Energy, Intensity & Radiation Pressure

Here is the reveal this module exists for: two formulas you learned as circuit facts were field facts all along. The capacitor chapter's energy density $\tfrac12\varepsilon_0E^2$ never mentioned plates in its final form — it prices any electric field, anywhere. The inductor chapter's $B^2/2\mu_0$ prices any magnetic field. An electromagnetic wave is both fields traveling together, so a beam of light is a moving warehouse of exactly those two stores — and, remarkably, they contribute equal shares, everywhere, always. From that warehouse picture the whole module unrolls with algebra. The amplitudes of the two fields are locked by the exchange rate $E_0/B_0 = c$ (which is why nobody talks about light's magnetic field — it is there, just numerically tiny in SI units). The intensity of a wave — watts crossing each square metre — is the energy density times the delivery speed, $I = \bar u c$ , and for a point source it dilutes over spheres as $I = P/4\pi r^2$ . And because the wave carries momentum as well as energy ( $p = U/c$ , quoted honestly), light pushes: absorb it and feel $I/c$ of pressure, mirror it back and feel twice that — the factor of two your momentum chapter already taught you with balls bouncing off walls. The payoff numbers are worth the trip: sunlight at Earth is a kilovolt-per-metre electric field riding with a few microtesla of magnetic field, pressing on your roof with micropascals — feeble here, yet strong enough to steer comet tails and sail spacecraft. This is JEE territory; the wave-numericals gym next door drills the chains.

Key takeaways

A light beam carries the two stores you already priced: $u_E = \tfrac12\varepsilon_0E^2$ (capacitor module's density) and $u_B = B^2/2\mu_0$ (inductor module's) — in a wave they are equal, always.
Amplitudes are locked: $E_0/B_0 = c$ — a 1 kV/m electric comb rides with only ~3.3 μT of magnetic comb.
Intensity: $I = \bar u\,c$ (energy density × delivery speed); from a point source, $I = P/4\pi r^2$ .
Fields in intensity formulas take rms values: $E_{\text{rms}} = E_0/\sqrt2$ — the AC chapter's rms discipline, reused verbatim.
Radiation pressure: $I/c$ if absorbed, $2I/c$ if reflected — the bounce doubles momentum transfer, exactly like a ball off a wall.
Key idea: light is a moving warehouse of equal electric and magnetic field energy, delivering power $I = \bar uc$ and momentum $I/c$ per square metre — so it pushes whatever it lands on.

The core insight

An electromagnetic wave transports energy with density $u = \tfrac12\varepsilon_0E^2 + B^2/2\mu_0$ (the two halves equal, since $E = cB$ and $c^2 = 1/\mu_0\varepsilon_0$ ), delivering intensity $I = \bar u c$ that dilutes as $P/4\pi r^2$ from a point source; because the wave also carries momentum $U/c$ , it exerts radiation pressure $I/c$ on an absorber and $2I/c$ on a mirror.

Where this lives today

Radiation pressure runs both the largest and smallest machines. Largest: inside stars, photon pressure props up the core against gravity, and for the most massive stars it sets the very limit of how bright a star can be before it blows itself apart (the Eddington limit). Smallest: optical tweezers — focused laser beams that hold and steer single cells and molecules by light pressure alone — earned the 2018 Nobel Prize and now run in thousands of biology labs. In between: solar sails (IKAROS flew, LightSail flew, missions are queued), laser cooling of atoms (the momentum kick per photon, harvested billions of times a second, chills atoms to microkelvins), and the LIGO mirrors, whose engineers must budget for the pressure of their own laser. Poynting's energy-flow bookkeeping, meanwhile, is the daily grammar of RF and antenna engineering.

The reveal — two old formulas were about light all along

Where does a wave keep its energy? In its fields — and you have already priced both. When the capacitor-energy module compressed $\tfrac12CV^2$ into a statement about the space between the plates, the punchline was a density: every cubic metre of electric field $E$ stores $u_E = \tfrac12\varepsilon_0E^2$ . When the inductor-energy module did the same for $\tfrac12LI^2$ , the density came out $u_B = B^2/2\mu_0$ . Nothing in either final formula refers to plates or coils — they are properties of fields, discovered via hardware. A traveling electromagnetic wave is nothing but $E$ and $B$ in flight; its energy content is the sum of the two densities, comb by comb, crest by crest. Now the elegant bookkeeping fact, and the flagship algebra of this module: the two stores are equal partners.

