Energy Density of an Electric and a Magnetic Field

Q: Why is the magnetic energy density so much larger for ordinary fields?

Because the two are equal only when E = cB, and c is 3 × 108. A 1.5 T field would need an electric field of 4.5 × 108 V/m to match it, and air breaks down at 3 × 106 V/m, a hundred and fifty times less. The row above gives you that comparison for whatever B you enter.

Empty space with a field in it holds energy. ½ε₀E² joules per cubic metre for an electric field and B²/2μ₀ for a magnetic one — and for fields a school laboratory can reach, the magnetic one wins by a factor of tens of thousands. This gives both, their total, and the split inside a light wave.

Last updated 11 Nov 2026

Energy per cubic metre in E and in B

½ε₀E² + B²/2μ₀

Electric field strength (V/m)Three million volts per metre is roughly the breakdown field of dry air — the point at which a spark jumps — so it is about the strongest electric field you can set up in a room. Inside a capacitor’s dielectric you can do better, which is exactly why dielectrics exist.

Magnetic flux density (T)1.5 T is a clinical magnetic resonance scanner. A fridge magnet manages about 5 × 10⁻³ T, the Earth about 5 × 10⁻⁵ T, the strongest continuous laboratory magnets about 45 T. Energy density goes as B², so a 3 T scanner stores four times as much per cubic metre as a 1.5 T one.

Volume you want the total energy for (cm³)1000 cm³ is one litre. The conversion is cubed: 1000 cm³ = 10⁻³ m³. The energy density rows do not depend on this at all — only the total-energy rows do.

Intensity of an electromagnetic wave (W/m²)For the third part of the page. 1361 W/m² is the solar constant, the power per square metre arriving at the top of the Earth’s atmosphere. A bright room is a few W/m²; a focused laser can be megawatts. In a wave the energy is shared exactly equally between the electric and magnetic fields, which is not true of the two fields you typed above.

Two cubic metres of empty space, one with your electric field in it and one with your magnetic field. The shading is the energy density: each extra line appears when the density passes another threshold, so the box fills in as you turn the field up and empties as you turn it down. The bar underneath splits the total between the two, to scale. At ordinary laboratory fields the electric sliver is too thin to see, which is the point.

895286.40J/m³Example

3.0 × 10⁶ V/m, 1.5 T, a 1000 cm³ volume, and sunlight at 1361 W/m²

Two densities, one unit, and the wave that splits them evenly

u_E = ½ε₀E² · u_B = B²/2μ₀ · equal when E = cB · in a wave: u = I/c and u_E = u_B · E_rms = √(u/ε₀)

J/m³ = Pa: a joule per cubic metre is a newton metre per cubic metre, which is a newton per square metre, which is a pascal. Energy density and pressure are the same quantity. B²/2μ₀ is therefore the magnetic pressure — at 1.5 T it is about 9 atmospheres, which is why big magnets need serious mechanical restraint
the squares: both densities go as the field squared, so doubling a field quadruples the energy it holds. That is why the step from a 1.5 T scanner to a 3 T one is a much bigger engineering problem than it sounds
ε₀ and μ₀: 8.854 187 8188(14) × 10⁻¹² F/m and 1.256 637 061 27(20) × 10⁻⁶ N/A² (CODATA 2022). Both have carried an uncertainty since the 2019 SI revision. Their product is 1/c², which is where the E = cB condition comes from and why light travels at the speed it does
u = I/c: the energy density in a travelling wave of intensity I. The reasoning is as simple as it looks: in one second a square metre of wavefront delivers I joules, and those joules were spread through a column c metres long, so there are I/c of them per cubic metre
half and half: inside an electromagnetic wave in a vacuum, E and B are locked together by E = cB at every point and every instant, so the two energy densities are exactly equal. That is a property of a wave, not of fields in general — the two static fields you type into this page need not be anywhere near balanced

