Electromagnetic waves — derivation from Maxwell's equations: Worked examples

Definition Symbols used on this page (defined once, so nothing is "assumed")

$E, B$ — the electric and magnetic field vectors; $E_{0}, B_{0}$ are their peak amplitudes (the biggest value the wiggle reaches).
$\hat{E}, \hat{B}, \hat{k}$ — unit vectors (length 1) giving the directions of $E$ , $B$ , and of travel.
$k$ — the wavenumber: how many radians of the wave fit into one metre, $k = 2 π / λ$ . Big $k$ = tightly packed crests.
$ω$ — the angular frequency: how many radians of oscillation happen per second, $ω = 2 π f$ . Big $ω$ = fast wiggling.
$J$ — the current density: how much electric current flows per unit area (in vacuum $J = 0$ ).
$u_{E} = \frac{1}{2} ε_{0} E^{2}$ — the electric energy density (joules stored per cubic metre by $E$ ); $u_{B} = \frac{1}{2 μ _{0}} B^{2}$ is the magnetic version.
$S$ — the Poynting vector: energy flowing per second through one square metre, $S = \frac{1}{μ _{0}} E \times B$ . Its time-average is written $⟨ S ⟩$ .

The scenario matrix

Every EM-wave problem you meet is really one of these cells. The column "Example" tells you which worked example below nails that cell.

#	Cell class	What makes it tricky	Example
C1	Amplitude conversion $B_{0} \to E_{0}$ or back	which one is "bigger", factor $c$	Ex 1
C2	Frequency ↔ wavelength, tiny/huge scales	$c = f λ$ across the spectrum	Ex 2
C3	Direction / triad geometry (all axis & sign combos of $\hat{E}, \hat{B}, \hat{k}$ )	right-hand rule, which axis is which	Ex 3 (figure)
C4	Zero / degenerate input ( $E ∥ k$ , $B_{0} = 0$ , $J = 0$ )	does a wave even exist?	Ex 4
C5	Energy density & Poynting flux (real-world power)	equal energies, unit traps	Ex 5
C6	Verify a candidate solution (dispersion $ω = c k$ )	is it actually a wave?	Ex 6
C7	Limiting behaviour ( $c$ if $μ_{0}$ or $ε_{0}$ scaled, medium slow-down)	how $c$ responds to constants	Ex 7
C8	Exam twist (given a full field, extract everything)	reading $k, ω, \hat{k}$ off an expression	Ex 8 (figure)

Two constants recur everywhere, so pin them once: $μ_{0} = 4 π \times 1 0^{- 7} T\cdotm/A = 1.2566 \times 1 0^{- 6}, ε_{0} = 8.854 \times 1 0^{- 12} F/m, c = 2.998 \times 1 0^{8} m/s .$

Ex 1 — Amplitude conversion (cell C1)

Ex 2 — Frequency ↔ wavelength across the spectrum (cell C2)

Ex 3 — The direction triad, every axis and sign (cell C3)

Figure — Electromagnetic waves — derivation from Maxwell's equations

Worked example Which way does

$B$ point — covering all axis and sign combos? Find $\hat{B}$ using the rule that $\hat{E}, \hat{B}, \hat{k}$ form a right-handed set. We walk all three coordinate axes as $\hat{k}$ , and both signs of $\hat{E}$ , so every axis/sign combination is represented: (A) $\hat{k} = + \overset{x}{^}, \hat{E} = + \overset{y}{^}$ ; (B) $\hat{k} = + \overset{x}{^}, \hat{E} = - \overset{y}{^}$ ; (C) $\hat{k} = + \overset{z}{^}, \hat{E} = + \overset{x}{^}$ ; (D) $\hat{k} = - \overset{x}{^}, \hat{E} = + \overset{y}{^}$ ; (E) $\hat{k} = + \overset{y}{^}, \hat{E} = + \overset{z}{^}$ ; (F) $\hat{k} = + \overset{z}{^}, \hat{E} = - \overset{x}{^}$ ; (G) $\hat{k} = - \overset{y}{^}, \hat{E} = + \overset{x}{^}$ .

Forecast: if you flip only the travel direction $\hat{k}$ (case A → case D), does $\hat{B}$ flip too?

