Theory

Maxwell's equations and electromagnetic waves

How the displacement current completes Ampère's law, what the four Maxwell equations are, why they imply waves that travel at the speed of light, and how much energy and momentum these waves carry.

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Displacement current
Ampère-Maxwell law

Displacement current

Throughout this topic the fields are in vacuum, unless stated explicitly, and currents are denoted by the lowercase letter $i$ , because the letter $I$ denotes the intensity of a wave.

Ampère's law and a charging capacitor

Consider a parallel-plate capacitor that is being charged through a wire carrying the current $i$ , and a closed curve $γ$ that encircles the wire. The flat surface $S_{1}$ is crossed by the wire, so the current through it is $i$ . The surface $S_{2}$ , which has the same boundary, passes between the plates, where there are no moving charges, so the current through it is zero. Ampère’s law would therefore give two different values for the same circulation $\oint_{γ} B \cdot d l$ , and in the form of the reminder it holds only for steady currents.

Between the plates, however, there is an electric field that grows while the capacitor charges, and its flux through $S_{2}$ varies in time. Maxwell added to Ampère’s law a term proportional to the rate of change of that flux.

Charging capacitor: the flat surface

S_{1}

is crossed by the current

i

, while the surface

S_{2}

with the same boundary

γ

passes between the plates, where only the displacement current flows

Let $S$ be an oriented surface and let $Φ_{E} = \iint_{S} E \cdot \overset{n}{^} d A$ be the flux of the electric field through it. The displacement current through $S$ is

i_{s} = ε_{0} \frac{d Φ _{E}}{d t} .

It is measured in amperes, like a conduction current, because the product $ε_{0} Φ_{E}$ is measured in $C^{2} / (N \cdot m^{2}) \cdot N \cdot m^{2} /C = C$ . The displacement current does not correspond to a motion of charges. For example, if the uniform field between the plates of a capacitor of area $A = 0.010 m^{2}$ grows at the rate $\frac{d E}{d t} = 1.0 \cdot 1 0^{12} V/ (m \cdot s)$ , the displacement current is $i_{s} = ε_{0} A \frac{d E}{d t} = 8.85 \cdot 1 0^{- 12} \cdot 0.010 \cdot 1.0 \cdot 1 0^{12} A \approx 0.089 A$ .

Displacement current in a parallel-plate capacitor

Consider a parallel-plate capacitor in vacuum, with plates of area $A$ , connected to a circuit carrying the current $i = \frac{d q}{d t}$ , where $q$ is the charge on the positive plate, and neglect edge effects. Then the displacement current through a surface that passes between the plates, parallel to them, is equal to the conduction current in the wires:

i_{s} = i .

The displacement current therefore continues between the plates the conduction current, which stops at the plates. Hence the surfaces $S_{1}$ and $S_{2}$ , which have the same boundary $γ$ , give the same enclosed current. A capacitor being charged with $i = 2.0 A$ is crossed by a displacement current of $2.0 A$ , whatever the area of the plates and their separation.

Proof

Neglecting edge effects, the field between the plates is uniform, perpendicular to the plates, with magnitude $E = \frac{σ}{ε _{0}} = \frac{q}{ε _{0} A}$ , while outside it is zero. The flux through the surface, oriented from the positive plate toward the negative one, is therefore

Φ_{E} = E A = \frac{q}{ε _{0}} .

Differentiating with respect to time gives $i_{s} = ε_{0} \frac{d Φ _{E}}{d t} = \frac{d q}{d t} = i$ .

Ampère-Maxwell law

For every oriented closed curve $γ$ and every surface $S$ bounded by $γ$ , oriented by the right-hand rule,

\oint_{γ} B \cdot d l = μ_{0} i_{enc} + μ_{0} ε_{0} \frac{d Φ _{E}}{d t} = μ_{0} (i_{enc} + i_{s}),

where $i_{enc}$ is the conduction current through $S$ and $Φ_{E}$ is the flux of the electric field through $S$ . It is a postulate, introduced by Maxwell to make Ampère’s law consistent with the conservation of charge and confirmed by the experimental discovery of electromagnetic waves.

The law states that the magnetic field is generated both by currents and by electric fields that vary in time. In the steady state the flux $Φ_{E}$ is constant and Ampère’s law is recovered. For example, if a capacitor is charged with $i = 2.0 A$ , the circulation of $B$ along a curve that encircles the wire is $μ_{0} i \approx 2.5 \cdot 1 0^{- 6} T \cdot m$ even with a surface that passes between the plates.

Magnetic field between the plates of a capacitor

Consider a parallel-plate capacitor in vacuum with circular plates of radius $R$ , being charged with current $i$ , and neglect edge effects. In the region between the planes of the plates the magnetic field lines are circles coaxial with the plates, and at distance $r$ from the axis the magnitude of the field is

$B = \frac{μ _{0} i r}{2 π R ^{2}}$ for $0 \leq r \leq R$ ;
$B = \frac{μ _{0} i}{2 π r}$ for $r \geq R$ .

The field grows linearly with $r$ up to the edge of the plates, where it reaches the maximum value $\frac{μ _{0} i}{2 π R}$ , and outside it coincides with the field of a straight wire carrying the current $i$ . With $R = 5.0 cm$ and $i = 2.0 A$ the maximum field is $\frac{4 π \cdot 1 0 ^{- 7} \cdot 2.0}{2 π \cdot 0.050} T = 8.0 μ T$ .

Top view: between the plates the field

E

points out of the page and grows; the magnetic field lines are coaxial circles traversed counterclockwise, inside and outside the edge of the plates

Magnitude of the magnetic field as a function of the distance from the axis, with

r

in units of

R

and

B

in units of

\frac{μ _{0} i}{2 π R}

: a straight line up to the edge of the plates, then a

\frac{1}{r}

behavior

Proof

The electric field between the plates is uniform and equal to $E = \frac{q}{ε _{0} π R ^{2}}$ , so $\frac{d E}{d t} = \frac{i}{ε _{0} π R ^{2}}$ . By the rotational symmetry about the axis, the magnetic field is tangent to the coaxial circles and has constant magnitude on each of them. Apply the Ampère-Maxwell law to the circle $γ$ of radius $r$ , in a plane parallel to the plates and lying between them, with the surface equal to the disk bounded by $γ$ . The disk is not crossed by conduction currents, so

2 π r B = μ_{0} ε_{0} \frac{d Φ _{E}}{d t} .

If $r \leq R$ , the flux is $Φ_{E} = E π r^{2}$ , hence $2 π r B = μ_{0} ε_{0} π r^{2} \frac{i}{ε _{0} π R ^{2}} = μ_{0} i \frac{r ^{2}}{R ^{2}}$ , which gives the first formula. If $r \geq R$ , the electric field is zero outside the plates, so $Φ_{E} = E π R^{2} = \frac{q}{ε _{0}}$ and $2 π r B = μ_{0} i$ , which gives the second.

The theory continues with 7 more sections (maxwells equations, local form of maxwells equations, wave equation and the speed of light…), plus all worked examples, the solutions and the PDF.

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