Poynting vector — energy flux in EM waves: Worked examples

The scenario matrix

Every problem this topic throws at you lands in one of these boxes. We work at least one example per row.

Cell	What varies	Example
A — measured intensity → amplitude	You know $I$ , want $E_{0}$ , $B_{0}$	Ex 1
B — amplitude → intensity & power	You know $E_{0}$ (or $B_{0}$ ), want $I$ , $P$	Ex 2
C — geometry / direction of $S$	angle between $d A$ and $S$ not zero	Ex 3 (figure)
D — degenerate: $E ∥ B$	cross product $\to 0$ , no energy flow	Ex 4 (figure)
E — limiting / instantaneous vs average	$S (t)$ vs $⟨ S ⟩$ , and $S = 0$ moments	Ex 5
F — real-world word problem	sun, distance, inverse-square	Ex 6
G — radiation pressure twist	absorb vs reflect, force on area	Ex 7
H — exam trap: the resistor / static fields	$S$ where you don't expect a wave	Ex 8 (figure)

Before each example, Forecast: — pause and guess the answer's size or sign. Guessing first is how the number sticks.

The constants we reuse (memorise these three):

Cell A — measured intensity → field amplitude

Worked example Ex 1. A radio station's field at your window

A radio transmitter produces intensity $I = 0.020 W/m^{2}$ at your house. Find the electric-field amplitude $E_{0}$ and the magnetic-field amplitude $B_{0}$ .

Forecast: $E_{0}$ in volts-per-metre — will it be a few, a few dozen, or thousands? (It's smaller than sunlight's $\sim 1000$ V/m because $I$ is much smaller.)

Start from the intensity formula $I = \frac{1}{2} ε_{0} c E_{0}^{2}$ . Why this step? Intensity is the time average of $S$ ; the parent note derived $⟨ cos^{2} ⟩ = \frac{1}{2}$ , so this is the bridge between the thing we measure ( $I$ ) and the thing we want ( $E_{0}$ ).
Invert for $E_{0}$ : $E_{0} = \frac{2 I}{ε _{0} c}$ . Why this step? $E_{0}$ is buried inside a square, so we undo it with a square root.
Plug in numbers: $E_{0} = \frac{2 ( 0.020 )}{( 8.85 \times 1 0 ^{- 12} ) ( 3.0 \times 1 0 ^{8} )} \approx 3.88 V/m$
Get $B_{0}$ from the wave lock $E_{0} = c B_{0}$ : $B_{0} = E_{0} / c = 3.88/ (3.0 \times 1 0^{8}) \approx 1.29 \times 1 0^{- 8} T$ . Why this step? In a plane wave $E$ and $B$ are not independent — the parent note showed $E = c B$ at every instant.

Verify: rebuild the intensity from $E_{0}$ : $\frac{1}{2} (8.85 \times 1 0^{- 12}) (3 \times 1 0^{8}) (3.88)^{2} \approx 0.020 W/m^{2}$ . ✓ Units: $(W/m^{2}) / (F/m \cdot m/s)$ collapses to V/m (trust the algebra; the check confirms it).

Cell B — amplitude → intensity and power

Worked example Ex 2. From magnetic amplitude to beam power

A laser beam has magnetic amplitude $B_{0} = 1.0 \times 1 0^{- 6} T$ and cross-sectional area $A = 3.0 mm^{2}$ . Find the intensity $I$ and the beam power $P$ .

Forecast: $B_{0} \sim 1 0^{- 6}$ T is comparable to sunlight's $B_{0}$ , so expect $I$ of order $1 0^{2}$ – $1 0^{3} W/m^{2}$ , and a milliwatt-scale power because the area is tiny.

Pick the $B_{0}$ form of intensity: $I = \frac{c B _{0}^{2}}{2 μ _{0}}$ . Why this step? We were handed $B_{0}$ , not $E_{0}$ . Using the $B$ -version avoids converting first — fewer chances to slip.
Plug in: $I = \frac{( 3.0 \times 1 0 ^{8} ) ( 1.0 \times 1 0 ^{- 6} ) ^{2}}{2 ( 4 π \times 1 0 ^{- 7} )} \approx 119 W/m^{2}$ .
Convert area: $A = 3.0 mm^{2} = 3.0 \times 1 0^{- 6} m^{2}$ . Why this step? $I$ is per square metre; mm² would give an answer $1 0^{6}$ times wrong.
Power = intensity × area (beam $∥$ area, so $S \cdot d A = S d A$ ): $P = I A = 119 \times 3.0 \times 1 0^{- 6} \approx 3.6 \times 1 0^{- 4} W \approx 0.36 mW$ .

Verify: cross-check via $E_{0} = c B_{0} = 300 V/m$ , then $I = \frac{1}{2} ε_{0} c E_{0}^{2} = \frac{1}{2} (8.85 \times 1 0^{- 12}) (3 \times 1 0^{8}) (300)^{2} \approx 119 W/m^{2}$ . Same number two ways. ✓

Cell C — geometry: $S$ tilted to the surface

Worked example Ex 3. A wave hitting a tilted panel

A wave carries $⟨ S ⟩ = 800 W/m^{2}$ travelling horizontally. A flat solar panel of area $A = 2.0 m^{2}$ is tilted so its normal makes an angle $θ = 6 0^{\circ}$ with the beam. What power does it collect?

