Physics · Electrodynamics

Poynting vector and electromagnetic energy flow

The Poynting vector gives the direction and instantaneous density of electromagnetic energy flow in vacuum.

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Formula

S = \frac{1}{μ _{0}} E \times B

S in W/m²; E in V/m; B in T; μ0 in H/m.

Variables & units

Variables & units
Symbol	Meaning	Unit
$S$	Directed instantaneous electromagnetic energy-flux density	W/m²
$E$	Electric field	V/m
$B$	Magnetic flux density	T
$μ0$	Vacuum magnetic constant	N/A²

LaTeX

\mathbf S=\frac{1}{\mu_0}\mathbf E\times\mathbf B

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At a fixed point in a wave travelling along +z, normalized x-E and y-B fields share the identical cosine curve. Below, S along +z follows cosine squared, stays nonnegative and has cycle average one half. Axes use separate normalized scales. — One travelling sinusoidal plane vacuum wave, shown at the same fixed point against time phase θ = ωt: E along x, B along y, energy along +z. Above, identical normalized cosine fields; below, normalized energy flux as cosine squared, mean one half. Field and energy-flux axes are normalized independently and do not share physical units. Both fields reverse together; directed energy flow remains along +z. Based on OpenStax §16.3.

Poynting vector and intensity calculator

Find the intensity of a sine wave, the power through an area, or the Poynting vector from the E and B fields in vacuum.

Calculation model

Find

Peak field strength E₀

Peak electric field strength, not the rms value, e.g. about 868 V/m.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for ⟨S⟩, substitute, evaluate

1. Rearrange for ⟨S⟩
$\langle S\rangle=E_0^2/(2\mu_0c)$
2. Substitute values (in SI units)
$\frac{868^2}{2\cdot 0.00000125664\cdot 299792458}$
3. Result
⟨S⟩ = 1000 W/m²

Rounded CODATA 2022 values: ε₀ = 8.8541878188·10⁻¹² F/m and μ₀ = 1.25663706127·10⁻⁶ H/m with measurement uncertainty. Numerical results are approximate.

Result: Intensity ⟨S⟩

⟨S⟩ = 1000 W/m²

With E₀ = 868 V/m, ⟨S⟩ = 1000 W/m².

Quick tips

Estimate: ⟨S⟩ ≈ E₀² ÷ (2 · 377 Ω), since μ₀c ≈ 377 Ω is the impedance of free space. A peak field of 868 V/m gives about 1000 W/m².
Twice the field amplitude gives four times the intensity. The direction is E × B: turn E the short way onto B and your right thumb shows the energy flow.

Assumptions and limits

Exactly one travelling sinusoidal plane wave in vacuum with perpendicular in-phase fields. E₀ is peak electric magnitude, not RMS. The result is the nonnegative complete-cycle mean flux magnitude along propagation, not area power. Not for a standing superposition or DC cable.

Derivation & background

The E/B expression uses the vacuum model; E×H is the usual macroscopic material expression. Evaluate both fields at the same place and time. Distinguish magnitude, direction and projection onto a surface normal. $E=cB$ and the half-peak sinusoidal average require one travelling plane vacuum wave, not arbitrary standing or DC fields.

Exam blueprint

Validity range

E/B form uses the vacuum model; E×H is the usual macroscopic material form. Evaluate E and B at the same place/time. Check magnitude, direction and surface-normal component separately. $E=cB$ and half-peak average intensity apply only to one travelling plane sine wave, not generally to standing or DC fields.

Derivation steps

The Poynting vector gives the direction and instantaneous density of electromagnetic energy flow in vacuum.

1Determine vector E and B at the same position and time in the vacuum model.
2Use $\nabla\cdot(\mathbf E\times\mathbf B)=\mathbf B\cdot(\nabla\times\mathbf E)-\mathbf E\cdot(\nabla\times\mathbf B)$ .
3Insert Faraday and Ampère–Maxwell; separate field change from work on charges.
4Identify $u=\frac{\varepsilon_0E^2}{2}+\frac{B^2}{2\mu_0}$ and $\mathbf S=\frac{\mathbf E\times\mathbf B}{\mu_0}$ .
5The balance is $\frac{\partial u}{\partial t}+\nabla\cdot\mathbf S=-\mathbf J\cdot\mathbf E$ .
6Check cross-product direction and integrate the actual normal component to obtain surface power.
7Only for a single travelling in-phase sine wave does the mean product become half its peak flux; treat counterpropagating waves and DC fields separately.

Rearrangements

Energy-flux magnitude for perpendicular fields

S=\frac{EB}{\mu_0}

Instantaneous magnitude for perpendicular fields; only for one in-phase travelling sine wave is cycle average half peak intensity.

Task variant

One travelling plane vacuum sine wave has rounded peak $E_0=0.87\,\mathrm{V/m}$ along +x and $B_0=2.9\,\mathrm{nT}$ along +y. Fields are in phase. Find Poynting direction, peak magnitude and cycle average.

Approach: $\hat{\mathbf x}\times\hat{\mathbf y}=\hat{\mathbf z}$ , $S_{\max}=\frac{E_0B_0}{\mu_0}$ and $\langle S\rangle=\frac{S_{\max}}{2}$ .
Substitution gives $S_{\max}\approx2.008\,\mathrm{mW/m^2}$ , cycle average $1.004\,\mathrm{mW/m^2}$ , both along +z.

Check: both fields reverse together; S remains along +z and vanishes at field zeros.

Common mistakes

Replace the mean of E cross B by the cross product of the mean fields.

Form the instantaneous cross product first. Average intensity follows by averaging that product in time.

Exam context

Wave intensity, oblique power surface and coaxial cable: distinguish field product, projection and area integration.

Formula cluster

Electrodynamics

Maxwell determines the fields; their energy transport connects the Poynting vector to field-energy density and induction.

