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Most electromagnetic waves questions ask for a wavelength or a field equation. A smaller group asks a harder thing: how much energy is in the wave, and how fast it arrives. Those questions need two quantities, energy density and intensity, and the NCERT chapter does not spell either of them out. This post builds both from the field formulas you already know, works three problems with numbers, and shows where students lose the factor of a half.
Is this on the syllabus?
The Syllabus for JEE (Main) 2026 lists "electromagnetic waves and their characteristics" in Unit 15, and the NEET (UG) 2026 syllabus uses the same wording. Neither names energy density or intensity.
The 2026-27 reprint of the NCERT Class 12 chapter Electromagnetic Waves says these waves "carry energy from one place to another" but has no section deriving energy density or intensity. Exercise 8.10(c) still asks you to "show that the average energy density of the E field equals the average energy density of the B field". So the idea is expected, even though the derivation is not in the text. Check the syllabus for your exam year.
Where does the energy in an EM wave sit?
Use two results from electrostatics and magnetism:
Energy per unit volume in an electric field: u_E = ½ε₀E².
Energy per unit volume in a magnetic field: u_B = B²/(2μ₀).
In a plane wave in vacuum, B = E/c and c² = 1/(μ₀ε₀). Substitute: u_B = E²/(2μ₀c²) = ½ε₀E². The two shares are equal at every instant, which is the result Exercise 8.10(c) asks for. The total is u = ε₀E².
The field oscillates, so E² averages to E₀²/2 over a cycle. The average total energy density is
⟨u⟩ = ½ε₀E₀² = B₀²/(2μ₀)
How does energy density become intensity?
Intensity is the energy crossing one square metre per second, at right angles to the beam. In one second, the energy in a column of length c and area 1 m² passes through. So
I = ⟨u⟩c = ½cε₀E₀²
The unit check works: J/m³ × m/s = W/m².
Worked example 1: a plane wave with sample numbers
A plane wave in vacuum has E₀ = 60 V/m (sample value). Take ε₀ = 8.85 × 10⁻¹² F/m and c = 3.0 × 10⁸ m/s.
B₀ = E₀/c = 60 / 3.0 × 10⁸ = 2.0 × 10⁻⁷ T.
Average electric energy density: ¼ε₀E₀² = 0.25 × 8.85 × 10⁻¹² × 3600 = 7.97 × 10⁻⁹ J/m³.
Average magnetic energy density: B₀²/(4μ₀) = (4.0 × 10⁻¹⁴) / (4 × 1.257 × 10⁻⁶) = 7.96 × 10⁻⁹ J/m³. Equal, as expected.
Total ⟨u⟩ = 1.59 × 10⁻⁸ J/m³.
I = ⟨u⟩c = 1.59 × 10⁻⁸ × 3.0 × 10⁸ = 4.78 W/m².
Worked example 2: how strong is the field in sunlight?
NASA's Earth fact sheet gives the solar irradiance at Earth as 1361.0 W/m². Take I = 1361 W/m² and treat sunlight as a single plane wave, as exam questions do (real sunlight is a mix of many waves, so this is an effective amplitude). Use the CODATA value from NIST, ε₀ = 8.854 × 10⁻¹² F/m, and c = 2.998 × 10⁸ m/s.
E₀ = √(2I / cε₀) = √(2 × 1361 / (2.998 × 10⁸ × 8.854 × 10⁻¹²)) ≈ 1013 V/m.
B₀ = E₀/c ≈ 3.38 × 10⁻⁶ T, about 3.4 μT.
⟨u⟩ = I/c ≈ 4.54 × 10⁻⁶ J/m³.
The rms field is E₀/√2 ≈ 716 V/m.
So the electric field of sunlight is about 1 kV/m, yet the energy in each cubic metre is only a few microjoules. Divide that same intensity by c and you also get the radiation pressure on an absorber, which our post on photon momentum and radiation pressure works through.
Worked example 3: a point source and the inverse square law
A small source radiates 10 W uniformly in all directions (sample value). Find I and E₀ at 2.0 m.
The power spreads over a sphere: I = P/(4πr²) = 10 / (4π × 4.0) = 0.199 W/m².
E₀ = √(2I / cε₀) ≈ 12.2 V/m.
At 4.0 m, I drops by a factor of 4 but E₀ only by a factor of 2, to about 6.1 V/m, because I goes as E₀².
Common mistakes and how to avoid them
Dropping the half. ε₀E² is the instantaneous total. The cycle average is ½ε₀E₀². If a question gives rms values, ⟨u⟩ = ε₀E_rms² with no half.
Thinking the magnetic share is small. B₀ is small in tesla, but μ₀ is small too. The two shares are always equal in a plane wave in vacuum.
Mixing power and intensity. Power is in watts. Intensity is watts per square metre. Divide by area first.
Halving E₀ when distance doubles. For a point source E₀ goes as 1/r, so doubling r halves E₀ and quarters I.
Forgetting the medium. In a medium the speed is 1/√(με), as NCERT notes, so do not use c without checking.
If the link between E, B and energy is still shaky, the Fermi Concept Library covers every concept in physics, each with its own lesson, so you can revise field energy before going back to waves. For the spectrum and field equations, see our guide to electromagnetic waves and board questions.
A practice routine for this topic
Derive u_B = u_E for a plane wave from B = E/c, on paper, without notes.
Solve NCERT Exercise 8.10 in full, including part (c).
Do three E₀ → I conversions and three I → E₀ conversions with different numbers.
Do two point-source problems that ask how E₀ and I change with distance.
Write one line for each answer saying whether you used peak or rms values.
Tonight, find E₀ for a beam of intensity 100 W/m² and check that your answer is about 274 V/m.
