substrate. This is why no material medium is needed, and why the idea of the luminiferous aether—a material medium for light waves—is physically wrong. The aether was searched for experimentally by Michelson and Morley (1887) and was not found. Maxwell's framework explains why: there is nothing to find. Exam implication: Questions that ask what “medium” electromagnetic waves require in vacuum have the answer: none. The wave is self-sustaining through the interplay of changing \(\vec{E}\) and changing \(\vec{B}\).

✓ Checkpoint · Why Light Must Be an Electromagnetic Wave

The speed of light in vacuum is approximately \(3.0 \times 10^8\,\mathrm{m/s}\). Maxwell derived an expression for the speed of electromagnetic waves in terms of \(\varepsilon_0\) and \(\mu_0\). Why is the precise numerical agreement between these two independently measured speeds not a coincidence?
The constants \(\varepsilon_0\) and \(\mu_0\) were measured purely from electrostatic and magnetostatic experiments. The speed of light was measured from optical and astronomical observations. The exact agreement of \(c = 1/\sqrt{\varepsilon_0\mu_0}\) with the independently measured optical speed of light means both are measuring the same physical phenomenon. Coincidence is excluded because the match holds to all available significant figures across multiple independent measurement methods.
A student says: “Electricity and magnetism were unified when Oersted showed that a current produces a magnetic field.” Give one reason why Maxwell's unification goes further than this.
Oersted's discovery established an experimental link between current and magnetism, but it remained a separate empirical fact layered onto electrostatics. Maxwell's equations showed that \(\vec{E}\) and \(\vec{B}\) are two components of a single invariant electromagnetic field, related by the geometry of spacetime. The decomposition into separate fields is observer-dependent; the electromagnetic field itself is the fundamental quantity. Maxwell also predicted a new phenomenon—electromagnetic waves—purely from mathematical consistency, without any experiment requiring it.
State two properties of a light wave that follow directly from it being an electromagnetic wave, without requiring any separate experimental determination.
(i) Its speed in vacuum is \(c = 1/\sqrt{\varepsilon_0\mu_0}\), determined entirely by electrical and magnetic constants. (ii) It is a transverse wave: the oscillating \(\vec{E}\) and \(\vec{B}\) fields are both perpendicular to the direction of propagation. Both follow from Maxwell's equations without additional hypotheses.
When Maxwell calculated the speed of electromagnetic waves from his equations, he obtained a number close to \(3 \times 10^8\,\mathrm{m/s}\). What conclusion did he draw, and why was it justified?
Maxwell concluded that light is an electromagnetic disturbance. The justification is that the value \(1/\sqrt{\varepsilon_0\mu_0}\) agreed with the independently measured speed of light within the experimental precision of the time, and no other known phenomenon propagated at that speed. The only consistent conclusion was that light and electromagnetic waves are the same phenomenon.

Worked Examples

Example 1 · The Speed Hidden in Two Constants

Given. The free-space constants \(\varepsilon_0 = 8.85\times10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}}\) and \(\mu_0 = 4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}}\), both measured with capacitors and coils—no optics. Find. The speed Maxwell's equations predict for a disturbance in the electromagnetic field, and what it should be compared against.

Solution. Setup. Maxwell's wave equations give the free-space speed \(c = 1/\sqrt{\varepsilon_0\mu_0}\). Solve. \[ c = \frac{1}{\sqrt{(8.85\times10^{-12})(4\pi\times10^{-7})}} = \frac{1}{\sqrt{1.11\times10^{-17}}} = \frac{1}{3.34\times10^{-9}}\,\mathrm{m\,s^{-1}}. \] Answer. \(\boxed{c \approx 3.00\times10^{8}\,\mathrm{m\,s^{-1}}}\), to be compared with the optically measured speed of light—also \(3.00\times10^{8}\,\mathrm{m\,s^{-1}}\). Check. The two inputs came from electrostatics and magnetostatics alone, yet the output matches an optical measurement to every available figure. That coincidence is the whole argument: light is the electromagnetic disturbance these constants govern. Dimensionally, \([\varepsilon_0\mu_0] = \mathrm{s^2\,m^{-2}}\), so \(1/\sqrt{\varepsilon_0\mu_0}\) carries units of \(\mathrm{m\,s^{-1}}\)—a speed, as required. ✓

Example 2 · A Universe with Stronger Permittivity

Given. A hypothetical universe in which the permittivity of free space is four times ours, \(\varepsilon_0' = 4\varepsilon_0\), while \(\mu_0\) is unchanged. Find. The speed of light there, as a multiple of our \(c\).