Derivation AThe Fifty Fifty Split

The goal is to show that in an electromagnetic wave the magnetic energy density equals the electric energy density — exactly, at every point and instant — using only the amplitude lock $E = cB$ and the speed identity $c^2 = 1/\mu_0\varepsilon_0$ . The payoff is threefold: a beloved one-marker ("the energy is shared equally"), a doubling shortcut for total density, and the reason the "weak" magnetic field is an illusion of units rather than of energy.

Step 1. Start with the magnetic density and substitute the amplitude lock $B = E/c$ (the wave's fields are tied at every instant, not just at peaks):

u_B = \frac{B^2}{2\mu_0} = \frac{E^2}{2\mu_0 c^2}

Why substitute toward $E$ : we want to compare with $u_E$ , which is written in terms of $E$ — one variable, one comparison.

Step 2. Replace $c^2$ by $1/\mu_0\varepsilon_0$ (the previous module's central result, inverted):

u_B = \frac{E^2}{2\mu_0} \times \mu_0\varepsilon_0 = \frac{1}{2}\varepsilon_0 E^2

Why the $\mu_0$ cancels: the same two constants that set the wave's speed also calibrate the two energy prices — the speed identity is secretly an energy-balance statement.

Step 3. Compare: $u_B = \tfrac12\varepsilon_0E^2 = u_E$ . Equal, with no approximation and no averaging — the split holds crest, flank and zero alike. Total density is therefore $u = 2u_E = \varepsilon_0E^2$ instantaneously, and averaging over a cycle (the mean of the squared sine is $\tfrac12$ — the rms charter from the AC chapter, cited) gives $\bar u = \tfrac12\varepsilon_0E_0^2 = \varepsilon_0E_{\text{rms}}^2$ . Verified: sunlight numbers — $E_{\text{rms}} = 716$ V/m gives $u_E = \tfrac12\varepsilon_0E_{\text{rms}}^2 = 2.27\times10^{-6}$ J/m³, and $B_{\text{rms}} = 2.39\ \mu$ T gives $u_B = B_{\text{rms}}^2/2\mu_0 = 2.26\times10^{-6}$ J/m³ — equal to rounding ✓ (recomputed).

The units illusion is worth one explicit paragraph, because it feeds a standard trap. Sunlight's $E_0 \approx 1000$ V/m sounds mighty; its $B_0 \approx 3.4\times10^{-6}$ T sounds negligible (Earth's own field is ten times bigger). Yet their energies are identical — volts-per-metre and tesla are different-sized units, and the exchange rate $E_0/B_0 = c$ is enormous. "The magnetic component of light is weak" is false in the only currency that matters; what is true is that magnetic forces on slow charges are down by $v/c$ (the Lorentz-force module's hierarchy), which is why detectors and matter respond mainly to the electric comb.

Intensity — energy density in delivery

A wave's practical rating is its intensity $I$ : energy crossing a unit area per second, in W/m². The picture that produces the formula: stand a one-square-metre window in the beam. In one second, all the energy within a column $c$ metres long marches through it — the wave delivers its warehouse at speed $c$ . So

I = \bar u\, c = \frac{1}{2}\varepsilon_0 E_0^2\, c

where $\bar u$ is the cycle-averaged energy density and the second form substitutes the derivation's result. Reading: intensity is density times speed — the same logic as traffic flow (cars per km × km per hour = cars per hour). Given any one of $I$ , $E_0$ , $B_0$ , the other two follow from this and the exchange rate; that three-way conversion is the wave-numericals gym's bread and butter. For a point source radiating power $P$ equally in all directions, the energy crosses ever-larger spheres: at distance $r$ , area $4\pi r^2$ , so

I = \frac{P}{4\pi r^2}

— the inverse-square dilution you have met twice before (gravity's field, Coulomb's field), now for light. Double the distance, quarter the intensity; it is geometry, not absorption.