Rule. Since $\hat{E}, \hat{B}, \hat{k}$ are mutually perpendicular unit vectors with $\hat{E} \times \hat{B} = \hat{k}$ , the same right-hand set gives $\hat{B} = \hat{k} \times \hat{E}$ . Why this step? Faraday's law $\nabla \times E = - \partial_{t} B$ locks $B$ perpendicular to both $E$ and $k$ ; the right-hand rule fixes the sign.
Case A ( $\hat{k} = \overset{x}{^}, \hat{E} = \overset{y}{^}$ ): $\hat{B} = \overset{x}{^} \times \overset{y}{^} = + \overset{z}{^}$ (the violet arrow in the figure). Why this step? Baseline case — plug straight into $\hat{B} = \hat{k} \times \hat{E}$ ; cross of the two "positive" unit vectors gives the third.
Case B ( $\hat{k} = \overset{x}{^}, \hat{E} = - \overset{y}{^}$ ): $\hat{B} = \overset{x}{^} \times (- \overset{y}{^}) = - \overset{z}{^}$ . Why this step? Flip $\hat{E}$ ⇒ $\hat{B}$ flips too, so the wave still travels $+ \overset{x}{^}$ — this is the sign variation on the same axis pair.
Case C (a different axis, $\hat{k} = \overset{z}{^}, \hat{E} = \overset{x}{^}$ ): $\hat{B} = \overset{z}{^} \times \overset{x}{^} = + \overset{y}{^}$ . Why this step? Covers the "which axis" confusion — the rule $\hat{B} = \hat{k} \times \hat{E}$ is identical no matter which axis plays which role; here the cyclic order $z \to x \to y$ gives $+ \overset{y}{^}$ .
Case D (flip travel, $\hat{k} = - \overset{x}{^}, \hat{E} = \overset{y}{^}$ ): $\hat{B} = (- \overset{x}{^}) \times \overset{y}{^} = - \overset{z}{^}$ . Why this step? Flipping only $\hat{k}$ (not $\hat{E}$ ) flips $\hat{B}$ — answering the forecast: yes, it flips.
Case E ( $\hat{k} = + \overset{y}{^}, \hat{E} = + \overset{z}{^}$ ): $\hat{B} = \overset{y}{^} \times \overset{z}{^} = + \overset{x}{^}$ . Why this step? Covers the last coordinate axis ( $+ \overset{y}{^}$ ) as travel direction, completing all three positive axes as $\hat{k}$ .
Case F ( $\hat{k} = + \overset{z}{^}, \hat{E} = - \overset{x}{^}$ ): $\hat{B} = \overset{z}{^} \times (- \overset{x}{^}) = - \overset{y}{^}$ . Why this step? Same $\hat{k}$ as case C but with the opposite sign of $\hat{E}$ — shows the sign flip carries through on this axis pair too.
Case G ( $\hat{k} = - \overset{y}{^}, \hat{E} = + \overset{x}{^}$ ): $\hat{B} = (- \overset{y}{^}) \times \overset{x}{^} = + \overset{z}{^}$ . Why this step? A negative axis as travel direction with a positive $\hat{E}$ — confirms negative- $\hat{k}$ cases follow the identical rule, closing the full axis/sign census.

Verify: check $\hat{E} \times \hat{B} = \hat{k}$ each time. A: $\overset{y}{^} \times \overset{z}{^} = \overset{x}{^}$ . ✓ B: $(- \overset{y}{^}) \times (- \overset{z}{^}) = \overset{x}{^}$ . ✓ C: $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ . ✓ D: $\overset{y}{^} \times (- \overset{z}{^}) = - \overset{x}{^}$ . ✓ E: $\overset{z}{^} \times \overset{x}{^} = \overset{y}{^}$ . ✓ F: $(- \overset{x}{^}) \times (- \overset{y}{^}) = \overset{z}{^}$ . ✓ G: $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ . ✓ Every coordinate axis (both signs) as $\hat{k}$ , and both signs of $\hat{E}$ , obey the same triad — this is the geometry behind Polarization.

