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Maxwell's equations and electromagnetic waves

Learn the Gauss, Faraday and Ampère–Maxwell laws in integral form, compute symmetric fields and derive the electromagnetic wave equation that gives c = 1/√(μ₀ε₀).

Check yourself
In this lesson you will learn
  • Write the four Maxwell equations in integral form and explain their meaning
  • Use the Gauss and Ampère laws to compute symmetric fields
  • Understand the role of the displacement current
  • Calculate the speed and fields of an electromagnetic wave

In the 1860s James Clerk Maxwell combined all known laws of electricity and magnetism into four equations and drew an astonishing conclusion from them: there must be electromagnetic waves that travel through empty space at the speed of light. So light itself is an electromagnetic wave. In the late 1880s Heinrich Hertz produced such waves in the laboratory. Radio, Wi-Fi, mobile networks and X-rays are all solutions of Maxwell's equations.

Flux and circulation

Maxwell's equations are written with two kinds of integral. The flux through a closed surface S, ∮ E · dA, is the net number of field lines “leaving” the surface (dA points along the outward normal). The circulation around a closed loop C, ∮ E · dl, measures how much the field “swirls” along the loop; for the E field it is the work done on a unit charge carried around the loop.

The four equations

∮ E · dA = Qenc / ε₀∮ E · dA = Qenc / ε₀
where:
  • Qenctotal charge enclosed by the closed surface, in C
  • ε₀electric constant, 8.85 · 10⁻¹² F/m
  • Eelectric field (vector), in V/m
  • dAsurface element — its size is the area, its direction the outward normal, in m²

I. Gauss's law: charges are the sources of the electric field.

∮ B · dA = 0
where:
  • Bmagnetic field (vector), in T
  • ∮ … dAintegral over any closed surface

II. Gauss's law for magnetism: there are no magnetic charges (monopoles); B field lines are closed.

∮ E · dl = −dΦB / dt∮ E · dl = −dΦB / dt
where:
  • ΦBmagnetic flux through a surface bounded by the loop, in Wb
  • dlelement of the closed loop (vector), in m; the left side is the EMF around the loop, in V

III. Faraday's law: a changing magnetic field creates a circulating electric field.

∮ B · dl = μ₀ · (I + ε₀ · dΦE / dt)∮ B · dl = μ₀ · (I + ε₀ · dΦE / dt)
where:
  • μ₀magnetic constant, 4π · 10⁻⁷ H/m
  • Iconduction current through the loop, in A
  • ε₀ · dΦE / dtε₀ · dΦE / dtMaxwell's displacement current, in A

IV. Ampère–Maxwell law: currents and changing electric fields create a circulating magnetic field.

With the Stokes and divergence (Gauss–Ostrogradsky) theorems the equations can be put in differential form: ∇ · E = ρ/ε₀, ∇ · B = 0, ∇ × E = −∂B/∂t, ∇ × B = μ₀j + μ₀ε₀ ∂E/∂t. The integral form is handy for symmetric problems, the differential form for waves and numerical computation.

Using symmetry: Gauss and Ampère

Example 1: a point charge and a charged plane

a) Derive the field of a point charge from Gauss's law. b) Find the field of an infinite plane with surface charge density σ = 1 μC/m².

Show solution
a) Take a sphere of radius r as the Gaussian surface: by symmetry E is the same everywhere on it and radial. E · 4πr² = q/ε₀ → E = q / (4πε₀r²) = kq/r² — Coulomb's law.
b) Gaussian surface: a cylinder (“pillbox”) crossing the plane, with end faces of area S. Flux passes only through the two ends: 2ES = σS/ε₀ → E = σ / (2ε₀).
E = 10⁻⁶ / (2 · 8.85 · 10⁻¹²) ≈ 5.6 · 10⁴ V/m — independent of distance!
Example 2: the field of a straight wire

A long straight wire carries 10 A. Find the magnetic field 5 cm from the wire.