∯E·dA=QV/ε₀; ∯B·dA=0; ∮E·dl=−dΦB/dt; ∮B·dl=μ₀(I_A+ε₀dΦE/dt)

Maxwell Equations in Integral Form

uE = 1/2 ε₀E²

Electric-field energy density in vacuum

U_ind = −L·(dI/dt)

Self-Induction of a Coil

Worked example

One travelling sinusoidal plane vacuum wave has consistently rounded peak fields $E_0=0.87\,\mathrm{V/m}$ and $B_0=2.9\,\mathrm{nT}$ . For perpendicular in-phase fields, $\langle S\rangle=\frac{E_0B_0}{2\mu_0}\approx1.00\,\mathrm{mW/m^2}$ along propagation. Peak flux is twice the mean. $\frac{E_0}{B_0}\approx3.00\times10^8\,\mathrm{m/s}$ checks the rounded wave relation; it is not universal.

Applications

Describe energy transport in light and radio waves and power radiated by antennas.

How do you calculate with Poynting Vector and Electromagnetic Energy Flux?

Here is how to work through a typical Poynting Vector and Electromagnetic Energy Flux (S = (E × B)/μ₀) task step by step:

1
Task
One travelling plane vacuum sine wave has rounded peak $E_0=0.87\,\mathrm{V/m}$ along +x and $B_0=2.9\,\mathrm{nT}$ along +y. Fields are in phase. Find Poynting direction, peak magnitude and cycle average.
Solution path
1. Approach: $\hat{\mathbf x}\times\hat{\mathbf y}=\hat{\mathbf z}$ , $S_{\max}=\frac{E_0B_0}{\mu_0}$ and $\langle S\rangle=\frac{S_{\max}}{2}$ .
2. Substitution gives $S_{\max}\approx2.008\,\mathrm{mW/m^2}$ , cycle average $1.004\,\mathrm{mW/m^2}$ , both along +z.
Check: both fields reverse together; S remains along +z and vanishes at field zeros.

Understand and apply the equation

The Poynting vector describes the direction of electromagnetic energy flow and the power crossing an area. It connects electric and magnetic fields to an energy balance. For a travelling sinusoidal wave it yields intensity; for a cable it shows that energy can also be transferred through the surrounding field without a radiated travelling wave.

From a cross product to power crossing an area

A capacitor can store energy in its field; a cable or light beam can carry energy through space. The Poynting vector describes this transport locally: its arrow gives direction and its magnitude gives power per area perpendicular to the flow. Stored energy and transported power are different quantities. We start with a pair of fields at one position, then calculate power crossing an area, before examining the complete energy balance.[1]

\begin{aligned}\mathbf S&=\frac{1}{\mu_0}\mathbf E\times\mathbf B\\|\mathbf S|&=\frac{EB\sin\alpha}{\mu_0}\\P_A(t)&=\iint_A\mathbf S\cdot d\mathbf A\end{aligned}

α is the angle between the fields. The component normal to a surface determines transmitted power.[1][3]

The symbol $\times$ denotes a cross product: it forms a third direction perpendicular to both inputs. Its magnitude contains the sine of their angle. The symbol $\cdot$ in a surface calculation denotes a dot product, selecting the part that actually crosses the area. This is why the field calculation uses a sine and the later surface projection a cosine.[3]

Both fields are evaluated at the same position and time. The cross product follows the right-hand rule: E along +x and B along +y give S along +z. S has units of watts per square metre and is a flux density, not stored energy per volume. Positive surface-normal projection denotes flow along the chosen normal; a closed enclosure uses the outward normal.[2]

Only perpendicular E and B permit the magnitude to be calculated by simply multiplying their magnitudes. Direction must still be checked. A large field alone does not ensure large energy flow: parallel E and B can give zero S while stored field energy remains nonzero.[3]

Explore further: How the energy balance follows from Maxwell

\begin{aligned}\nabla\cdot(\mathbf E\times\mathbf B)&=\mathbf B\cdot(\nabla\times\mathbf E)-\mathbf E\cdot(\nabla\times\mathbf B)\\\nabla\cdot\mathbf S&=-\frac{\partial}{\partial t}\left(\frac{\varepsilon_0 E^2}{2}+\frac{B^2}{2\mu_0}\right)-\mathbf J\cdot\mathbf E\end{aligned}

The vector identity together with Faraday and Ampère–Maxwell connects energy flow, field changes and work on charges.[3]

\begin{aligned}u&=\frac{\varepsilon_0E^2}{2}+\frac{B^2}{2\mu_0}\\\frac{\partial u}{\partial t}+\nabla\cdot\mathbf S&=-\mathbf J\cdot\mathbf E\\\frac{d}{dt}\iiint_Vu\,dV&=-\oiint_{\partial V}\mathbf S\cdot d\mathbf A-\iiint_V\mathbf J\cdot\mathbf E\,dV\end{aligned}

For a fixed volume, stored field energy decreases through net outward flow and positive field work on charges.[2]

The term J·E is initially electromagnetic power transferred to matter, rather than necessarily heat. In an ohmic conductor it becomes dissipated power; inside an active source its sign can reverse. The integral balance counts total contributions over the selected volume. S need not point outward everywhere, and negative net outward power describes energy entering.[3]

The stated energy densities and E/B expression use the vacuum model. In the usual macroscopic material formulation $\mathbf S=\mathbf E\times\mathbf H$ is used. Simple quadratic storage terms require suitable linear, nondispersive constitutive relations; losses, dispersion or nonlinearity require material response to be included in the energy accounting.[3]

Electric-field energy density in vacuum describes the electric storage term in this balance. The Poynting vector instead describes transport; magnetic storage and work on charges may also contribute. A large electric storage term alone therefore establishes neither a large nor a directed flux density.[3][2]

Separate instantaneous, peak and average values

\begin{aligned}\mathbf E(z,t)&=E_0\cos(kz-\omega t)\,\hat{\mathbf x}\\\mathbf B(z,t)&=\frac{E_0}{c}\cos(kz-\omega t)\,\hat{\mathbf y}\\\mathbf S(z,t)&=\frac{E_0^2}{\mu_0c}\cos^2(kz-\omega t)\,\hat{\mathbf z}\end{aligned}

