Electromagnetic Waves

This article follows the chapter's own two-part structure. The first half fixes Ampere's circuital law: the contradiction that arises when charging a capacitor, the displacement current that resolves it, the corrected Ampere-Maxwell law, and all four of Maxwell's equations in vacuum, which with the Lorentz force law complete classical electromagnetism. The second half turns to electromagnetic waves themselves — why only accelerated charges radiate, why the fields are transverse and locked together by $B_0 = E_0/c$, how the wave speed $c = 1/\sqrt{\mu_0\varepsilon_0}$ emerges from the fields' own equations, and how a medium slows the wave once $\mu_0, \varepsilon_0$ are replaced by $\mu, \varepsilon$. The chapter closes with the electromagnetic spectrum — the seven named bands from radio waves to gamma rays, their production, detection and everyday uses. Three derivations make the displacement current, the wave speed, and the E-B energy-sharing rigorous; three worked examples apply all of it numerically, in the style of NCERT's own Examples 8.1, 8.2 and its end-of-chapter exercises.

Ampere's circuital law runs into a contradiction

Ampere's circuital law from Chapter 4 states that the line integral of the magnetic field around any closed loop equals $\mu_0$ times the current threading that loop:

where: the integral is taken around a closed path, and $i(t)$ is the current passing through any surface bounded by that path. Ordinarily it does not matter which surface you pick — a flat disc or a bulging balloon-shaped surface with the same rim give the same threading current, because charge is conserved and current is continuous along a wire. Now consider a parallel-plate capacitor being charged by a time-dependent current $i(t)$. Draw a circular loop of radius $r$ around the connecting wire, well outside the plates, to find the magnetic field at a point $P$ on that loop. A flat disc bounded by this loop is pierced by the wire, so Equation (1) gives $B(2\pi r) = \mu_0 i(t)$. But bulge the same-rimmed surface into a pot-like or tiffin-box shape that dips between the plates without touching the wire: no conduction current pierces it at all, so Ampere's law now predicts $B = 0$ at the very same point $P$. Same loop, same physical setup, two different answers, depending purely on an arbitrary choice of surface. Something is missing from Equation (1), and it must be found between the plates, the only place the two calculations disagree.

The displacement current

The missing ingredient is the electric field between the plates. If the plates have area $A$ and instantaneous charge $Q$, the field between them is $E = Q/(\varepsilon_0 A)$ (uniform, perpendicular to the plates), so the electric flux through the bulging surface is

As the capacitor charges, $Q$ changes with time at the rate $i = dQ/dt$, so differentiating Equation (2) gives $\varepsilon_0\, d\Phi_E/dt = dQ/dt = i$. This changing electric flux, multiplied by $\varepsilon_0$, has exactly the units and the value of a current, and Maxwell proposed that it acts as a genuine source of magnetic field in its own right. He named it the displacement current:

where: $\Phi_E$ is the electric flux through the surface in question and $\varepsilon_0 = 8.854\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$ is the permittivity of free space. Outside the plates there is only conduction current ($i_c = i$, $i_d = 0$); between the plates it is the reverse ($i_c = 0$, $i_d = i$). Every surface bounded by the loop, however it bulges, now carries the same total current $i_c + i_d$, and the contradiction disappears.

The Ampere-Maxwell law and Maxwell's equations in vacuum

Generalising Equation (1) to include displacement current gives the corrected law, valid for any surface bounded by the loop:

where: $i_c$ is the conduction current (moving charges) and the second term is $\mu_0$ times the displacement current of Equation (3). This is the Ampere-Maxwell law: displacement current has exactly the same physical effect as conduction current, sourcing a magnetic field circulating around it by the right-hand rule, and it can exist entirely alone with no conduction current nearby — precisely the case between capacitor plates, and, as the next section shows, in the empty space an electromagnetic wave travels through. Maxwell's achievement was to collect four such equations, which together with the Lorentz force law $\vec F = q\vec E + q\vec v \times \vec B$ express the entire classical theory of electricity and magnetism:

where: the second equation states that magnetic field lines never begin or end, since isolated magnetic monopoles analogous to electric charge have never been found — the one asymmetry left in an otherwise symmetric set of four laws. The third and fourth equations together state the symmetry plainly: a changing magnetic field produces an electric field (Faraday), and, Maxwell's addition, a changing electric field produces a magnetic field (Ampere-Maxwell). Time-varying $E$ and $B$ thus continuously regenerate each other — the mechanism behind every electromagnetic wave.

