Visualize the electric field, magnetic field and Poynting vector of a plane electromagnetic wave in vacuum. Adjust amplitude, frequency, position and time to learn how much power an EM wave actually carries.
Parameters
Field amplitude E_0
V/m
Frequency f
MHz
Observation position z
m
Time t
ns
Vacuum assumed (c = 2.998×10⁸ m/s, Z_0 = 376.73 Ω). The wave propagates in the +z direction.
Results
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Time-averaged intensity ⟨S⟩
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Magnetic flux amplitude B_0
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Wavelength λ
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Instantaneous Poynting S(z,t)
Electric, Magnetic and Poynting Fields
Blue = electric field E_x / green = magnetic field H_y / red = Poynting S_z (yellow vertical line = observation z)
Theory & Key Formulas
A plane electromagnetic wave traveling in vacuum in the +z direction has perpendicular electric and magnetic fields oscillating in phase, and carries energy in the propagation direction.
Electric and magnetic fields (c is the speed of light in vacuum):
With k = 2π/λ and λ = c/f. For E_0 = 1000 V/m and f = 1 GHz, ⟨S⟩ ≈ 1327 W/m² — close to the solar constant 1361 W/m².
What is the Poynting Vector Simulator
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We learned that "radio waves are oscillating electric and magnetic fields", but in the end, how much energy does that actually carry?
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Roughly speaking, the energy that a plane electromagnetic wave carries through unit area per second (in W/m²) is the "Poynting vector": $S = E\times H$. The cross product points in the propagation direction. Set the field amplitude E_0 to 1000 V/m in the simulator above — the ⟨S⟩ card should read about 1327 W/m². That is essentially the same as sunlight at the Earth (the solar constant, about 1361 W/m²).
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Wait, the sunlight's electric field is just 1000 V/m? That seems small.
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It does. To get a picture of "thousands of watts streaming through one square meter", a mere 1000 V/m is enough. The reason is the vacuum impedance Z_0 = 376.73 Ω in the denominator. Since $\langle S\rangle = E_0^2/(2Z_0)$, doubling the field quadruples the power flux. Set E_0 to 2000 V/m in the simulator — ⟨S⟩ jumps four times to about 5300 W/m².
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There are two readings, "instantaneous" and "time-averaged". Are they different?
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An important distinction. The instantaneous value $S(z,t) = E_x(z,t)\cdot H_y(z,t)$ oscillates as $\cos^2(kz-\omega t)$ between 0 and $E_0^2/Z_0$. Averaged over a long enough time, $\cos^2$ averages to 1/2, giving ⟨S⟩ = E_0²/(2Z_0). When you press "Advance Time" in the simulator, you see the red S_z curve always above zero, pulsating. Even when E (blue) or H (green) go negative, their product is cos²-shaped and stays positive.
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So what changes when I change the frequency?
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Interestingly, ⟨S⟩ does not depend on the frequency — look at the formula, f does not appear. At 100 MHz or 10 GHz, for the same field amplitude the time-averaged power flux is the same. What changes is the wavelength λ = c/f. In the simulator, going from 1 GHz to 10 GHz shrinks the wavelength from 30 cm to 3 cm and the wave packs together more tightly. Picture it as the "fringe spacing" getting finer, not the energy transfer itself.
Frequently Asked Questions
As "energy (joules) crossing one square meter in one second". For example, a 1000 W/m² electromagnetic wave passing through a 1 m² antenna aperture delivers 1000 J = 1 kJ of energy per second. When sizing a photovoltaic system, multiplying the solar constant 1361 W/m² by the panel area and efficiency to estimate output is a typical engineering use of this quantity.
From Maxwell's equations one derives $\partial u/\partial t + \nabla\cdot\mathbf{S} = -\mathbf{J}\cdot\mathbf{E}$, meaning "change of EM energy density u + divergence of power flux = dissipated Joule heat". For a closed surface, the outward Poynting flux equals the decrease of internal energy plus the heat generated — exactly the conservation of energy. S is not just convenient: it is the fundamental quantity describing how energy flows through space.
