Poynting vector — energy flux in EM waves: Exercises

Definition Symbols and conventions used on this page

Before we start, we fix every symbol so nothing is ambiguous later.

$u$ = energy density of the fields (energy per unit volume, $J/m^{3}$ ), split into an electric part and a magnetic part: $u = u_{E}, electric \frac{1}{2} ε_{0} E^{2} + u_{B}, magnetic \frac{1}{2 μ _{0}} B^{2} .$ This comes straight from the parent note and Energy density of electric and magnetic fields. Think of $u$ as how much energy is stored in each little box of space.
$S$ = Poynting vector = energy flux ( $W/m^{2}$ ): how much energy crosses a unit area each second, and in which direction.
$P$ (upright, no subscript) always means power in watts, $P = \oint S \cdot d A$ .
$P_{pres}$ (with the subscript "pres") always means radiation pressure in pascals — we deliberately do not write a bare $P$ for pressure, to keep power and pressure from ever colliding.
Sign convention for surface integrals: in $\oint S \cdot d A$ , the area element $d A$ points outward from the closed surface. Then $\oint S \cdot d A > 0$ means net energy leaving the enclosed region, and $< 0$ means net energy entering it. Keep this in mind whenever a sign looks surprising.

Level 1 — Recognition

(Can you pick the right formula and units? No heavy algebra.)

Recall Solution L1·Q1

False. $S$ is an energy flux: energy per area per time.

$J/m^{3}$ is the unit of energy density $u$ — energy sitting in a volume.
$S$ carries energy across a surface, so it is $\frac{energy}{area \cdot time} = \frac{J}{m ^{2} \cdot s} = \frac{W}{m ^{2}}$ .

Answer: $W/m^{2}$ (watts per square metre).

Recall Solution L1·Q2

$S = \frac{1}{μ _{0}} E \times B$ must point along the propagation direction $+ x$ . Point fingers along $E$ ( $+ y$ ), curl toward $B$ , thumb must land on $+ x$ . We need $\overset{y}{^} \times \hat{B} = + \overset{x}{^}$ . Since $\overset{y}{^} \times \overset{z}{^} = \overset{x}{^}$ , we require $B$ along $== + z ==$ .

Check: $\overset{y}{^} \times \overset{z}{^} = \overset{x}{^}$ ✓ — so $E \times B$ points along $+ x$ , the direction the wave goes.

Figure below (s01): the three arrows form a mutually perpendicular triad — black $E$ pointing up ( $+ y$ ), black $B$ pointing "into the page/up-right" ( $+ z$ ), and the red $S$ pointing right ( $+ x$ ). Curl your right-hand fingers from $E$ to $B$ ; your thumb lands on the red arrow. That red arrow is the only thing energy transport cares about.

Figure — Poynting vector — energy flux in EM waves

Recall Solution L1·Q3

$I = \frac{1}{2} ε_{0} c E_{0}^{2}$ The instantaneous flux is $S = ε_{0} c E^{2} = ε_{0} c E_{0}^{2} cos^{2} (k x - ω t)$ . Averaging over time, $⟨ cos^{2} ⟩ = \frac{1}{2}$ , which drags in the == $\frac{1}{2}$ ==. Intensity is what a detector reads, and detectors report the average.

Level 2 — Application

(Plug numbers into one formula.)

Recall Solution L2·Q1

(a) Invert $I = \frac{1}{2} ε_{0} c E_{0}^{2}$ (we measure $I$ , we want $E_{0}$ ): $E_{0} = \frac{2 I}{ε _{0} c} = \frac{2 ( 1360 )}{( 8.85 \times 1 0 ^{- 12} ) ( 3 \times 1 0 ^{8} )} \approx 1.01 \times 1 0^{3} V/m .$ (b) In a wave $E = c B$ , so $B_{0} = \frac{E _{0}}{c} = \frac{1.01 \times 1 0 ^{3}}{3 \times 1 0 ^{8}} \approx 3.4 \times 1 0^{- 6} T .$ Tiny — this is why the magnetic push of light is normally negligible.

Recall Solution L2·Q2

$S$ is power per area; the beam flows straight through the aperture, so $S ∥ d A$ and $P = ⟨ S ⟩ A$ . Convert area: $A = 2.0 mm^{2} = 2.0 \times 1 0^{- 6} m^{2}$ . $P = 500 \times 2.0 \times 1 0^{- 6} = 1.0 \times 1 0^{- 3} W = 1.0 mW .$ (Here $P$ is power, per the conventions box.)