Sources of electromagnetic waves: only accelerated charges radiate

Neither a stationary charge nor a charge in uniform motion (a steady current) radiates an electromagnetic wave. A stationary charge produces only a static electric field; a steady current produces a magnetic field that does not change with time, and an unchanging field cannot regenerate anything through Faraday's or the Ampere-Maxwell law. Only when a charge accelerates — speeds up, slows down, or changes direction — does its field change with time in a way that keeps propagating outward. The simplest case is a charge oscillating harmonically at frequency $\nu$: it produces an oscillating electric field, which by the Ampere-Maxwell law produces an oscillating magnetic field, which by Faraday's law produces an oscillating electric field further out, and so on, the two fields regenerating each other as the disturbance spreads at speed $c$. The radiated frequency exactly equals the source's oscillation frequency, and the wave's energy is drawn from the accelerating charge's own energy — an oscillating electric dipole is the most basic radiating source, and the working principle behind every transmitting antenna. Demonstrating this with an oscillating circuit tuned to visible light is hopeless in practice: yellow light oscillates at about $6\times10^{14}\ \text{Hz}$, while even modern circuits struggle past $10^{11}\ \text{Hz}$. Heinrich Hertz instead worked in the radio-wave region in 1887, generating and detecting metre-wavelength electromagnetic waves whose reflection, refraction and speed matched Maxwell's predictions exactly — the first experimental proof of Maxwell's theory. J. C. Bose, in Calcutta seven years later, produced still shorter waves (25 mm to 5 mm), and Marconi transmitted electromagnetic waves over many kilometres, founding wireless communication.

Nature of electromagnetic waves: transverse and mutually perpendicular

For a plane electromagnetic wave travelling along the $z$-direction, the electric field oscillates along $x$ and the magnetic field oscillates along $y$, both varying sinusoidally with $z$ at any fixed instant:

where: $E_0$ and $B_0$ are the field amplitudes, $k = 2\pi/\lambda$ is the magnitude of the wave vector (its direction gives the direction of propagation), $\lambda$ is the wavelength, and $\omega$ is the angular frequency; the speed of propagation is $\omega/k$. The essential geometric fact — already visible in the capacitor picture, where the E field between the plates is perpendicular to the plates and the displacement-current B field circulates parallel to them — is that $\vec E$, $\vec B$ and the propagation direction are mutually perpendicular at every point: the wave is transverse, propagating along $\vec E \times \vec B$. Substituting Equation (5) into Maxwell's equations gives two further results. First, the wave speed is fixed by the electric and magnetic constants alone:

where: $\mu_0 = 4\pi\times10^{-7}\ \text{T m/A}$ is the permeability of free space. Written in terms of frequency $\nu = \omega/2\pi$ and wavelength $\lambda = 2\pi/k$, this is the familiar wave relation $\nu\lambda = c$. Second, the two field amplitudes are locked together, not independent:

Electromagnetic waves are self-sustaining oscillations of $\vec E$ and $\vec B$ that need no material medium at all — unlike sound or water waves, there is nothing physically vibrating in the usual sense, just the fields themselves. If a medium of permittivity $\varepsilon$ and permeability $\mu$ is actually present, these constants replace $\varepsilon_0$ and $\mu_0$ and the wave slows to

which is why light travels slower in glass or water than in vacuum, and is the physical basis of refractive index (Chapter 9). The vacuum speed $c$ is measured to be independent of wavelength to within a few metres per second — a constancy so well established that $c$ is now used to define the metre.

The electromagnetic spectrum

The classification of electromagnetic waves by frequency (equivalently wavelength) is the electromagnetic spectrum. There is no sharp cut-off between neighbouring bands; the names reflect mainly how the waves are produced and detected, not any fundamental physical difference — every band obeys exactly the same Equations (5) to (8).

A few details are worth holding onto. Infrared waves are called heat waves because water and $\text{CO}_2$ molecules absorb them readily, raising internal (thermal) energy — the mechanism behind the greenhouse effect and physiotherapy lamps. Ultraviolet in excess causes skin tanning and sunburn and is largely blocked by the ozone layer, so its depletion by chlorofluorocarbons is an international concern; welders wear UV-blocking goggles for this reason. X-rays, made by decelerating high-energy electrons at a metal target, damage living tissue, so medical use is carefully dose-controlled.

Energy carried by an electromagnetic wave

Electromagnetic waves matter technologically above all because they carry energy — radio and TV signals, and, most vitally, sunlight carrying energy from the Sun to the Earth. The instantaneous energy density in the electric field alone is $u_E = \tfrac{1}{2}\varepsilon_0 E^2$, and in the magnetic field alone $u_B = B^2/2\mu_0$, exactly as in Chapters 2 and 6. Because $B_0 = E_0/c$, the two are equal on time-average (proved below), so each field carries exactly half the wave's total energy — which also gives the wave's intensity (average energy per unit area per unit time), $I = \tfrac{1}{2}\varepsilon_0 c E_0^2$. Beyond this chapter, an electromagnetic wave also carries momentum, $p = U/c$ for absorbed energy $U$, which is why sunlight exerts a tiny but measurable radiation pressure.