A single travelling plane vacuum wave has in-phase perpendicular fields satisfying $E=cB$ .[1]

\begin{aligned}\langle\mathbf S\rangle&=\frac{E_0^2}{2\mu_0c}\,\hat{\mathbf z}\\&=\frac{E_0B_0}{2\mu_0}\,\hat{\mathbf z}\\&=\frac{E_{\mathrm{rms}}^2}{\mu_0c}\,\hat{\mathbf z}\end{aligned}

The factor one half belongs to sinusoidal peak values. RMS already contains it through the mean square.[1][2]

Both fields reverse sign together. Their cross product therefore points in the same propagation direction for this single travelling wave and vanishes at their common zero crossings. Flux density has two maxima per field period. Multiplying the mean fields instead would give zero and is not a substitute for averaging their cross product.[3]

For a uniformly illuminated planar area, average power follows from intensity, area and the cosine of the angle between S and the surface normal. This is not the angle between E and B. A nonuniform beam or curved surface requires integrating the local normal component.[3]

The RMS value of sinusoidal quantities explains why mean-square averaging already contains half the peak-squared contribution. Apply this averaging rule to the specified single travelling sinusoid; it does not permit multiplying separately averaged fields with arbitrary phase relationships.[1][5]

Account for oblique incidence on an area

Find incident energy from field amplitude

Given: A sinusoidal plane vacuum wave has $E_0=30\,\mathrm{V/m}$ . A planar area fully within the uniform beam has $A=20\,\mathrm{cm^2}$ ; propagation makes $60^\circ$ with the positive surface normal. Observation lasts 10 s, covering many cycles.
Find: Find peak B, average flux density, transmitted power and energy.

Approach: Relate the wave fields, average the squared sinusoid and project onto the area.
$\begin{aligned}B_0&=\frac{E_0}{c}\\I&=\frac{E_0^2}{2\mu_0c}\\\overline P&=IA\cos\theta\\W&=\overline P\,\Delta t\end{aligned}$
Using $c\approx2.998\times10^8\,\mathrm{m/s}$ and $\mu_0\approx1.256637\times10^{-6}\,\mathrm{H/m}$ gives field and flux values.
$\begin{aligned}B_0&\approx1{,}001\times10^{-7}\,\mathrm T\\I&\approx1{,}1945\,\mathrm{W/m^2}\\S_{\max}&=2I\approx2{,}3890\,\mathrm{W/m^2}\end{aligned}$
Convert the squared area unit and use the normal component.
$\begin{aligned}A&=20\times10^{-4}\,\mathrm{m^2}=0{,}0020\,\mathrm{m^2}\\\overline P&=(1{,}1945)(0{,}0020)\cos60^\circ\,\mathrm W\\&\approx1{,}1945\,\mathrm{mW}\\W&\approx0{,}011945\,\mathrm J\end{aligned}$

Result: Approximately 1.19 mW crosses the area; over 10 s this is 11.9 mJ. It is incident energy, not necessarily energy absorbed in full.

Check: Normal incidence would double power. Units check: watts per square metre times square metres times seconds gives joules. Doubling $E_0$ would quadruple intensity and energy.

Energy in the field of a DC cable

An ideal coaxial cable feeding a load

Given: A long ideal vacuum coaxial cable has inner conductor radius $a=2\,\mathrm{mm}$ , inner radius of outer conductor $b=5\,\mathrm{mm}$ , constant voltage $U=12\,\mathrm V$ and inner conductor current $I=2\,\mathrm A$ along +z. The inner conductor is positive; return current flows in the outer conductor. End effects and conductor losses are neglected.
Find: Find local fields and S at $r=3\,\mathrm{mm}$ , and total power crossing the annular gap.

Approach: Cylindrical symmetry gives radial E and azimuthal B; their cross product points toward the load.
$\begin{aligned}E_r&=\frac{U}{r\ln(b/a)}\\B_\varphi&=\frac{\mu_0I}{2\pi r}\\S_z&=\frac{UI}{2\pi r^2\ln(b/a)}\end{aligned}$
At $r=3\,\mathrm{mm}$ , local flux density exceeds the annular average; it increases further toward the inner conductor.
$\begin{aligned}E_r(3\,\mathrm{mm})&\approx4{,}365\times10^3\,\mathrm{V/m}\\B_\varphi(3\,\mathrm{mm})&\approx133{,}3\,\mathrm{\mu T}\\S_z(3\,\mathrm{mm})&\approx4{,}632\times10^5\,\mathrm{W/m^2}\end{aligned}$
Integrate over the full annulus; flux density is not constant there.
$\begin{aligned}P&=\int_a^b S_z\,2\pi r\,dr\\&=\frac{UI}{\ln(b/a)}\int_a^b\frac{dr}{r}\\&=UI=24\,\mathrm W\end{aligned}$

Result: 24 W flows toward the load in the field between ideal conductors. A steady DC supply can have a time-independent Poynting vector.

Check: The independent circuit calculation U·I also gives 24 W. $E=cB$ is not assumed here: these fields are not a single travelling plane vacuum wave. Real conductor losses and energy entering the metal belong to an expanded balance.