Worked example

3.0 × 10⁶ V/m, 1.5 T, a 1000 cm³ volume, and sunlight at 1361 W/m²
THE ELECTRIC PART. u_E = ½ε₀E² = 0.5 × 8.854 × 10⁻¹² × (3.0 × 10⁶)² = 0.5 × 8.854 × 10⁻¹² × 9.0 × 10¹² = 39.84 J/m³. Note how the 10⁻¹² and the 10¹² cancel; that is the usual shape of this sum.
THE MAGNETIC PART. u_B = B²/2μ₀ = 2.25/(2 × 1.2566 × 10⁻⁶) = 2.25/(2.513 × 10⁻⁶) = 895 247 J/m³. Twenty-two thousand times the electric figure, from a field that is nothing special while the electric one is already sparking.
ADD THEM. u = 39.84 + 895 247 = 895 286.40 J/m³, which is the headline. For practical purposes the electric contribution has vanished: it is 0.0045% of the total, and the bar in the diagram cannot draw it.
WHAT WOULD IT TAKE TO BALANCE THEM? The two are equal when E = cB = 3.0 × 10⁸ × 1.5 = 4.5 × 10⁸ V/m, which is 150 times the breakdown field of air. There is no way to set that up in a room, and that single fact is why energy storage in electrical engineering is magnetic.
THE TOTAL IN YOUR LITRE. 1000 cm³ = 10⁻³ m³ (cubed conversion), so the energy is 895 286 × 10⁻³ = 895.3 J. About the kinetic energy of a 2 kg brick thrown at 30 m/s, sitting invisibly in a litre of air inside a scanner's bore.
NOW THE SUNLIGHT, WHICH SPLITS EVENLY. u = I/c = 1361/(3.0 × 10⁸) = 4.54 × 10⁻⁶ J/m³, half electric and half magnetic. From u = ε₀E_rms² that gives E_rms = 716 V/m, a peak of 1013 V/m, and a peak B of 1013/c = 3.38 µT — a magnetic field weaker than the Earth's, carrying all the sunlight that falls on the planet.

What different fields are worth, per cubic metre

The row that surprises people is the fourth against the fifth. An electric field strong enough to tear air apart stores 40 joules per cubic metre; a magnetic field you can walk into stores 895 000. The reason is the factor of c² hidden in ε₀μ₀ — you need E = cB for the two to match, and c is very large. The last column is the same number read as a pressure, because a joule per cubic metre is a pascal.
Field	Energy density	Read as a pressure
Earth’s magnetic field, 5 × 10⁻⁵ T	9.9 × 10⁻⁴ J/m³	about 1 µPa
Fridge magnet at its face, 5 × 10⁻³ T	9.9 J/m³	about 10 Pa
Fair-weather electric field, 100 V/m	4.4 × 10⁻⁸ J/m³	negligible
Air about to spark, 3 × 10⁶ V/m	39.8 J/m³	about 40 Pa
Clinical MRI scanner, 1.5 T	895 000 J/m³	about 9 atmospheres
Sunlight at the top of the atmosphere	4.5 × 10⁻⁶ J/m³	4.5 µPa

Where the energy actually sits, and why the magnetic field usually wins

A field is not a bookkeeping device; it holds the energy. Charge a capacitor and the work you did is in the space between the plates, at ½ε₀E² joules per cubic metre. Build up a current in a coil and the work is in the field through it, at B²/2μ₀ per cubic metre. Both are densities of energy in a region, and you can locate them: the energy really is there, which is how an electromagnetic wave can carry energy across a vacuum with no charges and no currents anywhere along the way.

Those joules per cubic metre are also pascals. Check the units and you will see it: J/m³ = N·m/m³ = N/m². So B²/2μ₀ is a pressure, and at 1.5 T it is about nine atmospheres. That is not an analogy. It is why the windings of a large magnet have to be mechanically restrained, why a tokamak can squeeze plasma with nothing but a field, and why two parallel wires carrying current push each other about.

The magnetic field wins by a factor of c². Setting ½ε₀E² = B²/2μ₀ and using ε₀μ₀ = 1/c² gives E = cB. To match a single tesla you would need three hundred million volts per metre, a hundred times the field at which air tears itself apart into a spark. There is no way round that in a room, and it settles a practical question: energy in electrical engineering is stored magnetically, in inductors and transformers, not electrostatically. See self and mutual inductance for the coils that do it.

Inside a light wave the two are exactly equal, and that is a different statement. The two fields you type into this page are independent and need not balance at all. But in a wave travelling through a vacuum, E and B are locked together by E = cB at every point and every instant, so the two energy densities are identical and each carries half the total. The whole lot is u = I/c, which you can see without any algebra: a square metre receives I joules each second, and they arrive spread along a column c metres long. Absorb them and you feel a pressure of the same I/c — tiny, 4.5 micropascals for sunlight, and enough to push a solar sail or to shape a comet’s tail. For the field that stores the energy in the first place, see Gauss’s law and electric flux.

Frequently asked questions

Why is the magnetic energy density so much larger for ordinary fields?

Because the two are equal only when E = cB, and c is 3 × 10⁸. A 1.5 T field would need an electric field of 4.5 × 10⁸ V/m to match it, and air breaks down at 3 × 10⁶ V/m, a hundred and fifty times less. The row above gives you that comparison for whatever B you enter.

Is a joule per cubic metre really the same as a pascal?

