How EM waves are produced
The first thing to settle is what a source of electromagnetic radiation is not. A stationary charge sets up only an electrostatic field that does not change with time. A charge in uniform motion — a steady current — sets up a magnetic field, but that field, too, is constant in time. Neither situation produces a travelling wave.
It is a central result of Maxwell's theory that accelerated charges radiate electromagnetic waves. NCERT presents the standard example: a charge oscillating with some frequency. An oscillating charge is one kind of accelerating charge, and the frequency of the wave it sends out equals the frequency of oscillation of the charge. The energy that the wave carries away is supplied at the expense of the energy of the accelerated charge.
| State of the charge | Field produced | Radiates a wave? |
|---|---|---|
| Stationary charge | Static electrostatic field | No |
| Uniform motion (steady current) | Steady magnetic field | No |
| Accelerating / oscillating charge | Time-varying $\mathbf{E}$ and $\mathbf{B}$ | Yes |
Historically, this prediction could not first be tested with visible light, because the frequency of yellow light is about $6 \times 10^{14}$ Hz while the highest frequency reachable with electronic circuits is only around $10^{11}$ Hz. The experimental confirmation therefore came in the radio-wave region, in Hertz's experiment of 1887. Jagdish Chandra Bose, working in Calcutta, later produced waves of much shorter wavelength (25 mm to 5 mm), and Marconi transmitted such waves over many kilometres — the beginning of communication by electromagnetic waves.
The self-regenerating field
The mechanism that lets the wave sustain itself away from the source follows from the symmetry of Maxwell's equations. A time-varying electric field acts as a source of a magnetic field (the Ampère–Maxwell law, through the displacement current), and a time-varying magnetic field acts as a source of an electric field (Faraday's law). The oscillating fields therefore regenerate each other as the disturbance moves outward — the wave needs no charges or currents once it has left the source.
Each field, when it changes with time, generates the other — so the pair propagates as a self-sustaining wave (NCERT §8.3.1–8.3.2).
Transverse nature: E, B and direction
Maxwell's equations show that in an electromagnetic wave the electric and magnetic fields are perpendicular to each other and to the direction of propagation. NCERT motivates this from the displacement-current picture of a capacitor: $\mathbf{E}$ is perpendicular to the plates while the magnetic field it generates runs along a circle parallel to the plates, so $\mathbf{E} \perp \mathbf{B}$. This is a general feature of the wave.
For a plane wave travelling along $z$, NCERT writes the fields as
$$E_x = E_0 \sin(kz - \omega t), \qquad B_y = B_0 \sin(kz - \omega t)$$so the electric field oscillates along $x$, the magnetic field along $y$, and the wave moves along $z$. The two fields are in phase — they reach their peaks together. The propagation direction is fixed by the right-hand sense of $\mathbf{E} \times \mathbf{B}$.
Linearly polarised plane wave after NCERT Fig. 8.3: $E$ along $x$, $B$ along $y$, both transverse to the $z$-direction of travel.
"E is the dominant field because $E_0 \gg B_0$"
Since $B_0 = E_0/c$ and $c \approx 3\times10^{8}\ \text{m s}^{-1}$, the magnetic amplitude is numerically about $10^{8}$ times smaller than the electric amplitude. Students read this as "the electric field is stronger" and conclude it carries more energy. It does not. The two fields carry equal average energy density; the numbers differ only because $E$ and $B$ are measured in different units.
$B_0 = E_0/c$ makes $B_0$ tiny in tesla, but $u_E = u_B$ on average. Also remember: only accelerating charges radiate, and EM waves need no medium.
The speed relations
Substituting the plane-wave forms into Maxwell's equations gives the dispersion relation $\omega = ck$, with the speed in vacuum fixed by the two constants of free space:
$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 3 \times 10^{8}\ \text{m s}^{-1}$$The same analysis ties the field amplitudes together. The magnitudes of the electric and magnetic fields are related by
$$c = \frac{E_0}{B_0} \qquad \Longleftrightarrow \qquad B_0 = \frac{E_0}{c}$$Experiments on waves of widely different wavelengths confirm that this vacuum speed is the same to within a few metres per second, which is why $c$ is now used to define the standard of length.
A plane electromagnetic wave of frequency 25 MHz travels in free space along the $x$-direction. At a point, $\mathbf{E} = 6.3\,\hat{\jmath}\ \text{V/m}$. Find $\mathbf{B}$ at this point.
Magnitude: $B = E/c = \dfrac{6.3}{3\times10^{8}} = 2.1\times10^{-8}\ \text{T}$.
Direction: the wave travels along $x$ and $\mathbf{E}$ is along $y$, so $\mathbf{B}$ must be perpendicular to both. Since $\hat{\jmath}\times\hat{k} = \hat{\imath}$, $\mathbf{B}$ is along $z$: $\mathbf{B} = 2.1\times10^{-8}\,\hat{k}\ \text{T}$.
The "$\partial\mathbf{E}/\partial t$ produces $\mathbf{B}$" half of the regeneration loop comes from Displacement Current — read it first to see why a changing electric field has the same magnetic effect as a real current.
