14.2. Maxwell equations

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14.2. Maxwell equations

  1. General
  2. No charge and no current
  3. Field in the conductor


\begin{align} &\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}\label{eq-14.2.1}\\ &\nabla \cdot \mathbf{B} = 0\label{eq-14.2.2}\\ &\nabla \times \mathbf{E} = -\frac {\partial \mathbf{B}}{\partial t}\label{eq-14.2.3}\\ &\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0\varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\label{eq-14.2.4} \end{align} where $\rho$ and $\mathbf{J}$ are density of charge and current, respectively. See in Wikipedia

Remark 1.

  1. Equation (\ref{eq-14.2.1}) is a Gauss' law, (\ref{eq-14.2.2}) is a Gauss' law for magnetism, (\ref{eq-14.2.3}) is a Faraday's law of induction.

  2. Equations (\ref{eq-14.2.1}) and (\ref{eq-14.2.4}) imply continuity equation \begin{equation} \rho_t + \nabla \cdot \mathbf{J}=0. \label{eq-14.2.5} \end{equation}

No charge and no current

In absence of the charge and current we get \begin{equation*} \mathbf{E}_t=\mu^{-1}\varepsilon^{-1}\nabla \times \mathbf{B},\qquad \mathbf{B}_t =-\nabla \times \mathbf{E} \end{equation*} and then \begin{multline*} \mathbf{E}_{tt}=\mu^{-1}\varepsilon^{-1}\nabla \times \mathbf{B}_t= -\mu^{-1}\varepsilon^{-1}\nabla \times (\nabla \times \mathbf{E})=\\\mu^{-1}\varepsilon^{-1}\bigl( \Delta \mathbf{E}-\nabla (\nabla \cdot \mathbf{E})\bigr)=\mu^{-1}\varepsilon^{-1}\Delta \mathbf{E}; \end{multline*} so we get a wave equation \begin{equation} \mathbf{E}_{tt}=c^2\Delta \mathbf{E} \label{eq-14.2.6} \end{equation} with $c=1/\sqrt{\mu\varepsilon}$.

Field in the conductor

On the other hand, if we have a relatively slowly changing field in the conductor, then $\rho=0$, $\mathbf{J}=\sigma \mathbf{J}$ where $\sigma$ is a conductivity and we can neglect the last term in (\ref{eq-14.2.4}) and $\nabla \times \mathbf{B} = \mu\sigma \mathbf{E}$ and \begin{multline*} \mathbf{B}_t=-\nabla \times \mathbf{E}=\mu^{-1}\sigma^{-1}\nabla \times (\nabla \times \mathbf{B})=\\ \mu^{-1}\sigma^{-1}(\Delta \mathbf{B}-\nabla (\nabla\cdot\mathbf{B}))= \mu^{-1}\sigma^{-1}(\Delta \mathbf{B})\end{multline*} so $\mathbf{B}$ satisfies heat equation \begin{equation} \mathbf{B}_t=\mu_0^{-1}\sigma^{-1}\Delta \mathbf{B}. \label{eq-14.2.7} \end{equation}

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