Ex 4 — Zero and degenerate inputs (cell C4)

Worked example The three "does a wave even exist?" traps

Recall $J$ is the current density (current per unit area; zero in vacuum). Decide whether each configuration is a valid vacuum EM wave: (a) $E$ pointing along the travel direction $\hat{k}$ ; (b) a field with $E_{0} \neq = 0$ but $B_{0} = 0$ ; (c) using plain Ampère $\nabla \times B = μ_{0} J$ with $J = 0$ (drop the Displacement Current term).

Forecast: how many of the three survive?

(a) $\nabla \cdot E = 0$ gives $ik E_{0 k} = 0 \Rightarrow E_{0 k} = 0$ . Why this step? Gauss's law in vacuum forbids any field component along $\hat{k}$ . A longitudinal EM wave is impossible — it's degenerate, ruled out. Sound can be longitudinal; light cannot.
(b) Faraday for a plane wave gives $k E_{0} = ω B_{0}$ . If $E_{0} \neq = 0$ then $B_{0} = k E_{0} / ω \neq = 0$ . Why this step? You cannot have $E$ wiggling with no $B$ — a changing $E$ is a source of $B$ (Ampère–Maxwell). $B_{0} = 0$ is contradictory, not a wave.
(c) We repeat the parent note's derivation but with the crippled Ampère law. Take the curl of Faraday's law: $\nabla \times (\nabla \times E) = \nabla \times (- \frac{\partial B}{\partial t}) = - \frac{\partial}{\partial t} (\nabla \times B) .$ The left side, using the curl-of-curl identity and $\nabla \cdot E = 0$ , is $- \nabla^{2} E$ . The right side now uses the crippled law $\nabla \times B = μ_{0} J = 0$ (since $J = 0$ ), giving $- \frac{\partial}{\partial t} (0) = 0$ . Equating the two sides: $- \nabla^{2} E = 0$ , i.e. $\nabla^{2} E = 0$ — Laplace's equation, no time term, no wave. Why this step? Without the displacement current $μ_{0} ε_{0} \partial_{t} E$ , the right side has nothing left to become the $\partial_{t}^{2} E$ term. That missing term is literally what makes light.

Verify: all three degenerate cases fail, each for a different Maxwell equation: (a) Gauss, (b) Faraday, (c) Ampère–Maxwell. A genuine wave needs all four cooperating. ✓

Ex 5 — Energy density and Poynting power (cell C5, real-world)

Worked example Solar power at the top of the atmosphere

Recall (from the symbol box): $u_{E} = \frac{1}{2} ε_{0} E^{2}$ is the electric energy density (J/m³), $u_{B} = \frac{1}{2 μ _{0}} B^{2}$ the magnetic one, and $S = \frac{1}{μ _{0}} E \times B$ the Poynting vector (power per m²), with time-average $⟨ S ⟩$ . Sunlight arrives with $E_{0} = 1030 V/m$ . (a) Find $B_{0}$ . (b) Find $⟨ S ⟩ = \frac{1}{2} ε_{0} c E_{0}^{2}$ . (c) Confirm $u_{E} = u_{B}$ using amplitudes.

Forecast: will $⟨ S ⟩$ land near the famous "solar constant" of about $1360 W/m^{2}$ ?

(a) $B_{0} = E_{0} / c = 1030/2.998 \times 1 0^{8} = 3.435 \times 1 0^{- 6} T$ . Why this step? Same $E_{0} = c B_{0}$ relation as Ex 1, run backwards.
(b) $⟨ S ⟩ = \frac{1}{2} ε_{0} c E_{0}^{2} = \frac{1}{2} (8.854 \times 1 0^{- 12}) (2.998 \times 1 0^{8}) (1030)^{2} = 1408 W/m^{2}$ . Why this step? Time-averaging $S = E B / μ_{0}$ over a cycle gives the factor $\frac{1}{2}$ ; substituting $B = E / c$ and $c = 1/ μ_{0} ε_{0}$ yields the $\frac{1}{2} ε_{0} c E_{0}^{2}$ form. The result matches the real solar constant. ✓
(c) $u_{E} = \frac{1}{2} ε_{0} E_{0}^{2}$ and $u_{B} = \frac{1}{2 μ _{0}} B_{0}^{2}$ . Compute both. Why this step? The parent note claims equal energy; we test it numerically instead of trusting it.