Solution. Setup. The speed depends on the constants only through \(c = 1/\sqrt{\varepsilon_0\mu_0}\); nothing else enters. Solve. \[ c' = \frac{1}{\sqrt{\varepsilon_0'\mu_0}} = \frac{1}{\sqrt{4\varepsilon_0\mu_0}} = \frac{1}{2}\,\frac{1}{\sqrt{\varepsilon_0\mu_0}} = \frac{c}{2}. \] Answer. \(\boxed{c' = c/2 \approx 1.50\times10^{8}\,\mathrm{m\,s^{-1}}}\). Check. A stiffer response of the vacuum to electric fields (larger \(\varepsilon_0\)) slows the disturbance—consistent with \(c\) decreasing as either constant grows. The factor is exactly \(1/\sqrt{4}=1/2\), and no optical input was needed to obtain it. ✓

The Need for Displacement Current

◈ Displacement current

live bench

Charge a capacitor and watch the growing E-field between the plates act as a current: the conduction current flows in the wire, the equal displacement current threads the gap, and the Ampère–Maxwell loop closes.

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↗ open in the Simulations page

Ampère's circuital law, as you encountered it in Chapter 3 (Magnetism and Magnetic Effects of Current), states that the line integral of the magnetic field around any closed loop equals \(\mu_0\) times the conduction current threading through any surface bounded by that loop: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}. \]

The phrase “any surface bounded by that loop” is the key. For a given closed loop \(\mathcal{C}\), you are free to choose any surface whose boundary is \(\mathcal{C}\)—a flat disc, a curved balloon, a distorted sheet—and the enclosed current must give the same answer. For a straight wire carrying steady current, this freedom creates no problem: every surface threading the loop is crossed by the wire. But now consider a situation where the current is not steady.

The Capacitor Gap Paradox

A capacitor is being charged by a battery through a wire (Figure placeholder). The circuit carries a genuine conduction current \(I\) in the wire. Now choose a loop \(\mathcal{C}\) encircling the wire, and consider two surfaces bounded by \(\mathcal{C}\):

Surface 1: A flat disc that cuts through the wire. The conduction current \(I\) passes through this surface. Ampère's law gives: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I. \]

Surface 2: A surface that bulges out through the gap between the capacitor plates. No conduction current crosses this surface—no charges cross the dielectric. Ampère's law gives: \[ \oint \vec{B} \cdot d\vec{l} = 0. \]

The same loop \(\mathcal{C}\) now gives two different values of the line integral depending on which surface we choose. A law of physics cannot depend on an arbitrary choice of surface. This is not a subtle ambiguity or a boundary-case failure: it is a direct logical inconsistency. Ampère's law, as written, is incomplete.

What Is Different in the Gap

Look carefully at what happens between the plates while the capacitor charges. The conduction current \(I\) deposits charge on one plate and removes it from the other. As charge builds up on the plates, the electric field between them increases. The electric flux \(\Phi_E\) through the gap is: \[ \Phi_E = \frac{Q}{\varepsilon_0}, \] where \(Q\) is the charge on the positive plate. Since \(Q\) is increasing at rate \(I = dQ/dt\), we have: \[ \frac{d\Phi_E}{dt} = \frac{I}{\varepsilon_0}. \]

A changing electric flux in the gap. No conduction current, but a definite and calculable rate of change of the electric field.

Maxwell's resolution was direct: if a changing magnetic field can drive an electric field (Faraday's law), then by symmetry a changing electric field ought to drive a magnetic field. He proposed that the incomplete Ampère's law should be repaired by adding a term proportional to \(d\Phi_E/dt\), which he called the displacement current: \[ I_D = \varepsilon_0 \frac{d\Phi_E}{dt}. \]

The corrected Ampère's law becomes: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 \!\left(I_{\text{enc}} + \varepsilon_0 \frac{d\Phi_E}{dt}\right). \]

Why This Repairs the Paradox

With the displacement current included, Surface 2 (through the gap) is no longer empty. The displacement current through the gap is: \[ I_D = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \cdot \frac{I}{\varepsilon_0} = I. \]

The displacement current through the gap equals the conduction current in the wire. Both surfaces now give the same value for \(\oint \vec{B}\cdot d\vec{l}\), and the inconsistency is resolved. The magnetic field outside the capacitor and in its gap are part of a single consistent description.

◇ Figure 1

Fig 1. Inside a charging capacitor no charge crosses the gap, yet the growing electric flux supplies a displacement current \(I_D = \varepsilon_0\,d\Phi_E/dt\) exactly equal to the conduction current \(I\) in the wires. Any surface bounded by the same Ampèrian loop then encloses the same total current.