Figure

The delivery picture — a one-second column of energy, and the inverse-square dilution

Momentum and pressure — light pushes

The wave carries momentum along with energy. The relation, quoted honestly: for electromagnetic radiation, momentum delivered is energy delivered divided by $c$ ,

p = \frac{U}{c}

where $U$ is the absorbed energy. (Why exactly $1/c$ ? The full argument needs either the Lorentz force on the absorber's charges worked in detail or relativity's energy-momentum relation — both beyond the HS ceiling; the formula itself is the examinable fact.) Reading: momentum flux is intensity over $c$ — small, because $c$ is huge, which is why sunlight doesn't knock you over. Pressure follows from the momentum chapter's oldest routine — the ball against the wall. Radiation hitting an area $A$ for time $t$ brings momentum $IAt/c$ . If the surface absorbs (blackened), it swallows that momentum: force $= IA/c$ , pressure

p_{\text{abs}} = \frac{I}{c}

If the surface is a mirror, the wave bounces back with reversed momentum, so the surface must supply twice the momentum change — exactly the elastic-ball doubling from the momentum chapter, cited not re-derived:

p_{\text{ref}} = \frac{2I}{c}

Figure

The factor of two — the momentum chapter's wall bounce, replayed with light

Partial absorber/reflector? Weight the two terms by the fractions — a favourite JEE twist. The classic corollaries: comet tails stream away from the Sun partly because sunlight pushes the dust (Kepler's four-century-old hunch); solar sails are mirrors chosen precisely for the factor of two; and the measurement that made the formulas respectable is the story of this gym's neighbour module's physicist, Lebedev.

Sunlight, fully priced — the flagship numbers

The solar constant — sunlight's intensity just above Earth's atmosphere — is $I \approx 1.36$ kW/m². Run the whole module on it (each step recomputed):

quantity	chain	value
rms electric field	$E_{\text{rms}} = \sqrt{I/\varepsilon_0 c}$	$716$ V/m
peak electric field	$E_0 = \sqrt2\,E_{\text{rms}}$	$\approx 1.01\times10^3$ V/m
peak magnetic field	$B_0 = E_0/c$	$3.4\ \mu$ T
energy density (avg)	$\bar u = I/c$	$4.5\ \mu$ J/m³
pressure, absorbed	$I/c$	$4.5\ \mu$ Pa
pressure, mirrored	$2I/c$	$9.1\ \mu$ Pa

A kilovolt per metre of oscillating field, a magnetic comb thinner than Earth's field, an energy density of microjoules per cubic metre, and a push of micropascals — about $45\ \mu$ N on a $10$ m² roof, the weight of a few grains of sand. Feeble at human scale; decisive at astronomic scale, where it sculpts comet tails and can carry a sail to the outer planets fuel-free.

Worked examples

Board level — fields of a light bulb (JEE-Main pattern, ~2.5 minutes)

A 100 W bulb radiates uniformly in all directions (treat all the power as light). At 3.0 m, find (a) the intensity, (b) the peak electric field, (c) the peak magnetic field.

Step 1. Intensity by the swelling sphere: $I = \dfrac{P}{4\pi r^2} = \dfrac{100}{4\pi(3.0)^2} = \boxed{0.88\ \mathrm{W/m^2}}$ .

Step 2. Peak electric field from $I = \tfrac12\varepsilon_0E_0^2c$ : $E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = \sqrt{\dfrac{2\times0.88}{8.854\times10^{-12}\times3\times10^8}} = \boxed{26\ \mathrm{V/m}}$ .

Step 3. Exchange rate: $B_0 = \dfrac{E_0}{c} = \dfrac{26}{3\times10^8} = \boxed{8.6\times10^{-8}\ \mathrm{T}}$ . Sanity: a household bulb's light field is tens of volts per metre and a tenth of a microtesla — the anchor band worth remembering for option-killing.

Competitive level — the sail and the roof (JEE-Adv flavour, ~3.5 minutes)

Sunlight of intensity 1.36 kW/m² falls normally on (a) a 10 m² perfectly absorbing rooftop panel and (b) a 10 m² perfectly reflecting solar-sail panel of mass 20 g floating freely in space. Find the force on each, and the sail's acceleration. (c) Explain in one line why doubling the sail's area does not change its acceleration if the sail's mass is dominated by its area.

Step 1. Absorber: $F = \dfrac{IA}{c} = \dfrac{1360\times10}{3\times10^8} = \boxed{4.5\times10^{-5}\ \mathrm{N}}$ — the weight of a few sand grains, holding down your roof.

Step 2. Mirror: $F = \dfrac{2IA}{c} = \boxed{9.1\times10^{-5}\ \mathrm{N}}$ ; acceleration $a = F/m = \dfrac{9.1\times10^{-5}}{0.020} = \boxed{4.5\times10^{-3}\ \mathrm{m/s^2}}$ . Small — but applied for a month ( $2.6\times10^6$ s) it accumulates $\Delta v \approx 12$ km/s with zero fuel, which is the whole business case for sails.