Recall Solution L2·Q3

Why $u_{E} = u_{B}$ in a wave. The two energy densities are $u_{E} = \frac{1}{2} ε_{0} E^{2}$ and $u_{B} = \frac{1}{2 μ _{0}} B^{2}$ . In a plane wave $E = c B$ , so substitute $B = E / c$ and $c^{2} = 1/ (μ_{0} ε_{0})$ : $u_{B} = \frac{1}{2 μ _{0}} (\frac{E}{c})^{2} = \frac{E ^{2}}{2 μ _{0} c ^{2}} = \frac{E ^{2}}{2 μ _{0}} (μ_{0} ε_{0}) = \frac{1}{2} ε_{0} E^{2} = u_{E} .$ So the magnetic and electric "halves" are exactly equal at every instant, which is why we can write $u = u_{E} + u_{B} = 2 \cdot \frac{1}{2} ε_{0} E^{2} = ε_{0} E^{2} .$ Number: $u = (8.85 \times 1 0^{- 12}) (300)^{2} = 7.97 \times 1 0^{- 7} J/m^{3} .$ Flux: $S = u c = 7.97 \times 1 0^{- 7} \times 3 \times 1 0^{8} = 239 W/m^{2}$ . Cross-check: $ε_{0} c E^{2} = (8.85 \times 1 0^{- 12}) (3 \times 1 0^{8}) (300)^{2} = 239 W/m^{2}$ ✓.

Level 3 — Analysis

(Two or more ideas combined, or a "why" that needs reasoning.)

Recall Solution L3·Q1

For reflection the light reverses momentum, so pressure is doubled versus absorption: $P_{pres} = \frac{2 I}{c} = \frac{2 ( 1360 )}{3 \times 1 0 ^{8}} = 9.07 \times 1 0^{- 6} Pa .$ Force $= P_{pres} \cdot A = 9.07 \times 1 0^{- 6} \times 100 = 9.07 \times 1 0^{- 4} N$ ≈ $0.91 mN$ . (Note $P_{pres}$ = pressure, distinct from power $P$ ; see the conventions box.) See Radiation pressure — momentum flux is $S / c$ , and reflection returns it, giving the factor 2.

Recall Solution L3·Q2

Consider a face of area $A$ perpendicular to the wave. In time $d t$ , all the energy inside a slab of thickness $c d t$ behind the face crosses it (the wave moves at $c$ ).

Energy in the slab $= u \times (volume) = u (A c d t)$ .
Flux $S = \frac{energy crossed}{A d t} = \frac{u A c d t}{A d t} = u c .$ ✓

Figure below (s02): the red slab is the chunk of energy sitting just behind the face. Its thickness is $c d t$ (the distance light travels in time $d t$ ), so in that time the whole red slab sweeps across the black face of area $A$ . Dividing its energy $u (A c d t)$ by $A d t$ leaves $u c$ . This is the "river current = water density × flow speed" picture from the parent note.

Recall Solution L3·Q3

(a) Peak flux occurs when $cos^{2} = 1$ : $S_{m a x} = ε_{0} c E_{0}^{2} = (8.85 \times 1 0^{- 12}) (3 \times 1 0^{8}) (200)^{2} = 106.2 W/m^{2} .$ (b) Intensity is the average: $I = \frac{1}{2} ε_{0} c E_{0}^{2} = 53.1 W/m^{2} .$ Ratio $I / S_{m a x} = \frac{1}{2}$ , as expected from $⟨ cos^{2} ⟩ = \frac{1}{2}$ .

Level 4 — Synthesis

(Assemble a chain from Maxwell/energy-conservation ideas.)

Recall Solution L4·Q1

(a) Voltage across the wire is $V = I R$ , dropped over length $L$ , so the axial field is $E = \frac{V}{L} = \frac{I R}{L} (along the wire) .$ Ampère's law on a loop of radius $a$ : $B (2 π a) = μ_{0} I \Rightarrow B = \frac{μ _{0} I}{2 π a}$ (circling the wire).

(b) $E$ (axial) $\times B$ (azimuthal) points radially inward — energy flows into the wire from the surrounding field. Magnitude: $S = \frac{E B}{μ _{0}} = \frac{1}{μ _{0}} \cdot \frac{I R}{L} \cdot \frac{μ _{0} I}{2 π a} = \frac{I ^{2} R}{2 π a L} .$

(c) The outward area element $d A$ on the side points radially outward, while $S$ points radially inward, so $S \cdot d A = - S d A$ — a negative flux, meaning energy enters (consistent with the sign convention in the box). Taking magnitudes over the side area $A_{s i d e} = 2 π a L$ : $\oint S \cdot d A = S \cdot A_{s i d e} = \frac{I ^{2} R}{2 π a L} \cdot 2 π a L = I^{2} R .$ The dissipated heat enters through the sides, carried by the fields — Poynting's theorem exactly balances the Joule heating.

Figure below (s03): the grey bar is the resistor; the black axial arrow is $E$ along the current $I$ ; the red arrows are $S$ pointing straight into the wire from all around. The energy that heats the resistor arrives sideways from the field, not "down the copper."