Birth of the formula: The Ampère–Maxwell Law

Goal: repair Ampère's law so it gives the same magnetic field whether the bounding surface is pierced by the wire or threads the empty capacitor gap. Setup. The electric flux through the gap is \(\Phi_E = Q/\varepsilon_0\), with \(Q\) the plate charge. As the capacitor charges at the rate \(I = dQ/dt\), \[ \frac{d\Phi_E}{dt} = \frac{1}{\varepsilon_0}\frac{dQ}{dt} = \frac{I}{\varepsilon_0}. \] The added term. Maxwell defined the displacement current \(I_D = \varepsilon_0\,d\Phi_E/dt\), so through the gap \(I_D = \varepsilon_0(I/\varepsilon_0) = I\), exactly matching the conduction current in the wire. Result. Ampère's law, completed: \[ \boxed{\;\oint \vec{B}\cdot d\vec{\ell} = \mu_0\!\left(I_{\text{enc}} + \varepsilon_0\frac{d\Phi_E}{dt}\right)\;} \] Conditions. Always valid; the second term matters only where \(\vec{E}\) is changing in time. This is the term that makes light possible—a time-varying electric field is now a genuine source of magnetic field, on equal footing with a conduction current.

Core intuition: Displacement Current: What It Is and Is Not

The displacement current \(I_D = \varepsilon_0\,d\Phi_E/dt\) is not a flow of charge. No charged particle crosses the gap. It is a term that makes the magnetic field description self-consistent wherever the electric field is changing in time. The name “current” is historical: Maxwell introduced it by analogy with the polarisation current in dielectrics. For our purposes, the physical content is: \[ \boxed{ \text{a time-varying electric field } \frac{d\vec{E}}{dt} \text{ acts as a source of magnetic field,} } \] in exactly the same sense that a conduction current does.

Misconception alert: “Displacement Current Means Charges Flow Through the Dielectric”

The misconception: Since the capacitor gap has a “displacement current” and that current produces a magnetic field just like a real current, something must be flowing through the gap—perhaps through the insulating material. Why it is wrong: No charge crosses the gap. The displacement current \(I_D = \varepsilon_0\,d\Phi_E/dt\) is a rate of change of electric flux, not a motion of charge. It is a mathematical term added to Ampère's law to maintain self-consistency. The magnetic effect is entirely real: you would measure the field \(B\) computed above with a magnetometer. But the origin of this field is not moving charge—it is a time-varying electric field. This distinction is physically important because it is exactly what allows electromagnetic waves to propagate through vacuum: the changing \(\vec{E}\) produces \(\vec{B}\) without requiring any charge to be present. JEE trap: Some problems ask whether there is a current between the capacitor plates during charging. The answer is: conduction current—no; displacement current—yes. Failure to distinguish these leads to errors in problems about magnetic fields near capacitors.

✓ Checkpoint · The Need for Displacement Current

Ampère's circuital law states \(\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}\). Describe precisely how this law leads to a logical inconsistency when applied to a circuit containing a charging capacitor.
For a closed loop around the wire feeding the capacitor, one can choose two different surfaces bounded by the loop. Surface 1 (crossing the wire) encloses conduction current \(I\), giving \(\oint \vec{B}\cdot d\vec{l} = \mu_0 I\). Surface 2 (bulging through the capacitor gap) encloses no conduction current, giving \(\oint \vec{B}\cdot d\vec{l} = 0\). The same integral along the same loop cannot simultaneously equal two different values—the law, without modification, is self-contradictory whenever conduction current is discontinuous.
Define Maxwell's displacement current. State clearly what physical quantity it is, and what physical quantity it is not.
Displacement current is \(I_D = \varepsilon_0\,d\Phi_E/dt\), where \(\Phi_E\) is the electric flux through a surface. It is a rate of change of electric flux, not a flow of charge. No charged particle moves to produce it. It contributes to the magnetic field in exactly the same mathematical way as conduction current, but its physical origin is a time-varying electric field.
A capacitor is charged by a current \(I = 4\,\mathrm{A}\). The plates are circular with radius \(R\). Find the magnetic field magnitude at radial distance \(r = R/2\) from the axis, between the plates.
The displacement current density is uniform over the plate area \(\pi R^2\). The fraction enclosed within \(r = R/2\) is \((r/R)^2 = 1/4\), enclosing \(I_D/4 = 1\,\mathrm{A}\). By the corrected Ampère's law: \(B = \mu_0 I_D r / (2\pi R^2) = \mu_0 (4)(R/2)/(2\pi R^2) = \mu_0/( \pi R)\). Numerically: \(B = (4\pi\times10^{-7}) / (\pi R) = 4\times10^{-7}/R\) tesla, where \(R\) is in metres.
Why is Maxwell's correction to Ampère's law considered one of the most important modifications in the history of physics, even though the displacement current in ordinary capacitors is never large?
Although displacement current is tiny in practical circuits, the modification changes the structure of electromagnetism fundamentally. It creates a perfect symmetry: a changing \(\vec{B}\) drives \(\vec{E}\) (Faraday's law) and now a changing \(\vec{E}\) drives \(\vec{B}\) (modified Ampère's law). This mutual generation allows self-sustaining electromagnetic waves to exist in vacuum—which means light exists. Without the displacement current term, Maxwell's equations would predict no electromagnetic radiation at all.