Verify: $u_{E} = \frac{1}{2} (8.854 \times 1 0^{- 12}) (1030)^{2} = 4.696 \times 1 0^{- 6} J/m^{3}$ ; $u_{B} = \frac{1}{2 ( 1.2566 \times 1 0 ^{- 6} )} (3.435 \times 1 0^{- 6})^{2} = 4.696 \times 1 0^{- 6} J/m^{3}$ . Equal to 4 sig figs. ✓ Units of $⟨ S ⟩$ : $(F/m) (m/s) (V/m)^{2} = W/m^{2}$ . ✓

Ex 6 — Verify a candidate solution (cell C6)

Worked example Is this expression actually a wave?

Test whether $E_{y} = E_{0} cos (k x - ω t)$ solves $\frac{\partial ^{2} E}{\partial x ^{2}} = μ_{0} ε_{0} \frac{\partial ^{2} E}{\partial t ^{2}}$ , and find the condition on $ω, k$ . (Here $k$ is the wavenumber and $ω$ the angular frequency, defined in the symbol box.)

Forecast: any $ω$ and any $k$ , or only a special pair?

Space derivative twice. $\frac{\partial E}{\partial x} = - k E_{0} sin (k x - ω t)$ , then $\frac{\partial ^{2} E}{\partial x ^{2}} = - k^{2} E$ . Why this step? Two $x$ -derivatives of a cosine pull down $(- k^{2})$ and return the cosine — the shape survives.
Time derivative twice. $\frac{\partial E}{\partial t} = + ω E_{0} sin (k x - ω t)$ , then $\frac{\partial ^{2} E}{\partial t ^{2}} = - ω^{2} E$ . Why this step? Same mechanism in $t$ , giving $(- ω^{2})$ .
Impose the equation. $- k^{2} E = μ_{0} ε_{0} (- ω^{2}) E \Rightarrow k^{2} = μ_{0} ε_{0} ω^{2} \Rightarrow \frac{ω}{k} = \frac{1}{μ _{0} ε _{0}} = c$ . Why this step? The cosine cancels off both sides; what's left is the dispersion relation — it's only a wave if $ω = c k$ .

Verify: with $k = 2 π / λ$ , $ω = 2 π f$ , the condition $ω = c k$ becomes $f λ = c$ — the same rule used in Ex 2. Consistent. ✓ Numerically for $k = 1$ , $ω = c$ : $k^{2} = 1$ and $μ_{0} ε_{0} ω^{2} = μ_{0} ε_{0} c^{2} = 1$ . ✓

Ex 7 — Limiting behaviour of $c$ (cell C7)

Worked example What if the constants change?

(a) A student mistakenly doubles $ε_{0}$ (thought experiment). By what factor does $c$ change? (b) Separately, doubles $μ_{0}$ instead. By what factor now? (c) In glass with relative permittivity $ε_{r} = 2.25$ (and $μ_{r} = 1$ ), find the wave speed $v$ and the refractive index $n = c / v$ .

Forecast: does doubling either constant speed the wave up or slow it down — and does it matter which one you double?

(a) $c = \frac{1}{μ _{0} ε _{0}}$ , so $ε_{0} \to 2 ε_{0}$ gives $c \to \frac{1}{2} c$ . Why this step? $c$ depends on $ε_{0}^{- 1/2}$ ; scaling $ε_{0}$ scales $c$ by the inverse square root. Bigger $ε_{0}$ ⇒ slower wave.
(b) By the symmetric role of the two constants, $μ_{0} \to 2 μ_{0}$ also gives $c \to \frac{1}{2} c$ . Why this step? The formula $c = 1/ μ_{0} ε_{0}$ treats $μ_{0}$ and $ε_{0}$ identically — doubling either has the exact same effect. This is the edge case that shows neither constant is "special".
(c) In a medium $v = \frac{1}{μ _{0} μ _{r} ε _{0} ε _{r}} = \frac{c}{μ _{r} ε _{r}} = \frac{c}{2.25} = \frac{c}{1.5}$ . Why this step? A dielectric replaces $ε_{0} \to ε_{0} ε_{r}$ in Ampère–Maxwell; the derivation is identical but with the new constant.
Compute. $v = 2.998 \times 1 0^{8} /1.5 = 1.999 \times 1 0^{8} m/s$ , $n = c / v = 1.5$ . Why this step? $n = μ_{r} ε_{r} = 2.25 = 1.5$ — light slows and bends in glass.

