Maxwell's Equations

Ampere's Law with Maxwell's Correction

Since Ampere's Law is based on the assumption that current is steady $$\nabla\times \textbf{J} = -\frac{\partial\rho}{\partial t}$$ Applying divergence on Ampere's Law $$\nabla\cdot (\nabla\times \textbf{B}) = \nabla \cdot (\mu_0\textbf{J})$$ By definition of curl and divergence $$\nabla\cdot (\nabla\times \textbf{B}) = 0$$ When current is not steady \begin{align} \nabla \cdot (\mu_0\textbf{J}) &= \mu_0 (\nabla \cdot \textbf{J}) \\ &= -\mu_0 \frac{\partial\rho}{\partial t} \\ &= -\mu_0 \frac{\partial}{\partial t}\left( \epsilon_0 \nabla \cdot \textbf{E} \right) \\ &= -\mu_0 \epsilon_0 \frac{\partial}{\partial t}\left( \nabla \cdot \textbf{E} \right) \\ \end{align} In order to achieve \(\nabla\cdot (\nabla\times \textbf{B}) = 0\), the curl of magnetic field should be written as $$ \bbox[5px,border:2px solid #666] { \nabla\times \textbf{B} = \mu_0\textbf{J} + \mu_0 \epsilon_0 \frac{\partial\textbf{E}}{\partial t} } $$ in which $$\epsilon_0 \frac{\partial\textbf{E}}{\partial t}$$ is called the displacement current. Not that Ampere's Law is wrong, but Ampere's Law is applicable only when current is steady.

Example (Griffiths Third Edition Q 7.31)

A fat wire, radius \(a\), carries a constant current \(I\), uniformly distributed over its cross section. A narrow gap in the wire, of width \(w \ll a\), forms a parallel-plate capacitor. Find the magnetic field in the gap, at a distance \(s < a\) from the axis.

Solution: The displacement current density is $$\vec{J}_b=\epsilon_0\frac{\partial \vec{E}}{\partial t}=\frac{I}{A}=\frac{I}{\pi a^2}\hat{z}.$$ Drawing an amperian loop at radius \(s\), \begin{align} \oint \vec{B}\cdot d\vec{l} &= \mu_0 I_{\text{d}_{\text{enc}}}\\ B\cdot 2\pi s &= \mu_o\frac{I}{\pi a^2}\cdot \pi s^2 = \mu_0I\frac{s^2}{a^2}\\ B &= \frac{\mu_0Is^2}{2\pi s a^2} \end{align} So, $$\vec{B}=\frac{\mu_0Is}{2\pi a^2}\hat{\phi}$$

Checkpoint (Griffiths Third Edition Q7.32)

A thin wire connects to the centers of the plates. The current \(I\) is constant, the radius of the capacitor is \(a\), and the separation of the plates is \(w\ll a\). Assume that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at \(t=0\). Find the magnetic field at a distance \(s\) from the axis.

$$\sigma(t)=\frac{Q(t)}{\pi a^2}=\frac{It}{\pi a^2}$$ The electric field is $$\vec{E}=\frac{\sigma(t)}{\epsilon_0}\hat{z}=\frac{It}{\pi\epsilon_0 a^2}\hat{z}$$ The displacement current density is $$\vec{J}_b=\epsilon_0\frac{\partial \vec{E}}{\partial t}=\frac{I}{\pi\ a^2}\hat{z}.$$ Drawing an amperian loop at radius \(s\), \begin{align} \oint \vec{B}\cdot d\vec{l} &= \mu_0 I_{\text{d}_{\text{enc}}}\\ B\cdot 2\pi s &= \mu_o\frac{I}{\pi a^2}\cdot \pi s^2 = \mu_0I\frac{s^2}{a^2}\\ B &= \frac{\mu_0Is^2}{2\pi s a^2} \end{align} So, $$\vec{B}=\frac{\mu_0Is}{2\pi a^2}\hat{\phi}$$

Maxwell's Equations

The four electric and magnetic equations are called the Maxwell's equations. \begin{cases} \nabla\cdot\textbf{E} = \frac{\rho}{\epsilon} \\[2ex] \nabla\cdot\textbf{B} = 0 \\[2ex] \nabla\times\textbf{E} = -\frac{\partial \textbf{B}}{\partial t} \\[2ex] \nabla\times\textbf{B} = \mu_0\textbf{J} + \mu_0 \epsilon_0 \frac{\partial\textbf{E}}{\partial t} \end{cases} In terms of free charge and current, the Maxwell's equations can be written as \begin{cases} \nabla\cdot\textbf{D} = \rho_f \\[2ex] \nabla\cdot\textbf{B} = 0 \\[2ex] \nabla\times\textbf{E} = -\frac{\partial \textbf{B}}{\partial t} \\[2ex] \nabla\times\textbf{H} = \textbf{J}_f + \frac{\partial\textbf{D}}{\partial t} \end{cases}

Electromagnetic Wave

When charge density \(\rho\) and current density \(\rho\) are zero, the Maxwell's equations will be \begin{cases} \nabla\cdot\textbf{E} = 0 , &\text{(1)}\\[2ex] \nabla\cdot\textbf{B} = 0 , &\text{(2)}\\[2ex] \nabla\times\textbf{E} = -\frac{\partial \textbf{B}}{\partial t} , &\text{(3)}\\[2ex] \nabla\times\textbf{B} = \mu_0 \epsilon_0 \frac{\partial\textbf{E}}{\partial t} , &\text{(4)} \end{cases} Curl of curl of E-field is \begin{align} \nabla\times\left(\nabla\times\textbf{E}\right) &= \nabla\times\left(-\frac{\partial\textbf{B}}{\partial t}\right) \\ \nabla\left(\nabla\cdot\textbf{E}\right) - \nabla^2\textbf{E} &= -\frac{\partial}{\partial t} \left(\nabla\times\textbf{B}\right)\\ \nabla^2\textbf{E} &= \mu_0\epsilon_0\frac{\partial^2\textbf{E}}{\partial t^2} \end{align} Similarly, \begin{align} \nabla\times\left(\nabla\times\textbf{B}\right) &= \nabla\times\left(\mu_0\epsilon_0\frac{\partial\textbf{E}}{\partial t}\right) \\ \nabla\left(\nabla\cdot\textbf{B}\right) - \nabla^2\textbf{B} &= \mu_0\epsilon_0\frac{\partial}{\partial t} \left(\nabla\times\textbf{E}\right)\\ \nabla^2\textbf{B} &= \mu_0\epsilon_0\frac{\partial^2\textbf{B}}{\partial t^2} \end{align} So, the means electric field and magnetic field, satisfying the three dimensional wave equation $$\nabla^2f = \frac{1}{v^2}\frac{\partial f}{\partial t^2} $$ are waves, namely electromagnetic wave, with a wave speed $$ c = \frac{1}{\sqrt{\epsilon_0\mu_0}}$$ which is equal to the speed of light. This suggests light may be a type of electromagnetic wave.