Worked Examples

Example 3 · Displacement Current from a Rising Field

Given. A parallel-plate capacitor has plate area \(A = 100\,\mathrm{cm^2} = 1.0\times10^{-2}\,\mathrm{m^2}\). While it charges, the uniform electric field between the plates increases at the steady rate \(dE/dt = 5.0\times10^{12}\,\mathrm{V\,m^{-1}\,s^{-1}}\). Find. The displacement current threading the gap.

Solution. Setup. The electric flux through the gap is \(\Phi_E = EA\) (field uniform, area \(A\) perpendicular to \(\vec{E}\)), so \(d\Phi_E/dt = A\,dE/dt\) and \(I_D = \varepsilon_0\,d\Phi_E/dt = \varepsilon_0 A\,(dE/dt)\). Solve. \[ I_D = (8.85\times10^{-12})(1.0\times10^{-2})(5.0\times10^{12}) = 0.44\,\mathrm{A}. \] Answer. \(\boxed{I_D \approx 0.44\,\mathrm{A}}\). Check. No charge crosses the gap, yet the growing flux supplies a genuine \(0.44\,\mathrm{A}\) of source for \(\vec{B}\)—and this must equal the conduction current charging the plates. The units of \(\varepsilon_0 A\,(dE/dt)\) reduce to \(\mathrm{C\,s^{-1}} = \mathrm{A}\), as a current must. ✓

Example 4 · Charging Current as Displacement Current

Given. A capacitor of capacitance \(C = 200\,\mathrm{pF}\) is charged by a source that raises the plate voltage at \(dV/dt = 1.0\times10^{6}\, \mathrm{V\,s^{-1}}\). Find. The conduction current in the wire, and the displacement current in the gap; comment on their relation.

Solution. Setup. Conduction current \(I = dQ/dt = C\,dV/dt\) (since \(Q=CV\)). In the gap, \(\Phi_E = Q/\varepsilon_0\), so \(I_D = \varepsilon_0\,d\Phi_E/dt = dQ/dt = I\). Solve. \[ I = C\,\frac{dV}{dt} = (200\times10^{-12})(1.0\times10^{6}) = 2.0\times10^{-4}\,\mathrm{A} = 0.20\,\mathrm{mA}, \] and \(I_D = I = 0.20\,\mathrm{mA}\). Answer. \(\boxed{I = I_D = 0.20\,\mathrm{mA}}\). Check. The displacement current in the empty gap equals the conduction current in the wire identically—the whole point of Maxwell's term. Any Ampèrian surface, whether pierced by the wire or bulging through the gap, then encloses the same \(0.20\,\mathrm{mA}\), so \(\oint\vec{B}\cdot d\vec{\ell}\) is unambiguous. ✓

Origin and Nature of Electromagnetic Waves

◈ The travelling EM wave

live bench

A pseudo-3D E⊥B wave in phase: drag λ and E₀ and read off B₀=E₀/c, the energy density, and the intensity as the fields march forward at c.

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The modification to Ampère's law, introduced in Section , does more than patch an inconsistency. Together with Faraday's law, it creates a coupled pair of equations with a remarkable property: a time-varying electric field produces a time-varying magnetic field, which in turn produces a time-varying electric field, which again produces a time-varying magnetic field. The two fields feed each other. The disturbance does not stay in one place—it propagates.

How the Fields Sustain Each Other

The logic is straightforward once both laws are in place:

Faraday's law: A changing magnetic field \(d\vec{B}/dt \neq 0\) at some region of space produces a circulating electric field around that region. If \(\vec{B}\) oscillates, \(\vec{E}\) oscillates with it.
Modified Ampère's law: A changing electric field \(d\vec{E}/dt \neq 0\) at some region of space produces a circulating magnetic field around that region. If \(\vec{E}\) oscillates, \(\vec{B}\) oscillates with it.

Taken together: start with a region where the electric field is oscillating. It drives an oscillating magnetic field nearby. That oscillating magnetic field drives an oscillating electric field slightly further away. That drives a magnetic field further still. The disturbance propagates outward through empty space, carrying energy with it, needing neither charges nor a medium.

This is an electromagnetic wave.

Core intuition: The Self-Sustaining Mechanism

An electromagnetic wave is not driven by any external agent as it travels. Once the fields are set in oscillation—by an accelerating charge, for instance—the mutual induction of \(\vec{E}\) and \(\vec{B}\) perpetuates the oscillation forward through space. The wave carries energy in the oscillating fields themselves. In vacuum, that energy propagates without loss: the wave can travel across a billion light-years of empty space without weakening due to absorption (it does spread geometrically, reducing intensity as \(1/r^2\), but there is no dissipative loss).

