Electromagnetic Waves

A capacitor is charging. Draw an Amperian loop around the wire feeding it, then choose a surface bounded by that loop which passes between the plates, where no current flows at all. Ampère's law now gives $B=0$. Choose a flat surface cutting the wire and it gives $B={\mu }_{0}I/2\pi r$. Same loop, two answers. Which is right?

Neither, as Ampère wrote it. The law is incomplete.

Between the plates there is no charge flowing, but there is an electric field growing, and Maxwell's insight was that a changing electric flux does everything a current does. Define the displacement current

$i_d={\epsilon }_0\frac{d{\Phi }_E}{dt}$

and for a charging parallel-plate capacitor, with $E=\frac{q}{{\epsilon }_{0}A}$,

$i_d={\epsilon }_{0}A\frac{dE}{dt}=\frac{dq}{dt}=i_c$

exactly equal to the conduction current in the wire. Both surfaces now give the same answer, and the paradox evaporates.

One repaired equation produced light. The rest of the chapter is the consequences: a wave that needs no medium, travels at a speed built from two electrostatic constants, carries energy in equal electric and magnetic shares, and pushes on whatever absorbs it.

1. The Ampère-Maxwell law and the field inside a capacitor

The completed law reads

$\oint \vec{B}\cdot d\vec{l}={\mu }_0\left(i_c+{\epsilon }_0\frac{d{\Phi }_E}{dt}\right)$

Applying it inside a charging capacitor, where $i_c=0$, gives a genuine magnetic field between the plates. At radius $r$ from the axis of circular plates of radius $R$,

$B=\frac{{\mu }_0{i}_{d}r}{2\pi R^2}\left(r<R\right),B=\frac{{\mu }_0{i}_d}{2\pi r}\left(r>R\right)$

which is identical in form to the field of a thick current-carrying wire. Outside the plates you cannot tell the difference.

Illustration 1

A parallel-plate capacitor with circular plates of radius $6$ cm is charged so that the electric field between the plates grows at $10^{12}$ V m${}^{-1}$ s${}^{-1}$. Find the displacement current and the magnetic field at the edge of the plates.

$i_d={\epsilon }_{0}A\frac{dE}{dt}=\left(8.85\times 10^{-12}\right)\pi {\left(0.06\right)}^2\left(10^{12}\right)$

$i_d=8.85\times \pi \times 3.6\times 10^{-3}\times 10^{-1}$, giving $i_d=0.10$ A

$B=\frac{{\mu }_0{i}_d}{2\pi R}=\frac{\left(2\times 10^{-7}\right)\left(0.10\right)}{0.06}=3.3\times 10^{-7}$ T

A field of a third of a microtesla, from no moving charge at all. It is small, but it has been measured, and it is what closes Maxwell's equations.

Illustration 2

Show that the displacement current in a charging capacitor equals the conduction current, whatever the plate area.

$E=\frac{\sigma }{{\epsilon }_0}=\frac{q}{{\epsilon }_{0}A}$, so ${\Phi }_E=EA=\frac{q}{{\epsilon }_0}$

$i_d={\epsilon }_0\frac{d{\Phi }_E}{dt}=\frac{dq}{dt}=i_c$

The area cancels completely. That is what makes the current continuous: what flows as charge in the wire continues as changing flux in the gap, with no discontinuity anywhere.

2. Maxwell's equations, and where $c$ comes from

The four laws in integral form are

$\oint \vec{E}\cdot d\vec{A}=\frac{q_{enc}}{{\epsilon }_0},\oint \vec{B}\cdot d\vec{A}=0$

$\oint \vec{E}\cdot d\vec{l}=-\frac{d{\Phi }_B}{dt},\oint \vec{B}\cdot d\vec{l}={\mu }_0{i}_c+{\mu }_0{\epsilon }_0\frac{d{\Phi }_E}{dt}$

Read them as a chain. A changing $\vec{B}$ creates a circulating $\vec{E}$; a changing $\vec{E}$ creates a circulating $\vec{B}$. Each sustains the other, and the disturbance propagates without any charge or medium being needed. Combining the last two produces a wave equation whose speed is

$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=3.00\times 10^8{\text{m s}}^{-1}$

Both constants came from laboratory electrostatics and magnetostatics. Neither had anything to do with light, and yet the number that emerged matched the measured speed of light — which is why Maxwell concluded that light is an electromagnetic wave.