Electromagnetic waves also carry momentum. The radiation pressure on a perfect absorber at normal incidence is $p = \langle S\rangle/c$, twice that on a perfect reflector. Sunlight (⟨S⟩ ≈ 1361 W/m²) gives only about 4.5 μPa, but solar sails use it to propel spacecraft. Optical tweezers and laser cooling are also applications of this momentum transfer, capable of trapping and cooling nanoparticles and atoms.
Solving ⟨S⟩ = E_0²/(2Z_0) gives the field amplitude corresponding to the solar constant: E_0 = √(2·Z_0·1361) ≈ 1012 V/m. Setting E_0 = 1000 V/m in the simulator gives ⟨S⟩ ≈ 1327 W/m², very close. The corresponding magnetic flux amplitude is B_0 = E_0/c ≈ 3.4 μT — about one tenth of the geomagnetic field (~30 μT) — a useful way to compare EM-wave field amplitudes to everyday quantities.
Real-World Applications
Communications and antenna design: The received power is the product of the antenna's effective area A_eff and the incident power flux ⟨S⟩. From cellular base stations to smartphones, the Friis equation P_r = P_t·G_t·G_r·(λ/4πd)² is built on the Poynting vector as a fundamental quantity. Even Wi-Fi signal strength (dBm) is ultimately a quantity derived from W/m².
Solar power and thermal engineering: The time-averaged intensity of sunlight reaching the ground (about 1000 W/m² after the atmosphere) multiplied by the conversion efficiency gives the generated power. Concentrated solar furnaces use mirrors to focus ⟨S⟩ several hundred times, producing temperatures of thousands of degrees to melt metals or drive solar-thermal chemistry.
Laser engineering and material processing: The spot diameter and total power of a laser beam yield the Poynting flux W/m² across the beam cross-section. Focusing a 10 W laser to a 10 μm spot gives ⟨S⟩ ≈ 1.3×10¹¹ W/m² — surpassing the surface of the Sun in order of magnitude and enabling instantaneous vaporization of metal for machining.
EM safety standards and exposure assessment: Human-safety evaluation of wireless devices measures or computes the spatial Poynting flux in W/m² and compares it against SAR (specific absorption rate) limits or the ICNIRP guideline values. The quantity also plays a central role in controlling EM-field leakage around 5G base stations and microwave heating equipment.
Common Misconceptions and Cautions
The most common misconception is to think that "looking at the electric field alone tells you the wave's energy". Electric and magnetic fields oscillate in phase, but the energy is carried by the cross product $\mathbf{S} = \mathbf{E}\times\mathbf{H}$. Neither the electric field alone nor the magnetic field alone has any "power flux". The vacuum impedance Z_0 = 376.73 Ω is the coefficient that links them: $H = E/Z_0$ and $S = E^2/Z_0$. The magnetic-field stat card is shown independently in the simulator because the two fields carry power as "one coupled wave".
The next common mistake is to confuse instantaneous and time-averaged values. The instantaneous Poynting vector $S(z,t) = E_x H_y$ has cos²(kz - ωt) form: the peak is $E_0^2/Z_0$ and the average is half that, $E_0^2/(2Z_0)$. For E_0 = 1000 V/m and f = 1 GHz the instantaneous peak is 2654 W/m² but the time average is 1327 W/m² — a factor of 2 apart. "Signal strength" or "light intensity" almost always means the time average ⟨S⟩. Mixing it up with the instantaneous peak doubles the power estimate.
Finally, remember that this formula is for an idealized "plane wave in vacuum". Real radio waves spread as spherical waves, attenuating as 1/r² with distance. In a medium the impedance changes (Z = Z_0/n), and dielectric or conductive losses introduce damping. Add reflection, refraction and diffraction, and standing waves and interference fringes complicate the spatial distribution of ⟨S⟩. The simulator shows the simplest case — a single-frequency plane wave propagating through infinite vacuum — and from there, real problems are built by adding medium properties and boundary conditions one at a time.