Structure of the Wave

A plane electromagnetic wave propagating in the \(+x\) direction has the following structure. The electric field oscillates in one transverse direction—say the \(y\)-direction: \[ \vec{E} = E_0 \sin(kx - \omega t)\,\hat{j}. \] The magnetic field oscillates in the other transverse direction: \[ \vec{B} = B_0 \sin(kx - \omega t)\,\hat{k}. \] Both oscillate in phase—they reach their maxima and minima at the same point and at the same moment. Both are perpendicular to each other and to the direction of propagation.

◇ Figure 2

Fig 2. An electromagnetic wave travelling along \(+\hat{x}\). The electric field \(\vec{E}\) (blue) and magnetic field \(\vec{B}\) (green) oscillate in phase, perpendicular to each other and to the propagation direction, with \(E = cB\) at every point.

The key relations are:

\(\vec{E} \perp \vec{B} \perp \hat{x}\) (both fields transverse to propagation direction).
\(\vec{E} \times \vec{B}\) points in the direction of propagation.
The amplitudes satisfy: \[ \frac{E_0}{B_0} = c = \frac{1}{\sqrt{\varepsilon_0\mu_0}}. \]
The wave propagates at speed \(c = \omega/k = 1/\sqrt{\varepsilon_0\mu_0}\).

Birth of the formula: Speed of Electromagnetic Waves in Vacuum

Consider an electromagnetic wave propagating in the \(x\)-direction with: \[ E_y = E_0 \sin(kx - \omega t), \quad B_z = B_0\sin(kx-\omega t). \] Apply Faraday's law to a thin rectangular loop in the \(xy\)-plane, with width \(\Delta x\) and height \(h\). The induced EMF around the loop relates to the rate of change of magnetic flux through it. Taking the limit \(\Delta x \to 0\), this gives: \[ \frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{\partial t}. \] For our wave: \[ kE_0\cos(kx-\omega t) = \omega B_0\cos(kx-\omega t) \implies \frac{E_0}{B_0} = \frac{\omega}{k}. \] Apply the modified Ampère's law to a thin rectangular loop in the \(xz\)-plane. An analogous calculation gives: \[ \frac{\partial B_z}{\partial x} = -\mu_0\varepsilon_0\frac{\partial E_y}{\partial t}. \] \[ kB_0\cos(kx-\omega t) = \mu_0\varepsilon_0\,\omega E_0\cos(kx-\omega t) \implies \frac{B_0}{E_0} = \frac{\mu_0\varepsilon_0\,\omega}{k}. \] Multiply the two relations \(E_0/B_0 = \omega/k\) and \(B_0/E_0 = \mu_0\varepsilon_0\,\omega/k\): \[ 1 = \mu_0\varepsilon_0 \left(\frac{\omega}{k}\right)^2 \implies \frac{\omega}{k} = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \equiv c. \] The wave speed is determined entirely by \(\varepsilon_0\) and \(\mu_0\).

The \(E\)-to-\(B\) Amplitude Ratio

Birth of the formula: The Amplitude Relation \(E = cB\)

Goal: fix the ratio of the field amplitudes in a plane wave. Setup. Take a wave along \(+x\) with \(\vec{E} = E_0\sin(kx-\omega t)\,\hat{y}\) and \(\vec{B} = B_0\sin(kx-\omega t)\,\hat{z}\). Faraday's law applied to a thin rectangular loop in the \(xy\)-plane relates the spatial variation of \(E\) to the time variation of \(B\): \[ \frac{\partial E}{\partial x} = -\frac{\partial B}{\partial t}. \] Solve. Differentiating the assumed forms, \[ \frac{\partial E}{\partial x} = kE_0\cos(kx-\omega t), \qquad \frac{\partial B}{\partial t} = -\omega B_0\cos(kx-\omega t). \] Equating (the common cosine cancels): \[ kE_0 = \omega B_0 \quad\Longrightarrow\quad \frac{E_0}{B_0} = \frac{\omega}{k} = c, \] because \(\omega/k\) is the wave speed, and that speed is \(c\). Result. \[ \boxed{\,E_0 = c\,B_0\,} \qquad\text{(and }E = cB\text{ at every instant, since the fields are in phase).} \] Check. Dimensions: \([cB] = (\mathrm{m\,s^{-1}})(\mathrm{kg\,s^{-2}\,A^{-1}}) = \mathrm{kg\,m\,s^{-3}\,A^{-1}} = \mathrm{V\,m^{-1}} = [E]\). ✓

The relation \(E_0/B_0 = c\) is worth pausing on. The speed of light is \(3 \times 10^8\,\mathrm{m/s}\), so the electric amplitude is enormously larger than the magnetic amplitude when measured in SI units: for a wave in which \(E_0 = 300\,\mathrm{V/m}\), the magnetic amplitude is only \(B_0 = 10^{-6}\,\mathrm{T} = 1\,\mu\mathrm{T}\).