Why standing waves do not carry purely forward power

\begin{aligned}\mathbf E&=2E_0\cos(kx)\cos(\omega t)\,\hat{\mathbf y}\\\mathbf B&=\frac{2E_0}{c}\sin(kx)\sin(\omega t)\,\hat{\mathbf z}\\\mathbf S&=\frac{E_0^2}{\mu_0c}\sin(2kx)\sin(2\omega t)\,\hat{\mathbf x}\\\langle\mathbf S\rangle&=\mathbf0\end{aligned}

Superposition of two coherent equal-amplitude, equally polarized plane vacuum waves travelling in opposite directions. $E_0$ is the peak amplitude of each individual wave; $k=\frac{\omega}{c}>0$ . Local energy flow can oscillate while its cycle average is zero.[4]

View diagram: local standing-wave flux

Standing wave: directed local energy flux

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Original model curve at kx=π/4 and single-wave amplitude 10 V/m. Positive means +x, negative −x. Zero time average does not imply zero flux at every instant.[4]

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First add the fields of the travelling waves vectorially, then form the Poynting vector from the total fields. E and B in the stated standing wave are not in temporal phase. Consequently, the simple $E=cB$ relation between total field magnitudes does not hold everywhere and at every instant. Energy can be stored and transported back and forth locally even though a cross-section carries no net energy over a complete cycle.[4]

This matters for measurements: one field measurement in a region with reflections does not automatically give net transported power. Wave impedance, counterpropagating contributions and phase can matter. Depending on context, intensity denotes a directed cycle average or incident radiant power; state the intended meaning and measurement surface.[3]

Standing wave: finite flux now, zero mean

Given: The standing vacuum wave stated above has single-wave amplitude $E_0=10.0\,\mathrm{V/m}$ . At $kx=\frac\pi4$ , consider first $\omega t=\frac\pi4$ , then $\omega t=\frac{3\pi}{4}$ .
Find: Find both instantaneous energy fluxes and the cycle average.

Approach: use the superposed total fields first rather than adding individual wave intensities.
$S_x=\frac{E_0^2}{\mu_0c}\sin(2kx)\sin(2\omega t)$
Substitution: the position factor is one; the time factor changes from +1 to −1.
$\begin{aligned}S_x(t_1)&=+\frac{100}{\mu_0c}\,\mathrm{W\,m^{-2}}\approx+0.265442\,\mathrm{W\,m^{-2}}\\S_x(t_2)&\approx-0.265442\,\mathrm{W\,m^{-2}}\\\langle S_x\rangle_T&=0\end{aligned}$

Result: Energy flows along +x at first and later along −x; net flux vanishes over a complete cycle.

Check: Check: initially $E_y=10.0\,\mathrm{V/m}$ and $B_z=\frac{E_0}{c}\approx33.36\,\mathrm{nT}$ ; the cross product confirms positive flux. At the second instant only the electric field reverses. The travelling components each carry about 0.132721 W/m² in opposite directions on average.

Five exercises on direction, averaging and balance

Calculate · Foundation

At one point in vacuum, instantaneous field magnitudes are $E=2.0\times10^{-3}\,\mathrm{V/m}$ and $B=3.0\,\mathrm{\mu T}$ , separated by $30^\circ$ . Find instantaneous S magnitude.

Show hint

The cross product requires the sine of the field angle.

Learning goal: Distinguish instantaneous cross product from wave intensity.

Show solution and calculation

Approach: $S=\frac{EB\sin\alpha}{\mu_0}$ .
Approximately 2.39 mW/m².
These are general local fields; neither $E=cB$ nor an additional sinusoidal average was specified.

S=\frac{EB\sin30^\circ}{\mu_0}\approx2{,}387\times10^{-3}\,\mathrm{W/m^2}

[3]

Rearrange · Application

A single travelling plane vacuum wave has average intensity $I=0.170\,\mathrm{W/m^2}$ . Find RMS E and, for a sinusoid, its approximate peak value.

Show hint

For RMS values there is no additional factor one half.

Learning goal: Recover an RMS field correctly from average intensity.

Show solution and calculation

Approach: $I_{\mathrm{wave}}=\frac{E_{\mathrm{rms}}^2}{\mu_0c}$ .
RMS E is approximately 8.00 V/m; sinusoidal peak E is approximately 11.3 V/m.
Inserting RMS into the peak formula with factor one half would halve the intensity.

\begin{aligned}E_{\mathrm{rms}}&=\sqrt{I\mu_0c}\approx8{,}00\,\mathrm{V/m}\\E_0&=\sqrt2\,E_{\mathrm{rms}}\approx11{,}3\,\mathrm{V/m}\end{aligned}

[1]

Interpret · Application

E points along +x and B along −z at an instant. Which way does S point? What happens if both fields reverse simultaneously?

Show hint

Use the cyclic ordering of the x, y and z axes.

Learning goal: Determine energy-flow direction from vector orientation.

Show solution and calculation

Approach: $\mathbf S=\frac{\mathbf E\times\mathbf B}{\mu_0}$ .
S points along +y.
Reversing both fields leaves the cross product unchanged; reversing only one reverses S. This explains why sign changes of travelling-wave fields do not imply backward power.

\hat{\mathbf x}\times(-\hat{\mathbf z})=\hat{\mathbf y},\qquad(-\mathbf E)\times(-\mathbf B)=\mathbf E\times\mathbf B

[2]

Error analysis · Application

A travelling sinusoidal wave has zero mean E and zero mean B over a cycle. A calculation therefore concludes that average intensity is zero. Correct it.

Show hint

Form the cross product first, then average.

Learning goal: Detect incorrect averaging of field products.

Show solution and calculation

Approach: $\langle\mathbf S\rangle=\frac{\langle\mathbf E\times\mathbf B\rangle}{\mu_0}$ .
The mean of a product generally differs from the product of the means.
In-phase fields give a positive squared waveform whose mean contains a factor one half.
Mean fields alone do not retain this information.

\langle\mathbf S\rangle=\frac{1}{\mu_0}\langle\mathbf E\times\mathbf B\rangle=\frac{E_0B_0}{2\mu_0}\hat{\mathbf n}\ne\frac{\langle\mathbf E\rangle\times\langle\mathbf B\rangle}{\mu_0}

[3]

Transfer · Challenge

A fixed volume receives net electromagnetic inflow of 5 W. Simultaneously the field transfers 3 W to matter, entirely converted into heat there. How does stored field energy change?

Show hint

With an outward normal, net surface flux is −5 W.

Learning goal: Connect oriented flux balance with storage and dissipation.

