Electromagnetic Waves: Full Notes

8.1 Displacement Current & Maxwell’s Correction

The original Ampère’s law relates the magnetic field circulating around a loop to the current passing through any surface bounded by that loop. Maxwell noticed it breaks down for a charging capacitor: between the plates, no actual charge crosses the gap, yet a magnetic field is still observed there.

His fix was to recognise that a changing electric field between the plates acts just like a current for the purpose of producing a magnetic field. He called this the displacement current:

Displacement current

I_d = \varepsilon_0\dfrac{d\Phi_E}{dt}

where $\Phi_E$ is the electric flux through the surface. With this term added, Ampère’s law holds everywhere, even inside the gap of a charging capacitor where no real charge flows at all:

Ampère–Maxwell law

\oint \vec{B}\cdot d\vec{l} = \mu_0\left(I_{conduction} + I_d\right)

This single correction is what makes self-sustaining electric and magnetic field oscillations possible in empty space, with neither real charges nor real currents required.

8.2 Maxwell’s Equations

Taken together, four equations completely describe classical electricity and magnetism, and predict electromagnetic waves as a direct consequence:

8.2.1 Gauss’s law for electricity

Electric field lines originate on positive charge and terminate on negative charge; the net flux through any closed surface is proportional to the enclosed charge.

8.2.2 Gauss’s law for magnetism

Magnetic field lines never begin or end anywhere, isolated magnetic poles do not exist, so the net magnetic flux through any closed surface is always exactly zero.

8.2.3 Faraday’s law of induction

A changing magnetic field induces a circulating electric field (this is the EMF behind every generator and transformer).

8.2.4 Ampère–Maxwell law

A current, or a changing electric field (the displacement current), induces a circulating magnetic field.

The last two equations are the key: a changing $B$ creates an $E$ , and a changing $E$ creates a $B$ , so the two fields can regenerate each other and propagate outward indefinitely, with no medium and no source charges needed once the wave exists.

8.3 Properties of Electromagnetic Waves

They are transverse waves: both fields oscillate perpendicular to the direction the wave travels.
$\vec{E}$ , $\vec{B}$ , and the direction of propagation are mutually perpendicular, and the two fields oscillate in phase (they rise and fall together, not out of step).
The ratio of their peak magnitudes is fixed: $E_0/B_0 = c$ .
They require no medium, unlike sound: they travel through vacuum at $c = 1/\sqrt{\varepsilon_0\mu_0} \approx 3\times10^8$ m/s.
They carry both energy and momentum, which is why light exerts a (tiny) radiation pressure on anything it strikes.
Every EM wave, regardless of frequency, obeys the same universal relationship:

Universal relationship

c = f\lambda

8.3.1 Energy density and radiation pressure

Every field stores energy. The electric field alone stores an energy density $u_E=\tfrac{1}{2}\varepsilon_0E^2$ , and the magnetic field alone stores $u_B=\tfrac{B^2}{2\mu_0}$ . For an EM wave these two are not independent, since $E=cB$ and $c=1/\sqrt{\mu_0\varepsilon_0}$ :

u_B = \dfrac{B^2}{2\mu_0} = \dfrac{(E/c)^2}{2\mu_0} = \dfrac{E^2}{2\mu_0c^2} = \dfrac{E^2}{2\mu_0\cdot\frac{1}{\mu_0\varepsilon_0}} = \dfrac{\varepsilon_0E^2}{2} = u_E

the electric and magnetic fields of an EM wave always carry exactly equal energy density, so the total is simply twice either one:

Total energy density

\textcolor{#e08a1e}{u} = u_E+u_B = \varepsilon_0E^2

Since this energy moves at speed $c$ , the intensity (power per unit area) carried by the wave is $I=uc$ , so a wave of known intensity has average energy density $u=I/c$ . Momentum goes along with that energy, $p=u/c$ per unit volume, which is where radiation pressure comes from: a surface that absorbs the wave entirely picks up that momentum directly, giving pressure $\textcolor{#e08a1e}{P}=u$ ; a surface that reflects the wave straight back reverses the wave’s momentum rather than just stopping it, so by Newton’s third law the surface feels twice the push, $\textcolor{#e08a1e}{P}=2u$ , exactly the same doubling a mirror gives a ball bounced off it compared with one that simply sticks.

Worked example

Radiation pressure from sunlight

Sunlight arrives at the Earth’s surface with an intensity of about 1000 W/m². Find the average energy density of this radiation, and the radiation pressure it exerts on a surface that (a) absorbs it completely, (b) reflects it completely.

u = \dfrac{I}{c} = \dfrac{1000}{3\times10^8} \approx \textcolor{#e08a1e}{3.33\times10^{-6}\text{ J/m}^3}

\text{(a) } P_{\text{absorb}} = u \approx \textcolor{#e08a1e}{3.33\times10^{-6}\text{ Pa}}, \qquad \text{(b) } P_{\text{reflect}} = 2u \approx \textcolor{#e08a1e}{6.67\times10^{-6}\text{ Pa}}

Millionths of a pascal, immeasurably small next to ordinary atmospheric pressure (about 10⁵ Pa), which is exactly why radiation pressure is irrelevant for everyday objects but becomes the dominant force shaping comet tails and the sails of proposed solar-sail spacecraft, where nothing else is pushing back.