In a medium, ${\mu }_0{\epsilon }_0$ becomes $\mu \epsilon$, so

$v=\frac{1}{\sqrt{\mu \epsilon }}=\frac{c}{n},n=\sqrt{{\mu }_r{\epsilon }_r}\approx \sqrt{{\epsilon }_r}$

for non-magnetic materials — a striking bridge between a capacitor measurement and an optical one.

Illustration 3

A non-magnetic medium has relative permittivity $2.25$. Find the refractive index, the wave speed, and the wavelength there of light whose vacuum wavelength is $600$ nm.

$n=\sqrt{2.25}=1.5$, so $v=\frac{3\times 10^8}{1.5}=2\times 10^8$ m s${}^{-1}$

Frequency is unchanged on entering a medium, so ${\lambda }^{\prime }=\frac{\lambda }{n}=\frac{600}{1.5}=400$ nm.

Frequency is set by the source and cannot change at a boundary, since the fields on the two sides must stay in step. It is the wavelength that adjusts.

3. The structure of the wave

For a plane wave travelling along $x$,

$E_y=E_0\sin \left(kx-\omega t\right),B_z=B_0\sin \left(kx-\omega t\right)$

Four features are examined repeatedly.

$\vec{E}$ and $\vec{B}$ are in phase — both peak together, both vanish together. They are mutually perpendicular, and both are perpendicular to the direction of travel, so the wave is transverse. Their magnitudes are locked:

$\frac{E_0}{B_0}=c$

And the direction of propagation is $\hat{E}\times \hat{B}$, which fixes the sign conventions completely.

Illustration 4

An electromagnetic wave has $\vec{E}$ along $+\hat{j}$ and travels along $+\hat{i}$. In which direction does $\vec{B}$ point, and what is $B_0$ if $E_0=90$ V m${}^{-1}$?

Propagation is $\hat{E}\times \hat{B}$, so $\hat{i}=\hat{j}\times \hat{B}$, which requires $\hat{B}=\hat{k}$.

$B_0=\frac{E_0}{c}=\frac{90}{3\times 10^8}=3\times 10^{-7}$ T

The magnetic amplitude is always tiny in SI units, and it is tempting to conclude the magnetic part is unimportant. It is not — the two carry equal energy, as the next section shows.

4. Energy, intensity and the Poynting vector

The energy density has two parts,

$u=\frac{1}{2}{\epsilon }_0{E}^2+\frac{B^2}{2{\mu }_0}$

and because $E=cB$ with $c=1/\sqrt{{\mu }_0{\epsilon }_0}$, the two are equal at every instant. Averaging the sine squared over a cycle gives

$\bar{u}=\frac{1}{2}{\epsilon }_0{E}_0^2,I=\bar{u}c=\frac{1}{2}{\epsilon }_0{E}_0^{2}c=\frac{E_{rms}{B}_{rms}}{{\mu }_0}$

The direction and rate of energy flow together are given by the Poynting vector

$\vec{S}=\frac{\vec{E}\times \vec{B}}{{\mu }_0}$

whose magnitude is the instantaneous power per unit area and whose direction is the direction of propagation.

Illustration 5

Sunlight at the top of the atmosphere has an intensity of $1367$ W m${}^{-2}$. Find the peak electric and magnetic fields.

$E_0=\sqrt{\frac{2I}{{\epsilon }_{0}c}}=\sqrt{\frac{2\times 1367}{\left(8.85\times 10^{-12}\right)\left(3\times 10^8\right)}}=\sqrt{1.03\times 10^6}$

$E_0=1015$ V m${}^{-1}$, and $B_0=\frac{1015}{3\times 10^8}=3.4\times 10^{-6}$ T

A kilovolt per metre is a substantial field, comparable with what a charged comb produces — yet the accompanying magnetic field is a tenth of the Earth's. That mismatch is entirely an artefact of SI units, since the energies are equal.

Illustration 6

A lamp radiates $100$ W uniformly in all directions. Find the intensity and the peak electric field at $3$ m.

