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.