Show solution and calculation

Approach: $\dot W_{\mathrm{field}}=-P_{\mathrm{out}}-P_{\mathrm{matter}}$ .
Field energy grows by 2 J per second.
Incoming power is partly dissipated and partly stored; equating all incoming power with heating would be incorrect.

\begin{aligned}\frac{\mathrm dW_{\mathrm{field}}}{\mathrm dt}&=-P_{\mathrm{out}}-P_{\mathrm{matter}}\\&=-(-5\,\mathrm W)-3\,\mathrm W\\&=2\,\mathrm W\end{aligned}

[2]

Build a reliable energy-flow calculation

Choose vacuum or an appropriate material model, evaluate the fields at the same place and time, and record their directions. Then distinguish an instantaneous value, a full-cycle average or the energy of a finite pulse. $E=cB$ is a property of a single travelling plane vacuum wave, rather than a general law for every electromagnetic field. Reflected or superposed waves require finding the actual total fields first.[3][4]

The power surface must match the problem: a detector’s illuminated area for a beam, the complete field cross-section for a cable, or a closed enclosure for a local energy balance. Check angle, spatial uniformity and units before replacing an integral by multiplication with area. A complete energy balance also checks whether input power is stored, transmitted onward or transferred to matter.[3]

What to take away

First form the cross product of the actual total fields. Then distinguish an instantaneous value from a time average and project onto the chosen surface normal for power. Simple amplitude relations apply only to the appropriate wave model. Incoming energy can be stored, transmitted onward or transferred to matter.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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October 2, 2026Added relevant scientific connections; preserved existing field and energy balances, derivations, examples, exercises, figures and model limits.
October 1, 2026Added an original DE/EN article with assumptions, two original worked examples and five different solved exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.

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Quanta. Poynting Vector and Electromagnetic Energy Flux. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/poynting-vektor

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Frequently asked questions about Poynting Vector and Electromagnetic Energy Flux

What do the Poynting vector and electromagnetic energy flux describe physically?+

The Poynting vector describes instantaneous directed electromagnetic energy-flux density. Its units are watts per square metre, not joules per cubic metre. In vacuum it is calculated from the cross product of electric and magnetic fields at the same place and time. The right-hand rule determines direction, while the angle between fields determines magnitude. A single travelling plane wave transports energy along its propagation direction, but this is not a general property of every field. A steady DC cable can also transport energy through fields. A standing wave may have local back-and-forth energy flow with zero cycle average. Power crossing an area depends on the vector’s normal component.

How can I check a calculation involving the Poynting vector and electromagnetic energy flux?+

First record field directions and establish whether instantaneous, peak or RMS values are given. A single travelling sinusoidal plane vacuum wave has perpendicular, in-phase E and B. The existing consistently rounded peaks of 0.87 volts per metre and 2.9 nanoteslas give about one milliwatt per square metre average intensity. Peak instantaneous flux is about twice that value. As a check, the field-amplitude ratio approximately matches light speed. General local fields do not assume this wave relation, and no factor one half is added automatically. Also check whether energy flow crosses the specified area normally or whether projection is needed to obtain the requested power.

What assumptions and limits apply to the Poynting vector and electromagnetic energy flux?+

The displayed E/B Poynting expression uses the vacuum model. In the usual macroscopic material model the corresponding quantity is E cross H. Simple quadratic storage terms require suitable linear, nondispersive material assumptions; actual losses or dispersion need an expanded balance. The E equals c times B relationship and the simple sinusoidal mean assume one travelling plane vacuum wave. With counterpropagating waves or reflections, form the total fields first. The mean cross product is generally not the cross product of the mean fields. Even for perpendicular fields, the angle to the power surface must be determined separately. These conditions prevent a wave-intensity formula from being mistaken for a universal field formula.

How do I check units and scale for the Poynting vector and electromagnetic energy flux?+

Check units as power per area. Multiplying a uniform normal flux density by square metres gives watts; multiplying by time gives joules. An oblique planar area additionally requires the cosine of the angle to its normal. This is not the sine of the E-B field angle already contained in the cross product. Scale can be checked by an energy balance: input power must be stored, transmitted onward or transferred to charges. For an ideal DC coaxial cable, integrating across the complete annular gap gives voltage times current. A local S value cannot simply be multiplied by the whole cross-section there because it varies with radius.

When should I combine the Poynting vector and electromagnetic energy flux with another relation?+

To find power crossing an area, integrate the normal component of the Poynting vector. To find energy over an interval, additionally integrate that power in time. The complete Poynting balance connects net flux through a fixed enclosure to stored field energy and work on charges. This work is automatically heat only in an appropriate dissipative model; its sign can reverse inside a source. For a single travelling sinusoidal plane wave, field amplitudes can be related through light speed and average intensity. Those assumptions do not apply to a DC cable or standing superposition. Maxwell’s equations determine the underlying fields, while sources, geometry and boundary conditions specify the actual solution.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §16.3: Energy Carried by Electromagnetic Waves

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Retain Poynting Vector and Electromagnetic Energy Flux for exams

Create a curated FSRS exam set for S = (E × B)/μ₀: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Symbol

Meaning

Unit

S

Directed instantaneous electromagnetic energy-flux density

W/m²

E

Electric field

V/m

B

Magnetic flux density

μ0

Vacuum magnetic constant

N/A²

Derivation & background

Exam blueprint

Validity range

Derivation steps

The Poynting vector gives the direction and instantaneous density of electromagnetic energy flow in vacuum.

1Determine vector E and B at the same position and time in the vacuum model.
2Use $\nabla\cdot(\mathbf E\times\mathbf B)=\mathbf B\cdot(\nabla\times\mathbf E)-\mathbf E\cdot(\nabla\times\mathbf B)$ .
3Insert Faraday and Ampère–Maxwell; separate field change from work on charges.
4Identify $u=\frac{\varepsilon_0E^2}{2}+\frac{B^2}{2\mu_0}$ and $\mathbf S=\frac{\mathbf E\times\mathbf B}{\mu_0}$ .
5The balance is $\frac{\partial u}{\partial t}+\nabla\cdot\mathbf S=-\mathbf J\cdot\mathbf E$ .
6Check cross-product direction and integrate the actual normal component to obtain surface power.
7Only for a single travelling in-phase sine wave does the mean product become half its peak flux; treat counterpropagating waves and DC fields separately.