This numerical disparity does not mean the magnetic field is weaker or less important: the energy densities \(\tfrac{1}{2}\varepsilon_0 E^2\) and \(\tfrac{1}{2\mu_0}B^2\) are equal at every point in the wave. The fields are energetically balanced; they merely look unequal in SI units because \(\varepsilon_0\) and \(\mu_0\) have very different magnitudes.

Why No Medium Is Required

A mechanical wave—sound in air, a wave on a rope—is a disturbance of some material medium. The medium's particles oscillate; the pattern propagates. Remove the medium, and the wave cannot exist.

An electromagnetic wave is structurally different. There are no material particles oscillating. It is the fields themselves that oscillate. Fields exist in vacuum; they are not properties of matter. Therefore the wave exists in vacuum and requires no medium to support it.

When physicists of the nineteenth century searched for an aether—a material medium for light—they were implicitly assuming that light was mechanically analogous to sound. The Michelson–Morley experiment (1887) found no aether. The correct explanation is that electromagnetic waves are not mechanical: the question “what is oscillating?” is answered by “the electric and magnetic fields,” not “some material particles.”

Core intuition: Source Versus Self-Sustaining Propagation

To generate an electromagnetic wave, you need an accelerating charge (or equivalently, an oscillating current). The accelerating charge creates time-varying \(\vec{E}\) and \(\vec{B}\) fields that detach from the source and propagate as radiation. Once the wave is propagating, the source is no longer needed. The fields sustain themselves through mutual induction as described above. A photon emitted by a star a billion light-years away arrives at your eye having crossed empty space entirely on its own.

Misconception alert: “The \(\vec{E}\) Field Creates the \(\vec{B}\) Field and Then \(\vec{B}\) Creates \(\vec{E}\), Alternately”

The misconception: In an electromagnetic wave, \(\vec{E}\) oscillates first, then produces \(\vec{B}\), which then produces \(\vec{E}\) again, in a relay. The two fields take turns. Why it is wrong: \(\vec{E}\) and \(\vec{B}\) oscillate in phase: they are both maximum at the same instant and both zero at the same instant. There is no alternating creation—they are coupled and simultaneous. What Faraday's and Ampère's laws say is that the spatial variation of one field is related to the time variation of the other. It is a differential relationship, not a sequential one. The picture of fields “taking turns” producing each other is intuitive but physically incorrect, and it leads to wrong conclusions about phase relationships and energy transport.

✓ Checkpoint · Origin and Nature of Electromagnetic Waves

Explain qualitatively why an electromagnetic wave can propagate through vacuum without any material medium.
In a mechanical wave, oscillating material particles carry the disturbance. In an electromagnetic wave, it is the \(\vec{E}\) and \(\vec{B}\) fields that oscillate. Fields are properties of space, not of matter. Faraday's law ensures that a time-varying \(\vec{B}\) drives \(\vec{E}\), and the modified Ampère's law ensures that a time-varying \(\vec{E}\) drives \(\vec{B}\). These coupled differential relations sustain the wave in any region of space, whether or not matter is present.
For a plane electromagnetic wave propagating in the \(+z\) direction, the electric field is directed along \(+x\). In what direction does the magnetic field point? Justify your answer.
The magnetic field points along \(+y\). In an electromagnetic wave, \(\vec{E}\), \(\vec{B}\), and the propagation direction \(\hat{k}\) are mutually perpendicular, and \(\vec{E}\times\vec{B}\) must point in the propagation direction \(+z\). With \(\vec{E}\) along \(+x\), the only direction for \(\vec{B}\) that makes \(\hat{x}\times\hat{B}\) point along \(+z\) is \(+y\).
In a certain electromagnetic wave in vacuum, the peak electric field is \(E_0 = 600\,\mathrm{V/m}\). Find the peak magnetic field.
Using \(E_0/B_0 = c\): \(B_0 = E_0/c = 600/(3\times10^8) = 2\times10^{-6}\,\mathrm{T} = 2\,\mu\mathrm{T}\).
A student claims that electromagnetic waves are transverse because \(\vec{E}\) and \(\vec{B}\) are perpendicular to each other. Identify the precise error in this statement and correct it.
Being transverse means the oscillating quantities are perpendicular to the direction of propagation, not merely to each other. Both \(\vec{E}\) and \(\vec{B}\) are individually perpendicular to the propagation direction. They are also perpendicular to each other, but that is a separate fact. A wave could in principle have \(\vec{E}\perp\vec{B}\) without being transverse. The correct statement is: electromagnetic waves are transverse because both \(\vec{E}\) and \(\vec{B}\) are perpendicular to the direction of propagation.