Forecast: tilted away from face-on, so it should collect less than the face-on $800 \times 2 = 1600$ W. At $6 0^{\circ}$ we expect roughly half.

Write the flux as a dot product: $P = \oint S \cdot d A = S A cos θ$ . Why this step? Only the component of $S$ along the surface normal delivers energy through the surface. The dot product extracts exactly that component — look at the red arrow in the figure projecting onto the green normal.
Why cosine and not sine? Because $θ$ is measured from the normal; face-on ( $θ = 0$ ) should give the maximum, and $cos 0 = 1$ does. If we'd used sine, face-on would give zero — wrong.
Plug in: $P = 800 \times 2.0 \times cos 6 0^{\circ} = 800 \times 2.0 \times 0.5 = 800 W$ .

Verify: exactly half of the face-on $1600$ W, matching the forecast. Limiting checks: $θ = 0^{\circ} \Rightarrow 1600$ W (full), $θ = 9 0^{\circ} \Rightarrow 0$ W (edge-on, beam grazes past). ✓

Cell D — degenerate: parallel fields, zero flow

Worked example Ex 4. Crossed vs parallel fields

In region 1, $E$ points $+ \overset{x}{^}$ and $B$ points $+ \overset{y}{^}$ , both with magnitude giving $S = 50 W/m^{2}$ when perpendicular. In region 2, someone claims $E$ and $B$ both point $+ \overset{x}{^}$ (parallel). What is $S$ in each region?

Forecast: region 1 → energy flows along $+ \overset{z}{^}$ . Region 2 → the cross product of parallel vectors is zero, so no energy flow at all.

Region 1: $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ , so $S$ points $+ \overset{z}{^}$ with magnitude $50 W/m^{2}$ . Why this step? The right-hand rule: fingers along $E$ ( $\overset{x}{^}$ ), curl toward $B$ ( $\overset{y}{^}$ ), thumb points $\overset{z}{^}$ — that's the energy direction.
Region 2: $E \times B$ with both along $\overset{x}{^}$ : $\overset{x}{^} \times \overset{x}{^} = 0$ . Why this step? The cross-product magnitude is $E B sin ϕ$ where $ϕ$ is the angle between them; $ϕ = 0 \Rightarrow sin 0 = 0$ . This is exactly why the parent note insisted on a cross product — parallel fields carry no propagating energy, and only the cross product knows that.
General reminder: $∣ S ∣ = \frac{1}{μ _{0}} E B sin ϕ$ . It maxes when $ϕ = 9 0^{\circ}$ (a real wave) and vanishes when $ϕ = 0^{\circ}$ or $18 0^{\circ}$ .

Verify: region 2 magnitude $\propto sin 0 = 0$ . ✓ A dot product $E \cdot B$ would give $E B cos 0 \neq = 0$ here — that's precisely the mistake the parent warns against.

Cell E — instantaneous vs average, and the zero moments

Worked example Ex 5. When is

$S$ zero even in a bright wave? A wave has $E (t) = E_{0} cos (ω t)$ with $E_{0} = 100 V/m$ . Find (a) the peak instantaneous $S$ , (b) the time-averaged $S$ , and (c) at what phase $S = 0$ .

Forecast: the average should be exactly half the peak (the famous $\frac{1}{2}$ ). And $S$ momentarily hits zero twice per cycle, when the fields cross zero.

Instantaneous: $S (t) = ε_{0} c E (t)^{2} = ε_{0} c E_{0}^{2} cos^{2} (ω t)$ . Why this step? The parent showed $S = ε_{0} c E^{2}$ holds at each instant, before any averaging.
Peak (at $cos^{2} = 1$ ): $S_{m a x} = ε_{0} c E_{0}^{2} = (8.85 \times 1 0^{- 12}) (3 \times 1 0^{8}) (100)^{2} \approx 26.6 W/m^{2}$ .
Average: replace $cos^{2}$ by its average $\frac{1}{2}$ : $⟨ S ⟩ = \frac{1}{2} ε_{0} c E_{0}^{2} \approx 13.3 W/m^{2}$ . Why this step? Over a full cycle $⟨ cos^{2} ⟩ = \frac{1}{2}$ ; intensity is what a slow detector reads, so it's the average, not the peak.
Zeros: $S = 0$ when $cos (ω t) = 0$ , i.e. $ω t = \frac{π}{2}, \frac{3 π}{2}, \dots$ — twice per period. Why this step? $S \propto E^{2} \geq 0$ , so it never goes negative; it just dips to zero whenever the field passes through zero, then rises again.

Verify: $⟨ S ⟩ / S_{m a x} = 13.3/26.6 = 0.5$ exactly. ✓ Note $S$ oscillates at $2 ω$ (double frequency), because squaring a cosine doubles its frequency.

