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Topic 7 of 9

Maxwell's Equations

Maxwell's four equations — Gauss's law for E and B, Faraday's law, and the Ampere-Maxwell law — are the complete classical description of electromagnetism. The displacement current $\varepsilon_0\partial\vec E/\partial t$ that Maxwell added to Ampere's law was the key insight that predicted electromagnetic waves traveling at the speed of light.

Key Concepts

Displacement Current

\vec J_D=\varepsilon_0\partial\vec E/\partial t

. Maxwell added this to Ampere's law to ensure consistency with the continuity equation. Without it, Ampere's law would imply

\nabla\cdot\vec J=0

, forbidding charge accumulation.

Ampere-Maxwell Law

\oint\vec B\cdot d\vec l=\mu_0(I_{\rm enc}+\varepsilon_0\frac{d\Phi_E}{dt})

. The displacement current term is significant in capacitors and radiation fields where

\vec E

changes rapidly.

Continuity Equation

\nabla\cdot\vec J+\partial\rho/\partial t=0

. This local conservation of charge follows from Maxwell's equations and cannot be violated.

EM Wave Prediction

From Maxwell's equations in vacuum:

\nabla^2\vec E=\mu_0\varepsilon_0\partial^2\vec E/\partial t^2

. This is a wave equation with speed

c=1/\sqrt{\mu_0\varepsilon_0}=3\times10^8

m/s.

Key Equations

Maxwell's equations (vacuum)

\nabla\cdot\vec{E} = \frac{\rho}{\varepsilon_0},\quad \nabla\cdot\vec{B} = 0,\quad \nabla\times\vec{E} = -\frac{\partial\vec{B}}{\partial t},\quad \nabla\times\vec{B} = \mu_0\vec{J}+\mu_0\varepsilon_0\frac{\partial\vec{E}}{\partial t}

The four equations of classical electromagnetism.

Speed of light

c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} = 2.998\times10^8\text{ m/s}

Predicted by Maxwell from the values of μ₀ and ε₀.

Poynting vector

\vec{S} = \frac{1}{\mu_0}\vec{E}\times\vec{B}

Energy flux density (W/m²); direction of EM energy flow.

Worked Example

Displacement Current in a Capacitor

Problem

A parallel-plate capacitor ( $A=0.010$ m²) is charged with $I=2.0$ A. Find the displacement current density $J_D$ and the $B$ field at $r=0.05$ m inside the gap (same as Ampere's law with $I_{\rm enc}=I$ ).

Solution

Displacement current in gap equals conduction current in wire: $I_D=I=2.0$ A.

$J_D=I_D/A=2.0/0.010=200$ A/m².

B = \frac{\mu_0 I_D}{2\pi r}\cdot\frac{\pi r^2}{A} = \frac{\mu_0 J_D r}{2}

B = \frac{\mu_0\times200\times0.05}{2} = \frac{4\pi\times10^{-7}\times10}{2} = 2\pi\times10^{-6}\approx6.28\,\mu\text{T}

Answer

J_D=200

A/m²;

B\approx6.28

μT at

r=0.05

Practice

Exercises

7 problems

1 of 7

From $\mu_0=4\pi\times10^{-7}$ T·m/A and $\varepsilon_0=8.85\times10^{-12}$ C²/N·m², find $c=1/\sqrt{\mu_0\varepsilon_0}$ (in m/s, to 3 sig figs).

Your answer

m/s

2 of 7

A capacitor ( $A=0.020$ m²) is charged with $I=3.0$ A. Find the displacement current density $J_D=I/A$ (in A/m²).

Your answer

A/m²

3 of 7

An EM wave has $E_0=600$ N/C. Find $B_0=E_0/c$ (in μT).

Unlock Exercise 3