How do you calculate with Poynting Vector and Electromagnetic Energy Flux?

Here is how to work through a typical Poynting Vector and Electromagnetic Energy Flux (S = (E × B)/μ₀) task step by step:

Task

Solution path

Approach: $\hat{\mathbf x}\times\hat{\mathbf y}=\hat{\mathbf z}$ , $S_{\max}=\frac{E_0B_0}{\mu_0}$ and $\langle S\rangle=\frac{S_{\max}}{2}$ .
Substitution gives $S_{\max}\approx2.008\,\mathrm{mW/m^2}$ , cycle average $1.004\,\mathrm{mW/m^2}$ , both along +z.

Check: both fields reverse together; S remains along +z and vanishes at field zeros.

Understand and apply the equation

From a cross product to power crossing an area

\begin{aligned}\mathbf S&=\frac{1}{\mu_0}\mathbf E\times\mathbf B\\|\mathbf S|&=\frac{EB\sin\alpha}{\mu_0}\\P_A(t)&=\iint_A\mathbf S\cdot d\mathbf A\end{aligned}

α is the angle between the fields. The component normal to a surface determines transmitted power.[1][3]

Explore further: How the energy balance follows from Maxwell

\begin{aligned}\nabla\cdot(\mathbf E\times\mathbf B)&=\mathbf B\cdot(\nabla\times\mathbf E)-\mathbf E\cdot(\nabla\times\mathbf B)\\\nabla\cdot\mathbf S&=-\frac{\partial}{\partial t}\left(\frac{\varepsilon_0 E^2}{2}+\frac{B^2}{2\mu_0}\right)-\mathbf J\cdot\mathbf E\end{aligned}

The vector identity together with Faraday and Ampère–Maxwell connects energy flow, field changes and work on charges.[3]

\begin{aligned}u&=\frac{\varepsilon_0E^2}{2}+\frac{B^2}{2\mu_0}\\\frac{\partial u}{\partial t}+\nabla\cdot\mathbf S&=-\mathbf J\cdot\mathbf E\\\frac{d}{dt}\iiint_Vu\,dV&=-\oiint_{\partial V}\mathbf S\cdot d\mathbf A-\iiint_V\mathbf J\cdot\mathbf E\,dV\end{aligned}

For a fixed volume, stored field energy decreases through net outward flow and positive field work on charges.[2]

Separate instantaneous, peak and average values

\begin{aligned}\mathbf E(z,t)&=E_0\cos(kz-\omega t)\,\hat{\mathbf x}\\\mathbf B(z,t)&=\frac{E_0}{c}\cos(kz-\omega t)\,\hat{\mathbf y}\\\mathbf S(z,t)&=\frac{E_0^2}{\mu_0c}\cos^2(kz-\omega t)\,\hat{\mathbf z}\end{aligned}

A single travelling plane vacuum wave has in-phase perpendicular fields satisfying $E=cB$ .[1]

\begin{aligned}\langle\mathbf S\rangle&=\frac{E_0^2}{2\mu_0c}\,\hat{\mathbf z}\\&=\frac{E_0B_0}{2\mu_0}\,\hat{\mathbf z}\\&=\frac{E_{\mathrm{rms}}^2}{\mu_0c}\,\hat{\mathbf z}\end{aligned}

The factor one half belongs to sinusoidal peak values. RMS already contains it through the mean square.[1][2]

Account for oblique incidence on an area

Find incident energy from field amplitude

Given: A sinusoidal plane vacuum wave has $E_0=30\,\mathrm{V/m}$ . A planar area fully within the uniform beam has $A=20\,\mathrm{cm^2}$ ; propagation makes $60^\circ$ with the positive surface normal. Observation lasts 10 s, covering many cycles.
Find: Find peak B, average flux density, transmitted power and energy.

Approach: Relate the wave fields, average the squared sinusoid and project onto the area.
$\begin{aligned}B_0&=\frac{E_0}{c}\\I&=\frac{E_0^2}{2\mu_0c}\\\overline P&=IA\cos\theta\\W&=\overline P\,\Delta t\end{aligned}$
Using $c\approx2.998\times10^8\,\mathrm{m/s}$ and $\mu_0\approx1.256637\times10^{-6}\,\mathrm{H/m}$ gives field and flux values.
$\begin{aligned}B_0&\approx1{,}001\times10^{-7}\,\mathrm T\\I&\approx1{,}1945\,\mathrm{W/m^2}\\S_{\max}&=2I\approx2{,}3890\,\mathrm{W/m^2}\end{aligned}$
Convert the squared area unit and use the normal component.
$\begin{aligned}A&=20\times10^{-4}\,\mathrm{m^2}=0{,}0020\,\mathrm{m^2}\\\overline P&=(1{,}1945)(0{,}0020)\cos60^\circ\,\mathrm W\\&\approx1{,}1945\,\mathrm{mW}\\W&\approx0{,}011945\,\mathrm J\end{aligned}$

Result: Approximately 1.19 mW crosses the area; over 10 s this is 11.9 mJ. It is incident energy, not necessarily energy absorbed in full.

Check: Normal incidence would double power. Units check: watts per square metre times square metres times seconds gives joules. Doubling $E_0$ would quadruple intensity and energy.

Energy in the field of a DC cable

An ideal coaxial cable feeding a load

Given: A long ideal vacuum coaxial cable has inner conductor radius $a=2\,\mathrm{mm}$ , inner radius of outer conductor $b=5\,\mathrm{mm}$ , constant voltage $U=12\,\mathrm V$ and inner conductor current $I=2\,\mathrm A$ along +z. The inner conductor is positive; return current flows in the outer conductor. End effects and conductor losses are neglected.
Find: Find local fields and S at $r=3\,\mathrm{mm}$ , and total power crossing the annular gap.