Worked Examples

Example 5 · Reading Wave Parameters off the Phase

Given. A plane wave in vacuum has \(\vec{E} = (3.0\,\mathrm{V/m})\,\hat{y}\,\sin(kx-\omega t)\) with \(k = 2.0\times10^{6}\,\mathrm{rad\,m^{-1}}\) and \(\omega = 6.0\times10^{14}\,\mathrm{rad\,s^{-1}}\). Find. The propagation speed, wavelength, and frequency; identify the spectral region.

Solution. Setup. For \(\sin(kx-\omega t)\): speed \(v=\omega/k\), \(\lambda = 2\pi/k\), \(f = \omega/2\pi\). Solve. \[ v = \frac{\omega}{k} = \frac{6.0\times10^{14}}{2.0\times10^{6}} = 3.0\times10^{8}\,\mathrm{m\,s^{-1}},\qquad \lambda = \frac{2\pi}{k} = 3.14\times10^{-6}\,\mathrm{m}, \] \[ f = \frac{\omega}{2\pi} = 9.55\times10^{13}\,\mathrm{Hz}. \] Answer. \(\boxed{v = 3.0\times10^{8}\,\mathrm{m\,s^{-1}},\; \lambda \approx 3.1\,\mu\mathrm{m},\; f \approx 9.5\times10^{13}\,\mathrm{Hz}}\) —the infrared. Check. The speed came out to \(c\) with no adjustment, confirming a vacuum wave. Independently \(f\lambda = (9.55\times10^{13})(3.14\times10^{-6}) = 3.0\times10^{8}\,\mathrm{m\,s^{-1}} = v\). ✓

Example 6 · Propagation Direction and the Magnetic Partner

Given. In vacuum a plane wave has \(\vec{E} = (90\,\mathrm{V/m})\,\hat{z}\,\sin(ky-\omega t)\). Find. the direction of propagation and the full magnetic field \(\vec{B}\) (direction and amplitude).

Solution. Setup. The phase \(ky-\omega t\) advances along \(+\hat{y}\), so the wave travels in \(+\hat{y}\). The triad requires \(\hat{E}\times\hat{B} = \hat{k}_{\text{prop}} = \hat{y}\), and \(B_0 = E_0/c\), with \(\vec{B}\) in phase with \(\vec{E}\). Solve. With \(\hat{E} = \hat{z}\) we need \(\hat{z}\times\hat{B}=\hat{y}\); since \(\hat{z}\times\hat{x} = \hat{y}\), the magnetic direction is \(\hat{B} = \hat{x}\). The amplitude is \[ B_0 = \frac{E_0}{c} = \frac{90}{3.0\times10^{8}} = 3.0\times10^{-7}\,\mathrm{T}. \] Answer. \(\boxed{\text{propagation }+\hat{y};\quad \vec{B} = (3.0\times10^{-7}\,\mathrm{T})\,\hat{x}\,\sin(ky-\omega t)}\). Check. \(\hat{E}\times\hat{B} = \hat{z}\times\hat{x} = \hat{y}\), matching the propagation direction; \(\vec{E}\), \(\vec{B}\), and \(\hat{y}\) are mutually perpendicular, and \(B_0 = E_0/c\) as demanded by \(E=cB\). ✓

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Wavelength of an FM Signal

An FM radio station broadcasts at \(f = 100\,\mathrm{MHz}\). Find its wavelength in vacuum.

Solution. Given: \(f = 100\,\mathrm{MHz} = 1.0\times10^{8}\,\mathrm{Hz}\).
Find: \(\lambda\).
Setup: In vacuum every electromagnetic wave obeys \(c = f\lambda\).
Solve: \[ \lambda = \frac{c}{f} = \frac{3.0\times10^{8}}{1.0\times10^{8}} = 3.0\,\mathrm{m}. \] Answer: \(\boxed{\lambda = 3.0\,\mathrm{m}}\) — a few metres, as expected for radio. Check: Reverse it: \(f = c/\lambda = (3.0\times10^{8})/3.0 = 1.0\times10^{8}\,\mathrm{Hz} = 100\,\mathrm{MHz}\), the stated frequency. A few-metre wavelength is exactly the radio scale. ✓

Example 2 · Magnetic Amplitude of a Light Wave

A light wave in vacuum has electric field amplitude \(E_0 = 300\,\mathrm{V/m}\). Find its magnetic field amplitude \(B_0\).

