These notes aren’t anything official, more reference for equations when considering problems from Physics 115 or 122. They are not complete, nor are they guaranteed to be correct.

## Electrostatics, Magnetostatics

These are the infinitesimal versions of Gauss’s and Ampère’s laws \begin{align*} \nabla \cdot \textbf{E} & = \frac{\rho}{\epsilon_0},\\ \nabla\times {\bf B} & = \mu_0\textbf{J}. \end{align*} These are essentially the relations describing the electric and magnetic fields of source static and constant charges and currents. If we wanted the reaction of test charges and currents, we use the Lorentz Force Law $\textbf{F} = q(\textbf{E} + \textbf{v} \times \textbf{B}).$

If we combine the infinitesimal equations above with some calculus, especially versions of Stoke’s theorem, we get

\begin{align} \boldsymbol{\Phi_E} & = \iiint \nabla \cdot \textbf{E} \, dV\\ & = \iiint\frac{\rho}{\epsilon_0} \, dV\\ & = \frac{Q_{\text enc.}}{\epsilon_0}. \\ \oint\textbf{B} \cdot d\boldsymbol{\ell} & = \iint\left(\nabla\times\textbf{B}\right)\cdot d\textbf{A}\\ & = \iint\mu_0\textbf{J}\cdot d\textbf{A}\\ & = \mu_0 I_\text{enc.}. \end{align}

The former seems to do quite a bit for us, we can get Coloumb’s law, as well as the electric fields of a sphere, line and plate of charge.

The latter gives us the magnetic field of a wire of current. We will not try to derive the infinitesimal contribution to a magnetic field by each moving charge, but we give the contribution from segments of a wire with constant current,

$d\textbf{B} = \frac {\mu_0} {4\pi} \frac {I\,d\boldsymbol{\ell}\times \textbf{r}} {r^3}.$

If we have a loop of charge, it is of benefit to define the magnetic dipole moment, as it may make some calculations quite simple,

$\boldsymbol{\mu}=\frac{1}{2}\int\textbf{r}\times\textbf{J}\,dV$

As a result, we can write the torque of a magnetic field on a dipole moment as

$\begin{equation} \boldsymbol{\tau}=\boldsymbol{\mu}\times\textbf{B} \end{equation}$

Additionally we have the magnetic field of a dipole, the energy of a dipole, and the force on a dipole,

\begin{align*} \textbf{B} & = \frac{\mu_0}{4\pi}\frac{ 3\left(\boldsymbol{\mu}\cdot\hat{\textbf{r}}\right)\hat{\textbf{r}}-\boldsymbol{\mu}}{r^3},\\ U & = -\boldsymbol{\mu}\cdot\textbf{B},\\ \textbf{F} & = \nabla\left(\boldsymbol{\mu}\cdot\textbf{B}\right). \end{align*}

## Electrodynamics, Magnetodynamics

Now, what if things are moving? Well, the full dynamical equations for electromagnetism can be quite complicated, but here we will consider a few of the implications of further generality.

First we extend Maxwell’s equations with two more contributions, completing the set (I skipped $$=0$$).

\begin{align*} \nabla\times\textbf{E} & = -\textbf{B}',\\ \nabla\times\textbf{B} & = \mu_0\epsilon_0\textbf{E}'. \end{align*}

The integral versions, derived similarly as above, are

\begin{align*} \oint \textbf{E}\cdot d\boldsymbol{\ell} & = -\boldsymbol{\Phi}_B{}', \\ \oint \textbf{B}\cdot d\boldsymbol{\ell} & = \mu_0\epsilon_0\boldsymbol{\Phi}_E{}'. \end{align*}

The first, Faraday’s law, suggests that an EMF is induced in a current loop when the magnetic flux changes, and is used in a collection of introductory applications, including transformers (AC), motional emf, AC generators and moving a magnet near a wire loop Lenz’s law is a helpful mnemonic. The latter, Maxwell’s correction to Ampère’s law, is probably referenced in more subtle ways, for instance, charging a capacitor and EM waves, which is likely why it took some time to be discovered.

## Inductance

Then potential drop at an inductor in a circuit (self-inductance)

$\begin{equation} V=LI'. \end{equation}$

### Coil

Induced emf,

$\begin{equation} \mathcal{E}=-NAB'. \end{equation}$

Magnetic field due to solenoid, itself,

$\begin{equation} B=\mu\frac{N}{\ell}I. \end{equation}$

Thus, emf due to self is

\begin{align*} \mathcal{E} & = -NA\mu\frac{N}{\ell}I'\\ & \equiv -LI'. \end{align*}

### mutual inductance

Magnetic dipole for loops of first solenoid,

$\begin{equation} \vec{\mu}=N_1\pi r^2I\hat{\textbf{k}}. \end{equation}$

Magnetic field due to the first solenoid:

$\begin{equation} \textbf{B}_1 = \frac{\mu_0}{4\pi}\frac{ 3\left(\boldsymbol{\mu}\cdot\hat{\textbf{r}}\right)\hat{\textbf{r}}-\boldsymbol{\mu}}{r^3}. \end{equation}$

This sets up a magnetic flux in the second solenoid,

$\begin{equation} N_2\Phi_{21} = N_2B_1A_2. \end{equation}$

Then the emf induced in the second solenoid would be

\begin{align*} \mathcal{E}_{21} & = -N_2B_1'A_2\\ & = -N_2\mu\frac{N_1}{\ell}I_1'\\ & = -M_{21}I_1' \end{align*}