Step 3. If mass scales with area ( $m = \sigma A$ for areal density $\sigma$ ), then $a = 2IA/(c\sigma A) = 2I/c\sigma$ — area cancels. Sails improve by getting thinner (smaller $\sigma$ ), not bigger: the one-line insight JEE setters reward. Timing note: about 3.5 minutes. The factor-of-two word ("absorbing" vs "reflecting") is the first thing to underline; the area-cancellation in (c) is a limits-gate read (the capstone module's Gate 2 instinct) rather than a computation.

Exam traps

rms versus peak, again. $I = \tfrac12\varepsilon_0E_0^2c$ uses the peak; the $\tfrac12$ is the cycle average. Equivalently $I = \varepsilon_0E_{\text{rms}}^2c$ with no half. Mixing conventions costs a factor of 2 — same discipline as the AC chapter.

The magnetic field is not "weak". Equal energy share, tiny tesla-number. Options claiming light's energy is "mostly electric" die by the fifty-fifty derivation.

Inverse square is geometry. $I = P/4\pi r^2$ assumes a point source and no absorption; a laser beam does not dilute this way (its area is fixed) — read the source type before reaching for $4\pi r^2$ .

Pressure needs the right factor. $I/c$ , $2I/c$ , or $(1+\rho)I/c$ — underline the surface word first.

Force, not pressure, moves things. Multiply by the illuminated area, and for a sail at angle, only the normal component counts (setters love the cosine).

Boards and depth. This module is JEE-Main and JEE-Advanced territory, tagged accordingly: the field-intensity conversions, energy densities and radiation-pressure numericals are their standard fare (Advanced restored EM waves to its syllabus in 2023 and prices the momentum bookkeeping). CBSE, ISC and NEET syllabi keep EM-wave characteristics qualitative — NCERT states that waves carry energy and momentum and even quotes $p = U/c$ as an elementary fact, so a one-line conceptual question is fair there, but the quantitative chains above are not; those students should still read the sunlight table once (it recurs as general-knowledge bait). The Poynting vector — the full directional version of $I$ — is undergraduate machinery, named here only as the going-further label.

Origins · 1 minJohn Henry Poynting and how this was discovered

John Henry Poynting

1852–1914 · Mason College, Birmingham

Poynting answered a question nobody had thought to ask precisely: where does electrical energy actually travel? The obvious answer — "through the wires, with the current" — turned out to be wrong. In his 1884 paper "On the Transfer of Energy in the Electromagnetic Field," working directly from Maxwell's freshly minted equations, he showed that energy flows through the field around the apparatus, at a rate and direction fixed by the two fields jointly (the vector that now carries his name — this module's $I = \bar uc$ is its magnitude for a plane wave). A battery lighting a bulb pours energy out of the space surrounding the circuit into the filament, the wires acting more like rails than pipes — a picture so counterintuitive that it still startles engineering students, and so correct that antenna design has relied on it ever since. Poynting spent his career at Birmingham as the quietest kind of first-rate physicist: he also spent twelve years on a torsion-balance measurement of the gravitational constant (the "weighing the Earth" tradition — a mechanics-track story), chaired the committee that standardized his friend's — J. J. Thomson's — electrical units, and taught that energy bookkeeping, done honestly, is physics' sharpest tool. This module is one long exercise in taking him at his word.

History

1867

— Maxwell's theory implies light should press on matter; nobody can measure anything so small

1873

— Crookes' radiometer spins the wrong way (gas effects, not light pressure) and confuses a decade

1884

— Poynting derives the field's energy-flow law from Maxwell's equations

1900

— Lebedev's torsion balance in Moscow measures light pressure on mirrors, absorbed-vs-reflected factor included

1903

— Nichols and Hull at Dartmouth confirm to a few percent; radiation pressure becomes textbook fact

2010

— IKAROS deploys the first successful interplanetary solar sail — the factor of two, flown

Going further

The two energy densities come from Energy Stored in a Capacitor and Energy in an Inductor & the Magnetic Field — re-read their closing density paragraphs with this module's eyes. The rms discipline is AC Signals: Peak, Average & RMS; the bounce factor is the momentum chapter's wall collision. The wave anatomy and $c = 1/\sqrt{\mu_0\varepsilon_0}$ are The Nature of Electromagnetic Waves. Next door, the EM-Wave Numericals gym drills every chain in this module to exam speed, and the spectrum modules place the waves being priced. The directional energy-flow formalism — the Poynting vector proper — and the Lorentz-force derivation of $U/c$ are undergraduate fare (UG deep dive: energy density and the Poynting vector; momentum and radiation pressure). Provenance: downshifted from the undergraduate energy-density/Poynting and momentum/radiation-pressure articles (rewritten to HS algebra; vector machinery honestly flagged out); all numbers — solar-constant table, bulb fields, sail forces — recomputed for this article.