$I=\frac{P}{4\pi r^2}=\frac{100}{4\pi \left(9\right)}=0.884$ W m${}^{-2}$

$E_0=\sqrt{\frac{2\times 0.884}{\left(8.85\times 10^{-12}\right)\left(3\times 10^8\right)}}=\sqrt{666}=25.8$ V m${}^{-1}$

Intensity falls as the inverse square, so $E_0$ falls as $1/r$. Doubling the distance quarters the intensity but only halves the field amplitude.

5. Momentum and radiation pressure

Light carries momentum as well as energy, in the ratio

$p=\frac{U}{c}$

so a surface that absorbs an intensity $I$ feels a pressure $I/c$, while one that perfectly reflects it feels twice as much, because the momentum is reversed rather than merely stopped:

$P_{absorbed}=\frac{I}{c},P_{reflected}=\frac{2I}{c}$

The numbers are small — sunlight exerts about $5$ micropascals — but they are not negligible over large areas or long times, which is what makes a solar sail work and what shapes a comet's tail.

Illustration 7

A perfectly reflecting solar sail of area $10^4$ m${}^2$ is deployed where the solar intensity is $1367$ W m${}^{-2}$. Find the pressure and the total force.

$P=\frac{2I}{c}=\frac{2\times 1367}{3\times 10^8}=9.1\times 10^{-6}$ Pa

$F=PA=9.1\times 10^{-6}\times 10^4=0.091$ N

A tenth of a newton sounds hopeless, but it acts continuously and without fuel. Over a month it delivers the same impulse as a rocket burn, which is why sails are practical for slow interplanetary transfers.

Illustration 8

A laser of power $5$ W is shone onto a black surface. Find the force, and the force if the surface is replaced by a mirror.

Absorbing: $F=\frac{P}{c}=\frac{5}{3\times 10^8}=1.7\times 10^{-8}$ N

Reflecting: $F=\frac{2P}{c}=3.3\times 10^{-8}$ N

The area never appeared. Force depends on the total power intercepted, whereas pressure depends on how that power is spread — which is why focusing a laser raises the pressure enormously without changing the force.

6. Transverse nature, shown by polarisation

Nothing in the wave equation forces a wave to be transverse — sound is not. The decisive experimental evidence that electromagnetic waves are transverse is that they can be polarised, which a longitudinal wave never can.

An ideal polariser transmits only the field component along its pass axis. Unpolarised light contains all orientations equally, so averaging ${\cos }^2\theta$ over a full turn gives

$I_1=\frac{I_0}{2}$

and what emerges is fully polarised along the axis. A second polariser at angle $\theta$ to the first then obeys Malus's law:

$I_2=I_1{\cos }^2\theta$

Two consequences are examined constantly. Crossed polarisers pass nothing. Yet slide a third polariser at $45^{\circ }$ between them and light reappears — because each stage re-projects the field onto a new axis, and two projections of $45^{\circ }$ do what one of $90^{\circ }$ cannot.

Light reflected from a dielectric is also partly polarised, and at Brewster's angle it is completely so:

$\tan {\theta }_B=n$

At that angle the reflected and refracted rays are exactly perpendicular, which is why polarising sunglasses cut glare from water and roads so effectively.

Illustration 9

Unpolarised light of intensity $I_0$ passes through three polarisers, the second at $45^{\circ }$ to the first and the third at $90^{\circ }$ to the first. Find the emergent intensity.

After the first: $\frac{I_0}{2}$

After the second, at $45^{\circ }$: $\frac{I_0}{2}{\cos }^{2}45^{\circ }=\frac{I_0}{4}$

After the third, again at $45^{\circ }$ to the second: $\frac{I_0}{4}{\cos }^{2}45^{\circ }=\frac{I_0}{8}$

Removing the middle polariser gives zero. Inserting an extra absorbing element increases the transmitted light, which is impossible for a wave treated as a stream of blockable rays and entirely natural for one with a direction of oscillation.

Illustration 10

Find Brewster's angle for light reflecting off water of refractive index $1.33$, and state the polarisation of the reflected beam.

$\tan {\theta }_B=1.33\Rightarrow {\theta }_B=53.1^{\circ }$ from the normal

The reflected beam is fully polarised perpendicular to the plane of incidence, that is, horizontally for a horizontal water surface.