Worked example

FM radio: frequency to wavelength

An FM radio station broadcasts at 100 MHz. Find its wavelength.

\lambda = \dfrac{c}{f} = \dfrac{3\times10^{8}}{1\times10^{8}} = \textcolor{#e08a1e}{3\text{ m}}

3 m is comfortably above 10 cm, confirming this is a radio wave.

Worked example

X-rays: wavelength to frequency

A medical X-ray has wavelength 1 Å (1 × 10⁻¹⁰ m). Find its frequency.

f = \dfrac{c}{\lambda} = \dfrac{3\times10^{8}}{1\times10^{-10}} = \textcolor{#e08a1e}{3\times10^{18}\text{ Hz}}

3 × 10¹⁸ Hz sits between 3 × 10¹⁶ Hz and 3 × 10¹⁹ Hz, the X-ray band.

Worked example

How far apart are the extremes, really?

Compare a long-wave radio signal at 3 × 10⁵ Hz with a gamma ray at 3 × 10²⁰ Hz, a frequency ratio of 10¹⁵. Find both wavelengths and the ratio between them.

\lambda_{radio} = \dfrac{3\times10^{8}}{3\times10^{5}} = 1000\text{ m}, \qquad \lambda_{gamma} = \dfrac{3\times10^{8}}{3\times10^{20}} = 10^{-12}\text{ m}

\dfrac{\lambda_{radio}}{\lambda_{gamma}} = \dfrac{1000}{10^{-12}} = \textcolor{#e08a1e}{10^{15}}

Wavelength swings by a factor of a thousand trillion between these two waves, yet $f\lambda$ equals exactly $3\times10^8$ m/s for both of them, without exception.

8.4 The Electromagnetic Spectrum

The boundaries below are the standard textbook approximations; real transitions between bands are gradual, not sharp lines.

Radio waves (f < 3 × 10⁹ Hz): generated by accelerating charges in antennas; used for broadcasting and communication.
Microwaves (3 × 10⁹–3 × 10¹¹ Hz): produced by special vacuum tubes (klystrons, magnetrons); used in radar and microwave ovens.
Infrared (3 × 10¹¹–4 × 10¹⁴ Hz): radiated by warm objects; used in thermal imaging and remote controls.
Visible light (4 × 10¹⁴–7.5 × 10¹⁴ Hz): the narrow band human eyes detect; used for sight.
Ultraviolet (7.5 × 10¹⁴–3 × 10¹⁶ Hz): emitted by the Sun and UV lamps; used for sterilisation, also causes tanning and skin damage.
X-rays (3 × 10¹⁶–3 × 10¹⁹ Hz): produced when fast electrons are suddenly decelerated; used for medical and structural imaging.
Gamma rays (f > 3 × 10¹⁹ Hz): emitted by nuclear processes; used in cancer treatment, also hazardous due to very high photon energy.

See the interactive spectrum for the live version.

— From the NCERT exercises

Two of the chapter’s own exercise questions, with original worked solutions, the first directly exercising the energy-density derivation of §8.3.1.

NCERT Exercise 8.10

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0×10¹⁰ Hz and amplitude 48 V/m. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field.

Solution

\text{(a) } \lambda = \dfrac{c}{f} = \dfrac{3\times10^8}{2.0\times10^{10}} = \textcolor{#e08a1e}{1.5\times10^{-2}\text{ m}}

\text{(b) } B_0 = \dfrac{E_0}{c} = \dfrac{48}{3\times10^8} = \textcolor{#e08a1e}{1.6\times10^{-7}\text{ T}}

(c) This is exactly §8.3.1’s derivation: with $E=cB$ and $c=1/\sqrt{\mu_0\varepsilon_0}$ :

u_B=\dfrac{B^2}{2\mu_0}=\dfrac{(E/c)^2}{2\mu_0}=\dfrac{E^2}{2\mu_0c^2}=\dfrac{E^2}{2\mu_0/(\mu_0\varepsilon_0)}=\dfrac{\varepsilon_0E^2}{2}=\textcolor{#e08a1e}{u_E}

NCERT Exercise 8.7

The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave?

Solution

Directly from §8.3’s fixed ratio between the two amplitudes:

E_0 = cB_0 = (3\times10^8)(510\times10^{-9}) = \textcolor{#e08a1e}{153\text{ N/C}}