Cell F — real-world word problem (inverse square)

Worked example Ex 6. Power output of the Sun

Sunlight at Earth's orbit has intensity $I = 1360 W/m^{2}$ . Earth–Sun distance $r = 1.5 \times 1 0^{11} m$ . What is the Sun's total radiated power (luminosity)?

Forecast: the Sun is enormous — expect something like $1 0^{26}$ watts.

Spread the power over a whole sphere: the Sun radiates equally in all directions, so at radius $r$ the same total power $P$ is smeared over a sphere of area $4 π r^{2}$ . Thus $I = \frac{P}{4 π r ^{2}}$ . Why this step? $S$ is power per area; to get total power we multiply intensity by the area it's spread across — here a sphere, giving the inverse-square law $I \propto 1/ r^{2}$ .
Solve for $P$ : $P = I (4 π r^{2})$ .
Plug in: $P = 1360 \times 4 π (1.5 \times 1 0^{11})^{2} \approx 3.8 \times 1 0^{26} W$ .

Verify: the accepted solar luminosity is $\approx 3.85 \times 1 0^{26}$ W. ✓ Dimension check: $(W/m^{2}) (m^{2}) = W$ . Twice as far ( $r \to 2 r$ ) would quarter the intensity — inverse square confirmed.

Cell G — radiation-pressure twist

Worked example Ex 7. Push on a sail: absorb vs reflect

A solar sail of area $A = 100 m^{2}$ sits where $I = 1360 W/m^{2}$ . Find the force on it (a) if it perfectly absorbs the light, (b) if it perfectly reflects it.

Forecast: the force is tiny (light barely pushes), micronewton-ish, and reflection gives double the absorbing case.

Radiation pressure on an absorber: $P_{rad} = I / c$ . Why this step? Light carries momentum flux equal to its energy flux divided by $c$ ; absorbing it dumps that momentum onto the sail. (See Radiation pressure.)
Force = pressure × area: $F_{abs} = \frac{I}{c} A = \frac{1360}{3 \times 1 0 ^{8}} \times 100 \approx 4.5 \times 1 0^{- 4} N$ .
Reflection doubles it: $P_{rad} = 2 I / c$ , because the light's momentum is not just stopped but reversed, so the sail feels twice the kick. $F_{ref} = \frac{2 I}{c} A \approx 9.1 \times 1 0^{- 4} N$ . Why this step? Change in momentum for a bounce is $2 p$ vs $p$ for a catch — same reason a ball bouncing off a wall pushes harder than a ball sticking to it.

Verify: $F_{ref} / F_{abs} = 2$ exactly. ✓ Both $\sim 1 0^{- 4}$ N — a whisper of force over 100 m², matching the forecast.

Cell H — exam trap: energy into a resistor

Worked example Ex 8. Where does a resistor's heat come from?

A cylindrical resistor of radius $a$ , length $L$ , carries current $I$ with resistance $R$ . Using fields at its surface, show the Poynting flux into it equals $I^{2} R$ .

Forecast: it should equal the familiar Joule heating $I^{2} R$ — but the direction of energy flow will surprise you.

Electric field along the wire: the voltage drop is $V = I R$ over length $L$ , so $E = V / L = I R / L$ pointing along the axis (direction of current). Why this step? $E$ drives the current; its magnitude is the voltage gradient.
Magnetic field around the wire: Ampère's law at the surface gives $B = \frac{μ _{0} I}{2 π a}$ , circling the wire. Why this step? A straight current wraps $B$ around itself — this is the field just outside the surface.
Poynting vector direction: $S = \frac{1}{μ _{0}} E \times B$ . With $E$ axial and $B$ azimuthal, $E \times B$ points radially inward — energy flows into the wire through its curved side (red arrows in figure). Why this step? The right-hand rule from axial to azimuthal gives radial; the sign works out inward. This is the trap: energy enters through the sides, not "down the wire."
Magnitude and integrate: $S = \frac{E B}{μ _{0}} = \frac{1}{μ _{0}} \frac{I R}{L} \frac{μ _{0} I}{2 π a} = \frac{I ^{2} R}{2 π a L}$ . Multiply by the side area $2 π a L$ : $P = S (2 π a L) = \frac{I ^{2} R}{2 π a L} \times 2 π a L = I^{2} R .$

Verify: the $2 π a L$ cancels perfectly, leaving $P = I^{2} R$ — exactly Joule heating. ✓ (Poynting's theorem guarantees this bookkeeping.)

Recall Which cell is this problem?

Given only $I$ , asked for $E_{0}$ ? ::: Cell A — invert $I = \frac{1}{2} ε_{0} c E_{0}^{2}$ . Beam through a tilted surface? ::: Cell C — use $P = S A cos θ$ with $θ$ from the normal. $E$ and $B$ parallel? ::: Cell D — $S = 0$ , no energy flow. Difference between $S_{m a x}$ and $⟨ S ⟩$ ? ::: A factor of $\frac{1}{2}$ from $⟨ cos^{2} ⟩$ . Reflecting vs absorbing sail? ::: Cell G — reflection gives double the force ( $2 I / c$ vs $I / c$ ).

See also: Intensity and amplitude of waves, Energy density of electric and magnetic fields, EM wave equation, Vector calculus identities, Maxwell's equations.