Board level - electric field to magnetic field in a plane wave

A plane electromagnetic wave of frequency $25\ \text{MHz}$ travels in free space along the $x$-direction. At a particular point in space and time, $\vec E = 6.3\, \hat{j}\ \text{V/m}$. What is $\vec B$ at this point, and what is the wavelength of the wave?

Step 1. Find the magnitude of $B$ using $B_0 = E_0/c$: $B = \dfrac{6.3\ \text{V/m}}{3\times10^8\ \text{m/s}} = 2.1\times10^{-8}\ \text{T}$.

Step 2. Find the direction of $\vec B$. The wave propagates along $+x$, and $\vec E$ is along $+y$ ($\hat{j}$). Since $\vec E$, $\vec B$ and the propagation direction are mutually perpendicular with propagation along $\vec E\times\vec B$, and $\hat{j}\times\hat{k} = \hat{i}$, the magnetic field must point along $\hat{k}$, the $z$-direction.

Step 3. Find the wavelength from $c = \nu\lambda$: $\lambda = c/\nu = (3\times10^8)/(25\times10^6) = 12\ \text{m}$. Answer: $\vec B = 2.1\times10^{-8}\, \hat{k}\ \text{T}$, and the wavelength is $12\ \text{m}$.

Board level - full wave description from the electric field amplitude

The electric field amplitude of an electromagnetic wave in vacuum is $E_0 = 120\ \text{N/C}$ and its frequency is $\nu = 50.0\ \text{MHz}$. The wave travels along the $x$-direction with $E$ along $y$ and $B$ along $z$. Determine $B_0$, $\omega$, $k$ and $\lambda$, and write expressions for $E$ and $B$.

Step 1. Field amplitude: $B_0 = E_0/c = 120/(3\times10^8) = 4.0\times10^{-7}\ \text{T}$.

Step 2. Angular frequency: $\omega = 2\pi\nu = 2\pi\times(5.0\times10^7) = 3.14\times10^8\ \text{rad/s}$.

Step 3. Wavelength and wave number: $\lambda = c/\nu = (3\times10^8)/(5.0\times10^7) = 6.0\ \text{m}$, so $k = 2\pi/\lambda = 2\pi/6.0 = 1.05\ \text{rad/m}$ (a quick check: $\omega/k = 3.14\times10^8/1.05 = 3.0\times10^8\ \text{m/s} = c$, confirming consistency).

Step 4. Assemble the expressions, taking propagation along $x$ this time (so $kz-\omega t$ becomes $kx - \omega t$): $E_y = 120\sin(1.05x - 3.14\times10^8 t)\ \text{V/m}$ and $B_z = 4.0\times10^{-7}\sin(1.05x - 3.14\times10^8 t)\ \text{T}$. Answer: $B_0 = 4.0\times10^{-7}\ \text{T}$, $\omega = 3.14\times10^8\ \text{rad/s}$, $k = 1.05\ \text{rad/m}$, $\lambda = 6.0\ \text{m}$, with $E$ and $B$ as written above.

Competitive level (JEE / NEET) - displacement current and the magnetic field inside a charging capacitor

A parallel-plate capacitor made of circular plates, each of radius $R = 6.0\ \text{cm}$, has capacitance $C = 100\ \text{pF}$. It is connected to a $230\ \text{V}$ AC supply of angular frequency $\omega = 300\ \text{rad/s}$. Find (a) the rms conduction current, (b) whether it equals the displacement current, and (c) the amplitude of $B$ at a point $r = 3.0\ \text{cm}$ from the axis, between the plates.

Step 1. Find the rms conduction current using the capacitive relation $I_{\text{rms}} = V_{\text{rms}}\,\omega C$: $I_{\text{rms}} = 230 \times 300 \times (100\times10^{-12}) = 6.9\times10^{-6}\ \text{A} = 6.9\ \mu\text{A}$.

Step 2. For this circuit — a capacitor in series with the source, nothing else — the conduction current in the wires and the displacement current between the plates must be equal at every instant, since Kirchhoff's junction rule is only rescued by treating $i_c$ outside and $i_d$ between the plates as segments of one continuous total current. So yes, $i_d = i_c$ always, including rms and peak values.