Speed inside a medium
Electromagnetic waves are self-sustaining oscillations of $\mathbf{E}$ and $\mathbf{B}$ in free space, and unlike every mechanical wave studied earlier they involve no material medium for the vibration of the fields. They do, however, travel through matter — light passes through glass. Inside a medium of permittivity $\varepsilon$ and permeability $\mu$, those constants replace $\varepsilon_0$ and $\mu_0$, and the speed becomes
$$v = \frac{1}{\sqrt{\mu\varepsilon}}$$Writing $\varepsilon = \varepsilon_0 \varepsilon_r$ and $\mu = \mu_0 \mu_r$ casts this in terms of the vacuum speed and the relative constants of the medium:
$$v = \frac{1}{\sqrt{\mu_0\varepsilon_0}\,\sqrt{\mu_r\varepsilon_r}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$| Quantity | Relation | NCERT reference |
|---|---|---|
| Vacuum speed | c = 1/√(μ0ε0) | Eq. 8.9(a) |
| Amplitude relation | c = E0/B0 | Eq. 8.10 |
| Wave relation | c = νλ | Eq. 8.9(b) |
| Speed in medium | v = 1/√(με) = c/√(μrεr) | Eq. 8.11 |
Because $\mu_r \varepsilon_r \ge 1$ for ordinary matter, $v$ is always less than $c$. The refractive index of one medium relative to another is the ratio of the wave speeds in the two media — the bridge to the optics chapter.
Energy carried by the wave
The technological importance of electromagnetic waves comes from their ability to carry energy from one place to another: radio and television signals carry energy, and light carries energy from the Sun to the Earth, making life possible. The energy is stored jointly in the electric and magnetic fields, and a key result — set as NCERT Exercise 8.10(c) — is that the average energy density of the electric field equals the average energy density of the magnetic field.
Show that the electric and magnetic contributions to the energy density of a plane EM wave are equal, given $E_0 = c B_0$ and $c^2 = 1/(\mu_0\varepsilon_0)$.
Electric energy density: $u_E = \tfrac{1}{2}\varepsilon_0 E^2$. Magnetic energy density: $u_B = \dfrac{B^2}{2\mu_0}$.
Put $E = cB$: $u_E = \tfrac{1}{2}\varepsilon_0 c^2 B^2 = \tfrac{1}{2}\varepsilon_0 \cdot \dfrac{1}{\mu_0\varepsilon_0}\, B^2 = \dfrac{B^2}{2\mu_0} = u_B.$
So the wave shares its energy equally between the two fields, and the contributions to its intensity are in the ratio $1:1$.
This is precisely why the very different sizes of $E_0$ and $B_0$ do not mean an imbalance of energy: the factor $c$ that shrinks $B_0$ relative to $E_0$ is exactly the factor that, on squaring, restores the balance in the energy expressions.
Momentum and radiation pressure
Because the wave carries energy, it also carries momentum. When electromagnetic energy is absorbed by a surface, momentum is delivered to that surface, and the steady arrival of momentum is felt as a force per unit area — a radiation pressure. This is the same mechanism by which sunlight exerts a small outward push on dust grains and on the sails of proposed solar-sail spacecraft.
For a total energy $U$ carried in the direction of propagation, the momentum delivered to a fully absorbing surface is
$$p = \frac{U}{c}$$Dividing by $c$ again makes the momentum small for everyday light, which is why radiation pressure is negligible in daily life yet measurable with sensitive apparatus. A perfectly reflecting surface returns the wave and therefore receives up to twice this momentum.
The magnetic field of a plane EM wave is $B_y = (2\times10^{-7})\sin(0.5\times10^{3}x + 1.5\times10^{11}t)\ \text{T}$. Write the electric field.
Amplitude: $E_0 = B_0 c = 2\times10^{-7} \times 3\times10^{8} = 60\ \text{V/m}$.
$E$ is perpendicular to both the propagation direction and $\mathbf{B}$, so it lies along $z$: $E_z = 60\sin(0.5\times10^{3}x + 1.5\times10^{11}t)\ \text{V/m}$ — in phase with $B_y$.
Source and nature in one screen
- Only accelerating (oscillating) charges radiate; stationary charges and steady currents do not.
- A changing $\mathbf{E}$ makes $\mathbf{B}$ and a changing $\mathbf{B}$ makes $\mathbf{E}$ — the fields regenerate each other.
- The wave is transverse: $\mathbf{E} \perp \mathbf{B} \perp$ propagation; direction is along $\mathbf{E}\times\mathbf{B}$; $E$ and $B$ are in phase.
- $c = 1/\sqrt{\mu_0\varepsilon_0}$, $c = E_0/B_0$, $c = \nu\lambda$; in a medium $v = 1/\sqrt{\mu\varepsilon} = c/\sqrt{\mu_r\varepsilon_r} < c$.
- No medium needed; average $u_E = u_B$ (intensity ratio $1:1$); the wave carries momentum $p = U/c$ giving radiation pressure.