If light pushes things, why don't I feel sunlight pressing on me?

Scale. Full sunlight presses with about $4.5\ \mu$ Pa — atmospheric pressure is $10^5$ Pa, twenty-two billion times larger; the force on your whole body is under a tenth of a milligram-weight. You feel sunlight's energy (1.36 kW/m² is a serious heater) because your skin integrates watts efficiently, but its momentum arrives divided by $c$ , and $c$ is enormous. The push only matters where competing forces are feeble (dust grains in space, sails, levitated nanoparticles) or where the intensity is monstrous (laser fusion capsules, star interiors).

Where exactly is the energy in the wave — the crests, the zeros, or spread evenly?

Instantaneously, it rides the crests: $u = \varepsilon_0E^2$ is maximal where the combs peak and zero where they cross zero (both fields vanish together — the in-phase fact from the nature module). The pattern of energy clumps translates at $c$ , which is why the average over a cycle, $\bar u = \tfrac12\varepsilon_0E_0^2$ , is the number that enters intensity. This is the traveling-wave answer to the LC-instinct confusion: nothing sloshes between $E$ and $B$ locally; equal shares of energy arrive and leave together.

Why is momentum U/c and not something with the wave's speed squared, like kinetic energy?

Because light is not a slow ball — the nonrelativistic $p = mv$ , $K = \tfrac12mv^2$ ledger doesn't apply to something with no mass moving at $c$ . The honest sources of $p = U/c$ are (a) working the Lorentz force on the absorbing surface's charges through a full cycle — the fields conspire to push forward with exactly $1/c$ of the energy rate — or (b) relativity's general relation $E^2 = (pc)^2 + (mc^2)^2$ with $m = 0$ , giving $p = E/c$ directly. Both live beyond the HS ceiling, so the relation is quoted; the modern-physics track will earn it via photons, where $p = h/\lambda$ says the same thing packet by packet.

For the fields of a bulb at some distance, why do we use the point-source formula rather than the marching column?

You use both, in sequence — they answer different questions. $P/4\pi r^2$ tells you how much intensity survives the geometric dilution at your distance (the swelling-sphere picture). Then $I = \tfrac12\varepsilon_0E_0^2c$ converts that local intensity into local field amplitudes (the marching-column picture, inverted). The chain $P \to I \to E_0 \to B_0$ is the gym's canonical pipeline. Only for a beam (laser) does the first step change — divide by the beam's fixed area instead of the sphere's.

My board is CBSE/NEET — this module is tagged JEE-only, so can I skip it entirely?

Skim, don't skip. The rationalized CBSE/NEET syllabus keeps EM-wave characteristics qualitative, so the quantitative chains here (field-intensity conversions, pressure numericals) are genuinely out of your scope — that is why the tags say JEE-Main/JEE-Advanced. But NCERT's own chapter states, as elementary facts, that EM waves carry energy and momentum and quotes the momentum relation — so a one-line conceptual question ("do EM waves exert pressure? explain") is fair game on a board paper, and the fifty-fifty energy split plus the sunlight anchors make excellent assertion-reason armour. Twenty minutes on the reveal section, the warning box, and the sunlight table is the right investment; leave the worked chains to the JEE aspirants.

Do radio waves also exert pressure, or only light?

Every electromagnetic wave does — the ledger $p = U/c$ is band-blind. A 50 kW broadcast antenna pushes on the landscape with the same $I/c$ arithmetic; the numbers are just microscopic because broadcast intensities at any distance are tiny. At the other extreme, the pressure of intense radio-frequency fields is a real engineering quantity inside particle-accelerator cavities. The spectrum module's ladder changes photon energy per packet, but pressure cares only about total intensity — a good example of which properties are per-photon (ionization) and which are collective (heating, pressure).