Step 3. Find the peak (amplitude) current, since the requested $B$ is an amplitude, not an rms value: $i_0 = \sqrt{2}\, I_{\text{rms}} = 1.414 \times 6.9\times10^{-6} = 9.76\times10^{-6}\ \text{A}$.

Step 4. Apply the generalised Ampere's law to a circular loop of radius $r = 3.0\ \text{cm}$, which lies inside the plate radius $R = 6.0\ \text{cm}$. Because the displacement current is spread uniformly over the plate area (uniform $E$ field between the plates), the current enclosed by a loop of radius $r < R$ is only the fraction of the total that lies within that smaller circle, $i_{\text{enc}} = i_0\,(r^2/R^2)$. Ampere's law then gives $B(2\pi r) = \mu_0\, i_0\,(r^2/R^2)$, so $B = \dfrac{\mu_0\, i_0\, r}{2\pi R^2} = \dfrac{(4\pi\times10^{-7})(9.76\times10^{-6})(0.03)}{2\pi (0.06)^2} = 1.63\times10^{-11}\ \text{T}$. Answer: $I_{\text{rms}} = 6.9\ \mu\text{A}$; yes, $i_d = i_c$ at every instant; and $B \approx 1.63\times10^{-11}\ \text{T}$ at $r = 3.0\ \text{cm}$. Timing note: about three to four minutes. The two traps that cost the most marks here are, first, quietly using $I_{\text{rms}}$ where the question actually wants the amplitude $i_0 = \sqrt{2}I_{\text{rms}}$, and second, forgetting that for $r < R$ the enclosed current scales as $r^2/R^2$ (area fraction), not simply $i_0$ itself — that mistake would overestimate $B$ by a factor of $R^2/r^2 = 4$ here. Always check whether the field point lies inside or outside the plate radius before choosing the enclosed-current formula.

Treating displacement current as a flow of charge across the gap. No charge crosses between the plates of a capacitor. Displacement current is the name given to $\varepsilon_0\, d\Phi_E/dt$ because it has the same magnetic effect as a real current, not because charge is moving through the gap.

Forgetting that $i_d$ requires a changing electric field, not just a strong one. A capacitor holding a static charge, however large, has $E \ne 0$ but $d\Phi_E/dt = 0$, so $i_d = 0$ — only while charging or discharging does displacement current exist.

Assuming electromagnetic waves need a medium. Unlike sound (needs air) or a wave on a string (needs the string), electromagnetic waves are self-sustaining $E$-$B$ oscillations that propagate through vacuum; a medium, if present, only slows the wave via Equation (8).

Mixing up amplitude and rms in AC-driven displacement-current problems, exactly the trap in the competitive worked example above: always convert consistently, $i_0 = \sqrt{2}\,I_{\text{rms}}$.

Applying $i_{\text{enc}} = i_0(r^2/R^2)$ outside the plate radius. That formula only holds inside the plates ($r < R$); for $r \ge R$ the entire displacement current is enclosed and $B = \mu_0 i_0/(2\pi r)$, the long-straight-wire form.

Believing all electromagnetic waves are fundamentally different phenomena. Radio waves, visible light and gamma rays obey the very same Equations (5) to (8); only their wavelength, and hence how they are produced and detected, differs.

• State why Ampere's circuital law, in its original form, gives inconsistent results for a charging capacitor, and identify where the missing term comes from.
• Write down the displacement current $i_d = \varepsilon_0\, d\Phi_E/dt$ and the Ampere-Maxwell law, explaining why $i_d$ has the same magnetic effect as conduction current without being a flow of charge.
• List all four of Maxwell's equations in vacuum and state, in one sentence each, what each one says.
• Explain why only accelerated charges radiate, and why the radiated frequency equals the source's oscillation frequency.
• State that electromagnetic waves are transverse, with $\vec E$, $\vec B$ and the propagation direction mutually perpendicular, and find that direction from $\vec E \times \vec B$.
• Use $c = 1/\sqrt{\mu_0\varepsilon_0}$, $\nu\lambda = c$ and $B_0 = E_0/c$ fluently, finding any missing field, wavelength, frequency, $k$ or $\omega$ from two given quantities.
• Recall the seven named bands of the electromagnetic spectrum in order, with an approximate wavelength range, one production method and one use for each.
• Reproduce the logic proving that the time-averaged energy densities of the $E$ and $B$ fields in a plane electromagnetic wave are equal.
• Solve a displacement-current-inside-a-capacitor problem, distinguishing rms from amplitude values and applying $r^2/R^2$ correctly for field points inside the plate radius.

Ask the tutor: if you would like to work through the direction-finding logic for a wave propagating along an arbitrary axis, or practice more displacement-current-in-a-capacitor numericals step by step, just ask.