This is exactly why polarising sunglasses have a vertical pass axis: the glare they must remove is horizontally polarised.

7. The spectrum, and what each band is for

All electromagnetic waves travel at $c$ in vacuum and differ only in frequency. What changes across the spectrum is how each band is produced and how it interacts with matter.

Scattering across the visible band explains the sky. Particles much smaller than the wavelength scatter with intensity proportional to $1/{\lambda }^4$, so blue is scattered about ten times more strongly than red. Overhead you see scattered blue; at sunset the direct beam has lost its blue along a long slanting path and arrives red.

Radio waves come from oscillating currents in aerials; microwaves from klystrons and magnetrons and are absorbed by molecular rotation; infrared from molecular vibration, which is why it is felt as heat; visible and ultraviolet from outer-electron transitions; X-rays from inner-shell transitions and from decelerating electrons; and gamma rays from the nucleus itself.

Illustration 11

An FM station broadcasts at $100$ MHz. Find the wavelength, and compare with an X-ray of $10^{18}$ Hz.

${\lambda }_{FM}=\frac{3\times 10^8}{10^8}=3$ m

${\lambda }_X=\frac{3\times 10^8}{10^{18}}=3\times 10^{-10}$ m

Ten orders of magnitude apart, which is why one diffracts around buildings while the other resolves the spacing between atoms in a crystal.

Illustration 12

Why does a microwave oven use $2.45$ GHz rather than a frequency that water absorbs most strongly?

At $2.45$ GHz the absorption is deliberately moderate, so the wave penetrates several centimetres before being absorbed.

A frequency of peak absorption would deposit all its energy in the outermost millimetre, cooking the surface and leaving the interior raw.

The engineering choice is about penetration depth, not maximum absorption. This is a recurring theme wherever waves are used to deliver energy into a bulk material.

Summary

• Ampère's law is incomplete: add the displacement current $i_d={\epsilon }_0\frac{d{\Phi }_E}{dt}$, and the capacitor paradox disappears.
• For a charging capacitor $i_d=i_c$ exactly, with the plate area cancelling.
• Inside circular plates the field is $\frac{{\mu }_0{i}_{d}r}{2\pi R^2}$, identical in form to that of a thick wire.
• Maxwell's four equations chain together: changing $\vec{B}$ makes circulating $\vec{E}$, changing $\vec{E}$ makes circulating $\vec{B}$.
• $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$ — built from two electrostatic and magnetostatic constants, with no reference to light.
• In a medium $v=c/n$ with $n=\sqrt{{\mu }_r{\epsilon }_r}\approx \sqrt{{\epsilon }_r}$; frequency is fixed at a boundary, wavelength adjusts.
• $\vec{E}$ and $\vec{B}$ are in phase, mutually perpendicular, transverse, with $E_0/B_0=c$ and propagation along $\hat{E}\times \hat{B}$.
• Electric and magnetic energy densities are equal at every instant; $\bar{u}=\frac{1}{2}{\epsilon }_0{E}_0^2$.
• $I=\frac{1}{2}{\epsilon }_0{E}_0^{2}c=\frac{E_{rms}{B}_{rms}}{{\mu }_0}$, and $\vec{S}=\frac{\vec{E}\times \vec{B}}{{\mu }_0}$ gives both magnitude and direction of energy flow.
• Intensity falls as $1/r^2$ from a point source, so $E_0$ falls as $1/r$.
• Momentum $p=U/c$: pressure is $I/c$ on an absorber and $2I/c$ on a mirror.
• Force depends on total power intercepted; pressure depends on how that power is concentrated.
• Polarisation proves transversality: unpolarised light gives $I_0/2$, then Malus's law $I{\cos }^2\theta$; three polarisers give $I_0/8$ where two crossed give zero.
• Brewster: $\tan {\theta }_B=n$, reflected beam fully polarised, reflected and refracted rays perpendicular.
• Rayleigh scattering goes as $1/{\lambda }^4$ — blue sky overhead, red sun at the horizon.
• All bands travel at $c$ and differ only in how they are produced and how matter responds — from oscillating circuits at the radio end to nuclei at the gamma end.