Electromagnetic Waves

A capacitor is being charged by a current $I$. What is the magnetic field in the gap between the plates, where no charge crosses at all?

Most say zero. Nothing flows there.

It is exactly the field you would get from a wire carrying the same current $I$. Something is crossing the gap — a growing electric field — and it produces magnetism just as a current does.

The shortest chapter in the syllabus, carrying its largest single idea:

• A changing electric field is a source of magnetic field, exactly as a changing magnetic field is a source of electric field. Faraday gave the second half; Maxwell supplied the first.
• The two can therefore sustain each other and travel with nothing to carry them. Change in one makes the other, indefinitely.
• This was predicted, not discovered. Maxwell computed the wave's speed from two constants measured on charges and currents, got $3\times 10^8$ m s⁻¹, and recognised light — by arithmetic, twenty years before anyone made such a wave deliberately.

1. The Problem with Ampere's Law

Ampere's law relates the field round a closed loop to the current threading any surface bounded by that loop. That phrase is where it breaks.

Take a loop encircling the wire feeding one plate:

• A flat surface across the loop is pierced by the wire, so $I_{enc}=I$.
• A bag-shaped surface with the same boundary, bulging out between the plates, is crossed by no conduction current at all, so $I_{enc}=0$.

Two surfaces sharing one boundary give two answers. That is a contradiction, not an approximation — and the fault is in the physics, not the mathematics. Something must be crossing the gap that the law was not counting.

2. Displacement Current

What is happening in the gap is that the electric field is growing as the capacitor charges.

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

Trap. This is not a current in any conventional sense. Nothing flows. It is a changing electric flux given the units and the standing of a current.

The remarkable part is that it comes out exactly equal to the conduction current. Between the plates $E=\sigma /{\epsilon }_0=Q/{\epsilon }_{0}A$, so ${\Phi }_E=EA=Q/{\epsilon }_0$:

$I_d={\epsilon }_0\cdot \frac{1}{{\epsilon }_0}\frac{dQ}{dt}=\frac{dQ}{dt}=I$

The bag surface encloses a displacement current of $I$ after all. Both surfaces now give ${\mu }_{0}I$, and the continuity that seemed broken is restored exactly.

$\oint \vec{B}\cdot d\vec{l}={\mu }_0(I_c+I_d)$

This is the Ampere-Maxwell law, and the added term is the whole of what was new. Its physical content — a changing electric field produces a magnetic field — is what makes waves possible.

Illustration 1

A capacitor with circular plates of radius $R$ is charging. Find $B$ at a distance $r$ from the axis, inside the gap.

Only the displacement current inside radius $r$ counts, and it is spread uniformly over the plate area:

$I_{enc}=I_d\frac{r^2}{R^2}\Rightarrow B(2\pi r)={\mu }_0{I}_d\frac{r^2}{R^2}$

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

Rising linearly inside and falling as $1/r$ outside — the identical shape to $B$ inside a thick current-carrying wire. Displacement current behaves like current in every way that Ampere's law can see.

3. Maxwell's Equations

Four equations summarise all of classical electromagnetism. JEE Main treats them qualitatively.

Only the fourth was modified, and only by one added term. Read the last two together and the wave appears: each field's change creates the other, so a disturbance in one propagates as a disturbance in both.

$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}},v_{medium}=\frac{1}{\sqrt{\mu \epsilon }}=\frac{c}{n}$

Illustration 2

Compute $c$ from ${\mu }_0=4\pi \times 10^{-7}$ and ${\epsilon }_0=8.85\times 10^{-12}$, in SI units.

${\mu }_0{\epsilon }_0=(1.257\times 10^{-6})(8.85\times 10^{-12})=1.112\times 10^{-17}$

$\sqrt{{\mu }_0{\epsilon }_0}=3.335\times 10^{-9}\Rightarrow c=2.999\times 10^8{\text{m s}}^{-1}$

Do that arithmetic and you have repeated the moment that identified light. Neither constant has anything to do with light: ${\epsilon }_0$ comes from forces between static charges, ${\mu }_0$ from forces between currents. Their combination is the speed of light anyway.

Illustration 3

Yellow light of wavelength 589 nm in vacuum enters glass of refractive index 1.5. Find its speed, wavelength and frequency inside the glass.