Recall Solution L4·Q2

(a) $P = I A = 1360 \times 2 = 2720 W$ (power). (b) Why the momentum flux is $I / c$ . An EM wave carries momentum: a packet of field energy $U$ carries momentum $p = U / c$ (a standard result of electromagnetism, also what you get from relativity for massless energy). Now count what lands on one square metre in one second: the energy arriving is $I$ (that is the meaning of intensity), so the momentum arriving is $p = I / c$ per area per second. For full absorption that momentum is delivered once, giving a pressure $I / c$ and a force $F = \frac{I}{c} A = \frac{1360}{3 \times 1 0 ^{8}} \times 2 = 9.07 \times 1 0^{- 6} N .$ (c) Also $F = \frac{P}{c} = \frac{2720}{3 \times 1 0 ^{8}} = 9.07 \times 1 0^{- 6} N$ ✓. Since $p = U / c$ , differentiating in time gives $\frac{d p}{d t} = \frac{1}{c} \frac{d U}{d t} = \frac{P}{c}$ , i.e. force $=$ power $/ c$ .

Level 5 — Mastery

(Derive/prove something general, cover all cases.)

Recall Solution L5·Q1

Start: $I = \frac{1}{2} ε_{0} c E_{0}^{2}$ . Form 2: substitute one factor $E_{0} = c B_{0}$ : $I = \frac{1}{2} ε_{0} c E_{0} (c B_{0}) = \frac{1}{2} ε_{0} c^{2} E_{0} B_{0} .$ Use $c^{2} = \frac{1}{μ _{0} ε _{0}} \Rightarrow ε_{0} c^{2} = \frac{1}{μ _{0}}$ : $I = \frac{E _{0} B _{0}}{2 μ _{0}} . ✓$ Form 3: now replace $E_{0} = c B_{0}$ in Form 2: $I = \frac{( c B _{0} ) B _{0}}{2 μ _{0}} = \frac{c B _{0}^{2}}{2 μ _{0}} . ✓$ If you only know $B_{0}$ , use $I = \frac{c B _{0}^{2}}{2 μ _{0}}$ directly — no need to compute $E_{0}$ first.

Recall Solution L5·Q2

Why $E$ and $B$ are exactly in phase. Faraday's law for this wave, $\frac{\partial E}{\partial x} = - \frac{\partial B}{\partial t}$ , links the two fields. With $E = E_{0} cos (k x - ω t)$ , the left side is $- k E_{0} sin (k x - ω t)$ , so $\frac{\partial B}{\partial t} = k E_{0} sin (k x - ω t) \Rightarrow B = \frac{k E _{0}}{ω} cos (k x - ω t) = \frac{E _{0}}{c} cos (k x - ω t),$ using $ω / k = c$ . So $B$ has the same $cos (k x - ω t)$ as $E$ : they rise and fall together (in phase), and their amplitudes obey $B_{0} = E_{0} / c$ . Therefore $S = ε_{0} c E B = ε_{0} c E_{0}^{2} cos^{2} (k x - ω t) .$

Peaks ( $cos^{2} = 1$ ): $S = ε_{0} c E_{0}^{2}$ — max forward flux, when field is largest (either $+ E_{0}$ or $- E_{0}$ ).
Zeros ( $cos = 0$ ): $S = 0$ — instant when the field passes through zero; no energy crossing right then.
Never negative: a squared quantity can't be negative, so the wave always carries energy in $+ x$ ; it never flows backward. This is why a single travelling wave has a strictly non-negative flux.
Average: $⟨ S ⟩ = \frac{1}{2} ε_{0} c E_{0}^{2} = I$ .

(Contrast: a standing wave has $E$ and $B$ 90° out of phase, so $⟨ S ⟩ = 0$ — energy sloshes but doesn't propagate.)

Recall Solution L5·Q3

Incoming momentum flux (per area) $= I / c$ . The absorbed fraction $(1 - r)$ delivers its momentum once; the reflected fraction $r$ delivers twice (comes in and leaves reversed): $P_{pres} = \frac{I}{c} [(1 - r) + 2 r] = \frac{I}{c} (1 + r) .$

$r = 0$ (black): $P_{pres} = I / c$ ✓.
$r = 1$ (mirror): $P_{pres} = 2 I / c$ ✓. For $I = 1360 W/m^{2}$ , a $r = 0.5$ grey surface gives $P_{pres} = \frac{1360}{3 \times 1 0 ^{8}} (1.5) = 6.80 \times 1 0^{- 6} Pa$ .

Recall Self-test — reveal for the answer key

Use this as a quick quiz: read the question, answer aloud, then check.

Units of $S$ ? — $W/m^{2}$ .
Intensity from $E_{0}$ ? — $I = \frac{1}{2} ε_{0} c E_{0}^{2}$ .
Radiation pressure, reflecting? — $2 I / c$ ; partial: $\frac{I}{c} (1 + r)$ .
Why does a single travelling wave's flux never go negative? — $S \propto cos^{2} \geq 0$ because $E$ and $B$ are in phase.
Where does resistor heat enter from? — Radially through the sides, carried by $S$ ; the surface integral gives $I^{2} R$ .

Prerequisite links: Maxwell's equations · EM wave equation · Energy density of electric and magnetic fields · Radiation pressure · Vector calculus identities · Intensity and amplitude of waves · Poynting vector — energy flux in EM waves.