Show solution
Loop: a circle of radius r round the wire; by symmetry B is tangent to it and constant in size. For a steady current dΦE/dt = 0.
B · 2πr = μ₀I → B = μ₀I / (2πr) = 4π · 10⁻⁷ · 10 / (2π · 0.05) = 4 · 10⁻⁵ T.
That is comparable to the Earth's magnetic field — which is why a compass “feels” the current in a nearby wire.

The displacement current

Picture a charging capacitor and a loop around the wire. If the surface bounded by the loop cuts the wire, Ampère's law gives μ₀I; if a surface bounded by the same loop passes between the plates, no charge flows through it and the result is zero — a contradiction! Maxwell showed that the electric flux between the plates is changing and that the term ε₀ · dΦE/dt equals exactly I. This “displacement current” is not a flow of charge, yet it produces a magnetic field just like a conduction current.

Example 3: inside a capacitor

A parallel-plate capacitor with plates of area 0.01 m² is charged with a current of 2 A. How fast does the electric field between the plates grow?

Show solution
E = q / (ε₀A), so ΦE = E · A = q/ε₀ and ε₀ · dΦE/dt = dq/dt = I — the displacement current is 2 A.
dE/dt = I / (ε₀A) = 2 / (8.85 · 10⁻¹² · 0.01) ≈ 2.3 · 10¹³ V/(m · s).

Electromagnetic waves

In empty space (ρ = 0, j = 0) consider a plane wave travelling along x: E = Ey(x, t), B = Bz(x, t). Faraday's law on a small rectangle in the xy-plane gives ∂Ey/∂x = −∂Bz/∂t; the Ampère–Maxwell law on a rectangle in the xz-plane gives −∂Bz/∂x = μ₀ε₀ ∂Ey/∂t. Differentiating the first with respect to x and the second with respect to t and eliminating B yields the wave equation.

∂²E/∂x² = μ₀ε₀ · ∂²E/∂t², c = 1/√(μ₀ε₀) ≈ 3.0 · 10⁸ m/s, E = c · B∂²E/∂x² = μ₀ε₀ · ∂²E/∂t², c = 1/√(μ₀ε₀) ≈ 3.0 · 10⁸ m/s, E = c · B
where:
  • cspeed of electromagnetic waves in vacuum — the speed of light
  • E, Binstantaneous field values; E ⊥ B ⊥ direction of travel, in phase

Wave intensity: I = c · ε₀ · E₀² / 2 (W/m²).

Example 4: the speed of light and the field of sunlight

a) Calculate c from μ₀ and ε₀. b) Above the atmosphere sunlight has an intensity of about 1360 W/m². Find the amplitudes E₀ and B₀.

Show solution
a) c = 1 / √(4π · 10⁻⁷ · 8.85 · 10⁻¹²) ≈ 3.0 · 10⁸ m/s — the measured speed of light!
b) E₀ = √(2I / (cε₀)) = √(2 · 1360 / (3 · 10⁸ · 8.85 · 10⁻¹²)) ≈ 1.0 · 10³ V/m.
B₀ = E₀ / c ≈ 3.4 · 10⁻⁶ T.
Interactive
Loading simulation…
The graph shows the E field of the wave; the B field oscillates in phase in the perpendicular plane. In vacuum λ · f = c for every frequency.

Key points

  • ∮ E · dA = Qenc/ε₀ and ∮ B · dA = 0: charges are the sources of E; there are no magnetic monopoles.
  • ∮ E · dl = −dΦB/dt: a changing magnetic field creates an electric field.
  • ∮ B · dl = μ₀(I + ε₀ dΦE/dt): the displacement current completes Ampère's law.
  • In vacuum the fields obey the wave equation; c = 1/√(μ₀ε₀) ≈ 3 · 10⁸ m/s, E = cB.
  • Symmetric cases: point charge E = kq/r², plane E = σ/(2ε₀), straight wire B = μ₀I/(2πr).

Check yourself

10 questions. Every correct answer earns XP.

1 / 10
A closed surface encloses charges of +3 nC and −1 nC. What is the electric flux through it?