Yes, exactly, and it is worth checking once by hand: J = N·m, so J/m³ = N·m/m³ = N/m² = Pa. The consequence is physical rather than cosmetic — B²/2μ₀ is the magnetic pressure a field exerts, and a 1.5 T field pushing at nine atmospheres is a real engineering load on the magnet that makes it.

Does the energy in a wave really split evenly between E and B?

In a vacuum, yes, exactly and at every instant, because E = cB holds point by point in a travelling wave. Substituting that into the two expressions makes them identical. It is not true of static fields, and it is not true inside a medium with a refractive index, where the relation between E and B changes.

What happens if I set one of the fields to zero?

That contribution becomes exactly zero and the total is just the other one — which is correct, not a failure. Set the electric field to zero and the rows that divide by it (the ratio and the percentage share) vanish from the page rather than printing infinity, because the engine drops a figure it cannot evaluate.

Where is this energy stored in a capacitor or an inductor?

In the field, in the space itself. For a parallel-plate capacitor the field is between the plates, and multiplying ½ε₀E² by the volume between them gives you back the familiar ½CV². For a solenoid, multiplying B²/2μ₀ by the volume inside gives back ½LI². Those are two routes to the same number, and getting them to agree is a good exercise — see the parallel-plate capacitor page.

Related calculators

References

Fundamental Physical Constants — Frequently used constants (2022 CODATA wallet card), NIST Committee on Data for Science and Technology, physics.nist.gov/cuu/pdf/wallet_2022.pdf (read 11 November 2026). Every constant on this page comes from here, and the thing worth noticing is which numbers have a bracket after them. The elementary charge is printed 1.602 176 634 × 10⁻¹⁹ C and the speed of light 299 792 458 m s⁻¹, both with no bracket at all, because since 2019 they are exact by definition — the coulomb and the metre are now defined from them. But the vacuum electric permittivity is printed 8.854 187 8188(14) × 10⁻¹² F m⁻¹ and the vacuum magnetic permeability 1.256 637 061 27(20) × 10⁻⁶ N A⁻²: both now carry an uncertainty of about 1.6 parts in 10¹⁰, where before 2019 μ₀ was exactly 4π × 10⁻⁷ and ε₀ exactly 1/μ₀c². Almost every school book still prints the old exact forms. At three significant figures nothing changes; which constants are definitions changed completely.
The International System of Units (SI), 9th edition 2019, version V4.01 of June 2026, Bureau International des Poids et Mesures, bipm.org/documents/d/guest/si-brochure-9-en-pdf (read 11 November 2026). Cited for the units, from Table 4 of coherent derived units with special names: magnetic flux, weber (Wb) = kg m² s⁻² A⁻¹ = V s; magnetic flux density, tesla (T) = kg s⁻² A⁻¹ = Wb/m²; inductance, henry (H) = kg m² s⁻² A⁻² = Wb/A. Those three equalities are worth more than they look: a weber is a volt-second, so a flux change of one weber in one second is one volt — which is Faraday’s law written as a unit. And a henry is a weber per ampere, so an inductance is just how much flux linkage one ampere produces.
University Physics Volume 2, chapters 11 and 14, OpenStax, Rice University (openstax.org/books/university-physics-volume-2, key-equations pages read 11 November 2026). An openly licensed university text, cited here because the coaching sites that dominate searches for this topic are not citable and mostly copy one another. Chapter 11 gives the radius of a charged particle’s path as r = mv/(qB), its period as T = 2πm/(qB), the Hall potential as V = IBl/(neA), the drift velocity in crossed fields as v_d = E/B, the magnetic dipole moment as μ = NIA and the torque as τ = μ × B. Chapter 14 gives L_solenoid = μ₀N²A/l, NΦ = LI, ε = −L dI/dt, M = N₂Φ₂₁/I₁ = N₁Φ₁₂/I₂ and U = ½LI². Neither key-equations page lists a cyclotron frequency or a field energy density, so those two are stated here from the definitions and not attributed to this source.
Annual Syllabus, Class XII, Physics, Code 42, Session 2025-26, Central Board of Secondary Education, edustud.nic.in/edu/SYLLABUS_2025_26/12/12_Physics_EM.pdf (read 11 November 2026). The board’s own document, cited and not endorsed by it. Unit I Chapter 1 asks for “Electric flux, statement of Gauss’s theorem and its applications to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell (field inside and outside)”. Unit III Chapter 4 asks for “force on a moving charge in uniform magnetic and electric fields”, “torque experienced by a current loop in uniform magnetic field” and “Current loop as a magnetic dipole and its magnetic dipole moment”. Unit IV Chapter 6 is one line and covers three of this batch’s records: “Electromagnetic induction; Faraday’s laws, induced EMF and current; Lenz’s Law, Self and mutual induction”. The Hall effect is not in this document.