$v=\frac{c}{n}=\frac{3\times 10^8}{1.5}=2\times 10^8{\text{m s}}^{-1}$

$f=\frac{c}{{\lambda }_{vac}}=\frac{3\times 10^8}{589\times 10^{-9}}=5.09\times 10^{14}\text{Hz}$

The frequency is unchanged. It is set by the source, and the atoms at the boundary are driven at whatever frequency reaches them — nothing about entering the glass can alter how often they are pushed. The wavelength alone absorbs the difference:

${\lambda }_{glass}=\frac{v}{f}=\frac{2\times 10^8}{5.09\times 10^{14}}=393\text{nm}=\frac{{\lambda }_{vac}}{n}$

Holding the wavelength fixed and letting the frequency drop is the standard error, and it gets the colour wrong. Colour tracks frequency, which is why an object viewed under water does not change hue.

4. Nature of Electromagnetic Waves

Transverse. $\vec{E}$, $\vec{B}$ and the direction of travel are mutually perpendicular, and the wave travels along $\vec{E}\times \vec{B}$.

$E_y=E_0\sin (kx-\omega t),B_z=B_0\sin (kx-\omega t),\frac{E_0}{B_0}=c$

Trap. The two fields are in phase — peaking and vanishing at the same instants and places. They do not take turns. $B_0$ is numerically $3\times 10^8$ times smaller only because of SI units; the two carry exactly equal energy.

No medium is required. This is the deepest break from the mechanical waves of the earlier chapter, and it is why sunlight crosses interplanetary vacuum while sound cannot cross a room without air.

Being uncharged, EM waves are not deflected by electric or magnetic fields — which distinguishes them from cathode rays. They obey superposition, which makes interference and diffraction possible. And being transverse, they can be polarised: a longitudinal wave has no plane of vibration to select, which is exactly why sound cannot be polarised and light can.

Illustration 4

A plane wave travels along $+\hat{x}$ with its electric field along $+\hat{y}$. Find the direction of $\vec{B}$.

Propagation is along $\vec{E}\times \vec{B}$:

$\hat{y}\times \hat{B}=\hat{x}\Rightarrow \hat{B}=\hat{z}$

Check it: $\hat{y}\times \hat{z}=\hat{x}$. The cross product is not symmetric, so $\hat{B}=-\hat{z}$ would send the wave backwards. Always test your answer by re-forming $\vec{E}\times \vec{B}$.

How electromagnetic waves are produced

Only accelerating charges radiate. A stationary charge has a static field; a charge in uniform motion carries its field along; neither radiates. An oscillating charge radiates at exactly its own frequency, and this is the production mechanism for every band.

Illustration 5

An FM station broadcasts at 100 MHz. Find the length of a quarter-wave antenna.

$\lambda =\frac{c}{f}=\frac{3\times 10^8}{10^8}=3\text{m}\Rightarrow \frac{\lambda }{4}=0.75\text{m}$

Antenna length scales with wavelength, which is why a broadcast mast is metres tall and a phone antenna — working at around 2 GHz, $\lambda =15$ cm — fits inside the case.

Illustration 6

A plane wave in vacuum has $E_y=60\sin (500x-1.5\times 10^{11}t)$ V m⁻¹. Find the wavelength, frequency, speed, magnetic amplitude, intensity, direction of travel and band.

Match against $E_0\sin (kx-\omega t)$, giving $k=500$ rad m⁻¹ and $\omega =1.5\times 10^{11}$ rad s⁻¹:

$\lambda =\frac{2\pi }{k}=1.26\text{cm},f=\frac{\omega }{2\pi }=2.39\times 10^{10}\text{Hz}$

$v=\frac{\omega }{k}=\frac{1.5\times 10^{11}}{500}=3\times 10^8{\text{m s}}^{-1}$

The speed check confirms vacuum. A different value would have meant a medium, with $n=c/v$ following at once.

$B_0=\frac{E_0}{c}=2\times 10^{-7}\text{T},I=\frac{1}{2}{\epsilon }_0{E}_0^{2}c=4.78{\text{W m}}^{-2}$

The sign inside the bracket is $kx-\omega t$, so the wave travels along $+\hat{x}$; with $\vec{E}$ along $\hat{y}$, $\vec{B}$ must lie along $\hat{z}$. At 1.26 cm this is a microwave, near the 12 cm band a domestic oven uses.