Verify: (a) and (b) both give factor $\frac{1}{2} = 0.7071$ ; squaring recovers $1/ (μ_{0} \cdot 2 ε_{0})$ and $1/ (2 μ_{0} \cdot ε_{0})$ respectively — identical. ✓ (c) $n \cdot v = 1.5 \times 1.999 \times 1 0^{8} = 2.998 \times 1 0^{8} = c$ . ✓

Ex 8 — Exam twist: read everything off one expression (cell C8)

Worked example The "extract all seven quantities" question

Given (SI units) $E = (120 V/m) cos [(1.00 \times 1 0^{7} m^{- 1}) z - (3.00 \times 1 0^{15} s^{- 1}) t] \overset{x}{^}$ . Find: (1) $E_{0}$ , (2) $k$ (wavenumber), (3) $ω$ (angular frequency), (4) direction of travel, (5) wave speed, (6) $B_{0}$ , (7) direction of $B$ .

Forecast: the speed you get — will it come out as $c$ , confirming it's a vacuum wave?

Read amplitudes and shape. $E_{0} = 120 V/m$ ; $k = 1.00 \times 1 0^{7} m^{- 1}$ ; $ω = 3.00 \times 1 0^{15} s^{- 1}$ . Why this step? In $cos (k z - ω t)$ the number multiplying the position is $k$ , the number multiplying $t$ is $ω$ (both defined in the symbol box).
Direction of travel. The phase is $k z - ω t$ (a $z$ -term minus a $t$ -term) ⇒ travels along $+ \overset{z}{^}$ . Why this step? A point of constant phase needs $z$ increasing as $t$ increases, so the wave moves toward $+ z$ — the orange travel arrow in the figure.
Speed. $v = ω / k = 3.00 \times 1 0^{15} /1.00 \times 1 0^{7} = 3.00 \times 1 0^{8} m/s = c$ . Why this step? $ω / k$ is the phase velocity; getting $c$ confirms vacuum.
$B_{0}$ . $B_{0} = E_{0} / c = 120/2.998 \times 1 0^{8} = 4.003 \times 1 0^{- 7} T$ . Why this step? Amplitude relation from Faraday's law, exactly as Ex 1.
Direction of $B$ . $\hat{E} = \overset{x}{^}$ , $\hat{k} = \overset{z}{^}$ , so $\hat{B} = \hat{k} \times \hat{E} = \overset{z}{^} \times \overset{x}{^} = \overset{y}{^}$ . Why this step? Triad rule (Ex 3); $B$ oscillates along $\pm \overset{y}{^}$ , in phase with $E$ .

Verify: $v = ω / k = 3.00 \times 1 0^{8}$ matches $c$ . ✓ $\hat{E} \times \hat{B} = \overset{x}{^} \times \overset{y}{^} = \overset{z}{^} = \hat{k}$ . ✓ $E_{0} / B_{0} = 120/ (4.003 \times 1 0^{- 7}) = 2.998 \times 1 0^{8} = c$ . ✓

Recall Rapid self-test (reveal after answering)

Which Maxwell equation forbids a longitudinal EM wave? ::: Gauss's law $\nabla \cdot E = 0$ (kills the component along $\hat{k}$ ). If $B_{0} = 5 \times 1 0^{- 8}$ T, what is $E_{0}$ ? ::: $E_{0} = c B_{0} = 15.0 V/m$ . A wave has $\hat{E} = + \overset{y}{^}$ , $\hat{k} = + \overset{z}{^}$ ; which way is $B$ ? ::: $\hat{B} = \hat{k} \times \hat{E} = \overset{z}{^} \times \overset{y}{^} = - \overset{x}{^}$ . In glass with $n = 1.5$ , how fast does light go? ::: $v = c /1.5 \approx 2.00 \times 1 0^{8} m/s$ . Removing the displacement current does what to the wave equation? ::: Collapses it to $\nabla^{2} E = 0$ (Laplace's equation — no wave).