Approach: Cylindrical symmetry gives radial E and azimuthal B; their cross product points toward the load.
$\begin{aligned}E_r&=\frac{U}{r\ln(b/a)}\\B_\varphi&=\frac{\mu_0I}{2\pi r}\\S_z&=\frac{UI}{2\pi r^2\ln(b/a)}\end{aligned}$
At $r=3\,\mathrm{mm}$ , local flux density exceeds the annular average; it increases further toward the inner conductor.
$\begin{aligned}E_r(3\,\mathrm{mm})&\approx4{,}365\times10^3\,\mathrm{V/m}\\B_\varphi(3\,\mathrm{mm})&\approx133{,}3\,\mathrm{\mu T}\\S_z(3\,\mathrm{mm})&\approx4{,}632\times10^5\,\mathrm{W/m^2}\end{aligned}$
Integrate over the full annulus; flux density is not constant there.
$\begin{aligned}P&=\int_a^b S_z\,2\pi r\,dr\\&=\frac{UI}{\ln(b/a)}\int_a^b\frac{dr}{r}\\&=UI=24\,\mathrm W\end{aligned}$

Result: 24 W flows toward the load in the field between ideal conductors. A steady DC supply can have a time-independent Poynting vector.

Why standing waves do not carry purely forward power

\begin{aligned}\mathbf E&=2E_0\cos(kx)\cos(\omega t)\,\hat{\mathbf y}\\\mathbf B&=\frac{2E_0}{c}\sin(kx)\sin(\omega t)\,\hat{\mathbf z}\\\mathbf S&=\frac{E_0^2}{\mu_0c}\sin(2kx)\sin(2\omega t)\,\hat{\mathbf x}\\\langle\mathbf S\rangle&=\mathbf0\end{aligned}

View diagram: local standing-wave flux

Standing wave: directed local energy flux

QUANTA STUDYquanta-study.de

Original model curve at kx=π/4 and single-wave amplitude 10 V/m. Positive means +x, negative −x. Zero time average does not imply zero flux at every instant.[4]

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Standing wave: finite flux now, zero mean

Given: The standing vacuum wave stated above has single-wave amplitude $E_0=10.0\,\mathrm{V/m}$ . At $kx=\frac\pi4$ , consider first $\omega t=\frac\pi4$ , then $\omega t=\frac{3\pi}{4}$ .
Find: Find both instantaneous energy fluxes and the cycle average.

Approach: use the superposed total fields first rather than adding individual wave intensities.
$S_x=\frac{E_0^2}{\mu_0c}\sin(2kx)\sin(2\omega t)$
Substitution: the position factor is one; the time factor changes from +1 to −1.
$\begin{aligned}S_x(t_1)&=+\frac{100}{\mu_0c}\,\mathrm{W\,m^{-2}}\approx+0.265442\,\mathrm{W\,m^{-2}}\\S_x(t_2)&\approx-0.265442\,\mathrm{W\,m^{-2}}\\\langle S_x\rangle_T&=0\end{aligned}$

Result: Energy flows along +x at first and later along −x; net flux vanishes over a complete cycle.

Five exercises on direction, averaging and balance

Calculate · Foundation

At one point in vacuum, instantaneous field magnitudes are $E=2.0\times10^{-3}\,\mathrm{V/m}$ and $B=3.0\,\mathrm{\mu T}$ , separated by $30^\circ$ . Find instantaneous S magnitude.

Show hint

The cross product requires the sine of the field angle.

Learning goal: Distinguish instantaneous cross product from wave intensity.

Show solution and calculation

Approach: $S=\frac{EB\sin\alpha}{\mu_0}$ .
Approximately 2.39 mW/m².
These are general local fields; neither $E=cB$ nor an additional sinusoidal average was specified.

S=\frac{EB\sin30^\circ}{\mu_0}\approx2{,}387\times10^{-3}\,\mathrm{W/m^2}

[3]

Rearrange · Application

A single travelling plane vacuum wave has average intensity $I=0.170\,\mathrm{W/m^2}$ . Find RMS E and, for a sinusoid, its approximate peak value.

Show hint

For RMS values there is no additional factor one half.

Learning goal: Recover an RMS field correctly from average intensity.

Show solution and calculation

Approach: $I_{\mathrm{wave}}=\frac{E_{\mathrm{rms}}^2}{\mu_0c}$ .
RMS E is approximately 8.00 V/m; sinusoidal peak E is approximately 11.3 V/m.
Inserting RMS into the peak formula with factor one half would halve the intensity.

\begin{aligned}E_{\mathrm{rms}}&=\sqrt{I\mu_0c}\approx8{,}00\,\mathrm{V/m}\\E_0&=\sqrt2\,E_{\mathrm{rms}}\approx11{,}3\,\mathrm{V/m}\end{aligned}

[1]

Interpret · Application

E points along +x and B along −z at an instant. Which way does S point? What happens if both fields reverse simultaneously?

Show hint

Use the cyclic ordering of the x, y and z axes.

Learning goal: Determine energy-flow direction from vector orientation.

Show solution and calculation

Approach: $\mathbf S=\frac{\mathbf E\times\mathbf B}{\mu_0}$ .
S points along +y.
Reversing both fields leaves the cross product unchanged; reversing only one reverses S. This explains why sign changes of travelling-wave fields do not imply backward power.

\hat{\mathbf x}\times(-\hat{\mathbf z})=\hat{\mathbf y},\qquad(-\mathbf E)\times(-\mathbf B)=\mathbf E\times\mathbf B

[2]

Error analysis · Application

A travelling sinusoidal wave has zero mean E and zero mean B over a cycle. A calculation therefore concludes that average intensity is zero. Correct it.

Show hint

Form the cross product first, then average.

Learning goal: Detect incorrect averaging of field products.

Show solution and calculation

Approach: $\langle\mathbf S\rangle=\frac{\langle\mathbf E\times\mathbf B\rangle}{\mu_0}$ .
The mean of a product generally differs from the product of the means.
In-phase fields give a positive squared waveform whose mean contains a factor one half.
Mean fields alone do not retain this information.