5. Energy, Intensity and Momentum

$u_E=\frac{1}{2}{\epsilon }_0{E}^2,u_B=\frac{B^2}{2{\mu }_0}$

Averaged over a cycle these are exactly equal, which follows from $E_0/B_0=c$ together with $c^2=1/{\mu }_0{\epsilon }_0$. The total average energy density is $\frac{1}{2}{\epsilon }_0{E}_0^2$, counting both.

Illustration 7

Show that the electric and magnetic energy densities are equal at every instant, not merely on average.

Start from the magnetic density and substitute $B=E/c$:

$u_B=\frac{B^2}{2{\mu }_0}=\frac{E^2}{2{\mu }_0{c}^2}$

Now use $c^2=1/{\mu }_0{\epsilon }_0$, so that $1/c^2={\mu }_0{\epsilon }_0$:

$u_B=\frac{E^2{\mu }_0{\epsilon }_0}{2{\mu }_0}=\frac{1}{2}{\epsilon }_0{E}^2=u_E$

No averaging entered anywhere, because $E$ and $B$ are in phase: they peak together and vanish together. The energy is therefore split exactly down the middle at every point at every instant, so the instantaneous total is ${\epsilon }_0{E}^2$ and its cycle average is $\frac{1}{2}{\epsilon }_0{E}_0^2$.

$I=\frac{1}{2}{\epsilon }_0{E}_0^{2}c$

Intensity goes as the square of the amplitude, as for every wave.

Illustration 8

A 100 W lamp radiates uniformly in all directions. Find the intensity, peak electric field and peak magnetic field 2 m away.

$I=\frac{P}{4\pi r^2}=\frac{100}{4\pi (4)}=1.99{\text{W m}}^{-2}$

$E_0=\sqrt{\frac{2I}{{\epsilon }_{0}c}}=\sqrt{\frac{3.98}{2.66\times 10^{-3}}}=38.7{\text{N C}}^{-1}\Rightarrow B_0=\frac{E_0}{c}=1.3\times 10^{-7}\text{T}$

A field of 39 volts per metre from an ordinary lamp — and a magnetic field a few hundred times weaker than the Earth's, yet carrying exactly half the energy.

Momentum and radiation pressure

$p=\frac{U}{c}\text{(absorbed)},p=\frac{2U}{c}\text{(reflected)}$

Reflection delivers twice the momentum, for the same reason a bouncing ball delivers twice the impulse of one that sticks.

Illustration 9

A perfectly reflecting solar sail of area $10^4$ m² sits above the atmosphere where $I=1370$ W m⁻². Find the force on it.

$F=\frac{2IA}{c}=\frac{2(1370)(10^4)}{3\times 10^8}\Rightarrow F=0.091\text{N}$

A hectare of sail yields about a tenth of a newton — the weight of a small apple. But it never stops, needs no fuel, and over months of continuous thrust that beats any chemical rocket for a light enough craft. It is also why comet tails point away from the Sun whatever direction the comet is travelling.

Illustration 10

For a spherical dust grain of radius $a$ and density $\rho$ at distance $r$ from the Sun, compare the outward radiation force with the inward gravitational pull.

The Sun's luminosity $L$ spreads over a sphere, so the intensity at $r$ is $L/4\pi r^2$, and the grain intercepts its own cross-section $\pi a^2$:

$F_{rad}=\frac{L}{4\pi r^{2}c}\cdot \pi a^2\propto \frac{a^2}{r^2},F_{grav}=\frac{GM\left(\frac{4}{3}\pi a^3\rho \right)}{r^2}\propto \frac{a^3}{r^2}$

$\frac{F_{rad}}{F_{grav}}\propto \frac{1}{a}\text{, independent of }r$

Both forces fall as $1/r^2$, so the ratio is the same everywhere in the solar system. Moving a grain closer to the Sun or further from it changes nothing about which force wins; only size decides.

Radiation acts on area while gravity acts on volume, so below a critical grain size — roughly a micrometre — light wins and the grain is swept outward. That is why a comet's dust tail always points away from the Sun, and why the inner solar system is swept clear of fine dust.

