- 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
- 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.
- 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.
- 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.
- 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
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.
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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!
A long straight wire carries 10 A. Find the magnetic field 5 cm from the wire.
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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.
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?
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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.
- 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).
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.
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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.
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
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