Solution. Given: \(E_0 = 300\,\mathrm{V/m}\).
Find: \(B_0\).
Setup: In vacuum the amplitudes are locked by \(E_0 = cB_0\).
Solve: \[ B_0 = \frac{E_0}{c} = \frac{300}{3.0\times10^{8}} = 1.0\times10^{-6}\,\mathrm{T}. \] Answer: \(\boxed{B_0 = 1.0\,\mu\mathrm{T}}\). The magnetic amplitude looks tiny only because of the factor \(c\); the two fields carry equal energy. Check: Recover \(E_0 = cB_0 = (3.0\times10^{8})(1.0\times10^{-6}) = 300\,\mathrm{V/m}\), the given value. The factor of \(c\) makes \(B_0\) numerically minute, yet \(u_E = u_B\): the fields are energetically equal. ✓

Example 3 · Speed in a Dielectric

A non-magnetic material (\(\mu_r = 1\)) has relative permittivity \(\varepsilon_r = 2.25\). Find its refractive index and the speed of electromagnetic waves in it.

Solution. Given: \(\varepsilon_r = 2.25\), \(\mu_r = 1\).
Find: \(n\), \(v\).
Setup: \(n = \sqrt{\varepsilon_r \mu_r}\) and \(v = c/n\).
Solve: \[ n = \sqrt{2.25\times1} = 1.5, \qquad v = \frac{c}{n} = \frac{3.0\times10^{8}}{1.5} = 2.0\times10^{8}\,\mathrm{m/s}. \] Answer: \(\boxed{n = 1.5,\quad v = 2.0\times10^{8}\,\mathrm{m/s}}\). Check: \(v = 2.0\times10^{8}\,\mathrm{m/s} \lt c\), as any material medium demands. The Maxwell relation is self-consistent: \(n^2 = 2.25 = \varepsilon_r\), so \(n=\sqrt{\varepsilon_r}\) holds for this non-magnetic material. ✓

Example 4 · Wave Intensity

A plane wave in vacuum has peak electric field \(E_0 = 100\,\mathrm{V/m}\). Find its average intensity.

Solution. Given: \(E_0 = 100\,\mathrm{V/m}\).
Find: \(\langle I\rangle\).
Setup: \(\langle I\rangle = \tfrac12\varepsilon_0 c E_0^2\).
Solve: \[ \langle I\rangle = \tfrac12 (8.85\times10^{-12})(3.0\times10^{8})(100)^2 = 13.3\,\mathrm{W\,m^{-2}}. \] Answer: \(\boxed{\langle I\rangle \approx 13\,\mathrm{W\,m^{-2}}}\). Check: The magnetic form must agree. With \(B_0 = E_0/c = 3.3\times10^{-7}\,\mathrm{T}\), \(\langle I\rangle = cB_0^2/2\mu_0 = 13\,\mathrm{W\,m^{-2}}\)—the same number by a different route. ✓

Example 5 · Placing a Wave in the Spectrum

An electromagnetic wave has wavelength \(\lambda = 1.0\,\mathrm{cm}\). (a) Find its frequency. (b) Which band does it belong to? (c) Is its photon energy larger or smaller than that of visible light?

Solution. Given: \(\lambda = 1.0\,\mathrm{cm} = 1.0\times10^{-2}\,\mathrm{m}\).
Find: \(f\), band, photon-energy comparison.
Setup: \(f = c/\lambda\); band from the spectrum table; photon energy \(E = hf\) rises with frequency.
Solve: \[ f = \frac{c}{\lambda} = \frac{3.0\times10^{8}}{1.0\times10^{-2}} = 3.0\times10^{10}\,\mathrm{Hz} = 30\,\mathrm{GHz}. \] This lies in the microwave band. Its frequency is far below visible (\(\sim10^{15}\,\mathrm{Hz}\)), so its photon energy is much smaller.
Answer: \(\boxed{30\,\mathrm{GHz},\ \text{microwave, lower photon energy than visible}}\). Check: \(c = f\lambda = (3.0\times10^{10})(1.0\times10^{-2}) = 3.0\times10^{8}\,\mathrm{m/s}\) recovers \(c\). A centimetre wavelength sits between millimetre and metre—the microwave band—consistent with \(30\,\mathrm{GHz}\). ✓

Electromagnetic Waves

Perplexing Questions

Why Light Must Be an Electromagnetic Wave

The Predicted Speed

What Unified Electricity and Magnetism

What This Means for Light

Worked Examples

The Need for Displacement Current

The Capacitor Gap Paradox

What Is Different in the Gap

Why This Repairs the Paradox

Worked Examples

Origin and Nature of Electromagnetic Waves

How the Fields Sustain Each Other

Structure of the Wave

The \(E\)-to-\(B\) Amplitude Ratio

Why No Medium Is Required

Worked Examples

Solved examples

Problem bank

Chapter test

The chapter continues.