6. The Electromagnetic Spectrum

All electromagnetic waves travel at the same speed in vacuum and differ only in frequency. The band names are historical, reflecting how each was found and used, not any physical discontinuity.

The visible band spans less than one octave of an enormously wide spectrum. Nearly everything happening electromagnetically around you is invisible.

The atmosphere is transparent in only two windows — visible light and much of the radio band. That is precisely why optical and radio astronomy can be done from the ground while X-ray and most infrared astronomy need satellites. Ozone absorbs the shorter ultraviolet, which makes its depletion a health issue rather than an atmospheric curiosity. And greenhouse gases pass incoming visible light while absorbing outgoing infrared, which is the entire greenhouse mechanism stated in one line.

Discovery and penetrating power

The bands were found in an order unrelated to their position in the spectrum.

Illustration 11

Compare the photon energy of a 100 MHz radio wave with that of a 1 nm X-ray. (Photon energy is $E=hf$, taken up properly in the next chapter.)

$E_{radio}=hf=(6.63\times 10^{-34})(10^8)=6.6\times 10^{-26}\text{J}=4\times 10^{-7}\text{eV}$

$E_X=\frac{1240\text{eV nm}}{1\text{nm}}=1240\text{eV}$

A ratio of about $3\times 10^9$. That gulf is why hazard rises with frequency and not with intensity alone: radio photons can only heat tissue however many arrive, while a single X-ray photon carries enough to break a chemical bond. The threshold sits inside the ultraviolet band, which is why UV is split into A, B and C.

Illustration 12

A source emits at $3\times 10^{12}$ Hz. Identify the band and give the wavelength. Then do the reverse for a wavelength of 3 cm.

$\lambda =\frac{c}{f}=\frac{3\times 10^8}{3\times 10^{12}}=10^{-4}\text{m}=100\mu \text{m}$

That lies between 700 nm and 1 mm, so it is infrared — the far infrared, close to the microwave boundary.

$f=\frac{c}{\lambda }=\frac{3\times 10^8}{0.03}=10^{10}\text{Hz}$

Three centimetres sits between 1 mm and 0.1 m, making this a microwave, in the radar band.

Both answers came from the same relation, applied in opposite directions. The band boundaries are conventions rather than physics, so a value sitting on an edge — 1 mm can fairly be called either far infrared or microwave — is genuinely ambiguous, and questions keep away from those points.

Summary

• A changing electric field produces a magnetic field. That is Maxwell's addition and the whole chapter.
• Ampere's law was inconsistent for a charging capacitor: two surfaces on one boundary gave different answers.
• $I_d={\epsilon }_{0}d{\Phi }_E/dt$ resolves it and equals the conduction current exactly. Nothing flows.
• Between the plates $B\propto r$ inside and $1/r$ outside — the same profile as a thick wire.
• Maxwell's four equations: Gauss electric, Gauss magnetic, Faraday, Ampere-Maxwell. Only the last was modified.
• $c=1/\sqrt{{\mu }_0{\epsilon }_0}$ from two constants unconnected with light — which is how light was identified.
• In a medium, $v=c/n$. That is the definition of refractive index.
• Waves are transverse, travel along $\vec{E}\times \vec{B}$, need no medium, and are undeflected by fields.
• $\vec{E}$ and $\vec{B}$ are in phase, not alternating, with $E_0/B_0=c$ and equal average energy.
• Transverse nature is what allows polarisation — and why sound cannot be polarised.
• Only accelerating charges radiate, at their own frequency; antenna length scales with $\lambda$.
• $u=\frac{1}{2}{\epsilon }_0{E}_0^2$ and $I=\frac{1}{2}{\epsilon }_0{E}_0^{2}c$.
• Momentum $U/c$ absorbed, $2U/c$ reflected — hence radiation pressure, comet tails and solar sails.
• Spectrum by increasing wavelength: gamma, X-ray, UV, visible (400 to 700 nm), IR, microwave, radio.
• The atmosphere has only two windows, visible and radio, which is why X-ray astronomy needs satellites.
• Photon energy, penetrating power and biological hazard all rise with frequency; the bond-breaking threshold sits inside the UV band.
• Frequency never changes on entering a medium, so colour is preserved while speed and wavelength both fall by $n$.