\langle\mathbf S\rangle=\frac{1}{\mu_0}\langle\mathbf E\times\mathbf B\rangle=\frac{E_0B_0}{2\mu_0}\hat{\mathbf n}\ne\frac{\langle\mathbf E\rangle\times\langle\mathbf B\rangle}{\mu_0}

[3]

Transfer · Challenge

A fixed volume receives net electromagnetic inflow of 5 W. Simultaneously the field transfers 3 W to matter, entirely converted into heat there. How does stored field energy change?

Show hint

With an outward normal, net surface flux is −5 W.

Learning goal: Connect oriented flux balance with storage and dissipation.

Show solution and calculation

Approach: $\dot W_{\mathrm{field}}=-P_{\mathrm{out}}-P_{\mathrm{matter}}$ .
Field energy grows by 2 J per second.
Incoming power is partly dissipated and partly stored; equating all incoming power with heating would be incorrect.

\begin{aligned}\frac{\mathrm dW_{\mathrm{field}}}{\mathrm dt}&=-P_{\mathrm{out}}-P_{\mathrm{matter}}\\&=-(-5\,\mathrm W)-3\,\mathrm W\\&=2\,\mathrm W\end{aligned}

[2]

Build a reliable energy-flow calculation

What to take away

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant scientific connections; preserved existing field and energy balances, derivations, examples, exercises, figures and model limits.
October 1, 2026Added an original DE/EN article with assumptions, two original worked examples and five different solved exercises.
October 1, 2026Revised the introduction and learning sequence, added a third model example and quantitative diagram; reviewed existing exercises, assumptions and mathematical notation.

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Quanta. Poynting Vector and Electromagnetic Energy Flux. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/poynting-vektor

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Retain Poynting Vector and Electromagnetic Energy Flux for exams

Create a curated FSRS exam set for S = (E × B)/μ₀: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Poynting vector and electromagnetic energy flow

Variables & units

Poynting vector and intensity calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrodynamics

Worked example

Applications

How do you calculate with Poynting Vector and Electromagnetic Energy Flux?

From a cross product to power crossing an area

Explore further: How the energy balance follows from Maxwell

Separate instantaneous, peak and average values

Account for oblique incidence on an area

Find incident energy from field amplitude

Energy in the field of a DC cable

An ideal coaxial cable feeding a load

Why standing waves do not carry purely forward power

Standing wave: directed local energy flux

Standing wave: finite flux now, zero mean

Five exercises on direction, averaging and balance

At one point in vacuum, instantaneous field magnitudes are E=2.0×10−3 V/mE=2.0\times10^{-3}\,\mathrm{V/m}E=2.0×10−3V/m and B=3.0 μTB=3.0\,\mathrm{\mu T}B=3.0μT, separated by 30∘30^\circ30∘. Find instantaneous S magnitude.

A single travelling plane vacuum wave has average intensity I=0.170 W/m2I=0.170\,\mathrm{W/m^2}I=0.170W/m2. Find RMS E and, for a sinusoid, its approximate peak value.

E points along +x and B along −z at an instant. Which way does S point? What happens if both fields reverse simultaneously?

A travelling sinusoidal wave has zero mean E and zero mean B over a cycle. A calculation therefore concludes that average intensity is zero. Correct it.

A fixed volume receives net electromagnetic inflow of 5 W. Simultaneously the field transfers 3 W to matter, entirely converted into heat there. How does stored field energy change?

Build a reliable energy-flow calculation

What to take away

Sources for this explanation

Frequently asked questions about Poynting Vector and Electromagnetic Energy Flux

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Poynting Vector and Electromagnetic Energy Flux for exams

Poynting vector and electromagnetic energy flow

Variables & units

Poynting vector and intensity calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrodynamics

Worked example

Applications

How do you calculate with Poynting Vector and Electromagnetic Energy Flux?

From a cross product to power crossing an area

Explore further: How the energy balance follows from Maxwell

Separate instantaneous, peak and average values

Account for oblique incidence on an area

Find incident energy from field amplitude

Energy in the field of a DC cable

An ideal coaxial cable feeding a load

Why standing waves do not carry purely forward power

Standing wave: directed local energy flux

Standing wave: finite flux now, zero mean

Five exercises on direction, averaging and balance

At one point in vacuum, instantaneous field magnitudes are E=2.0×10−3 V/mE=2.0\times10^{-3}\,\mathrm{V/m}E=2.0×10−3V/m and B=3.0 μTB=3.0\,\mathrm{\mu T}B=3.0μT, separated by 30∘30^\circ30∘. Find instantaneous S magnitude.

A single travelling plane vacuum wave has average intensity I=0.170 W/m2I=0.170\,\mathrm{W/m^2}I=0.170W/m2. Find RMS E and, for a sinusoid, its approximate peak value.

E points along +x and B along −z at an instant. Which way does S point? What happens if both fields reverse simultaneously?

A travelling sinusoidal wave has zero mean E and zero mean B over a cycle. A calculation therefore concludes that average intensity is zero. Correct it.

A fixed volume receives net electromagnetic inflow of 5 W. Simultaneously the field transfers 3 W to matter, entirely converted into heat there. How does stored field energy change?

Build a reliable energy-flow calculation

What to take away

Sources for this explanation

Frequently asked questions about Poynting Vector and Electromagnetic Energy Flux

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Poynting Vector and Electromagnetic Energy Flux for exams

At one point in vacuum, instantaneous field magnitudes are $E=2.0\times10^{-3}\,\mathrm{V/m}$ and $B=3.0\,\mathrm{\mu T}$ , separated by $30^\circ$ . Find instantaneous S magnitude.

A single travelling plane vacuum wave has average intensity $I=0.170\,\mathrm{W/m^2}$ . Find RMS E and, for a sinusoid, its approximate peak value.

At one point in vacuum, instantaneous field magnitudes are $E=2.0\times10^{-3}\,\mathrm{V/m}$ and $B=3.0\,\mathrm{\mu T}$ , separated by $30^\circ$ . Find instantaneous S magnitude.

A single travelling plane vacuum wave has average intensity $I=0.170\,\mathrm{W/m^2}$ . Find RMS E and, for a sinusoid, its approximate peak value.