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\subsubsection*{Lenz's law}

The direction of the induced current is always
such as to produce a magnetic field that opposes
the change in flux that caused the emf.

\begin{example*}
    A circular wire in the $xy$-plane. If $B_z$
    inside the loop increases in time then
    $\mathcal{E} = -\dfrac{\mathcal{F}}{t} < 0$.
    This induces a clockwise current ($I < 0$)
    that generates a magnetic field with $B_z < 0$
    inside the loop.
    \begin{center}
        \includegraphics[width=0.4\linewidth]
        {images/5a7e1c20adda11ed.png}
    \end{center}
    Hence the minus sign in Faraday's law. This
    avoids an unstable situation in which the flux
    grows indefinitely.
\end{example*}

\subsubsection*{Inductance and magnetic energy}

If a current $I$ around a circuit $C$ generates a
magnetic field with flux $\mathcal{F}$, then the
\emph{inductance} of the circuit is defined by
\[ L = \frac{\mathcal{F}}{I} \]
and depends only on the geometry.

\begin{example*}
    An ideal solenoid with cross-sectional area
    $A$ and $N$ turns per unit length. The uniform
    field $B = \mu_0 NI$ inside the solenoid
    produces a flux $BA$ per turn, so the
    inductance per unit length of the solenoid is
    $\mu_0 N^2 A$.
\end{example*}

\noindent
Exercise: show that the magnetic flux through a
thin wire $C_i$ due to a current $I_j$ around
another thin wire $C_j$ is $\mathcal{F}_{ij} =
L_{ij} I_J$, where the \emph{mutual inductance} is
\[ L_{ij} = \frac{\mu_0}{4\pi} \int_{C_i}
\int_{C_j} \frac{\dd \bf{x}_i \cdot \dd
\bf{x}_j}{|\bf{x}_i - \bf{x}_j} = L_{ji} \]
(Hint: Use $\bf{A}$)

\myskip
When the current $I$ around a circuit $C$ is
varied, an emf
\[ \mathcal{E} = -\dfrac{\mathcal{F}}{t} = -L
\dfrac{I}{t} \]
is induced. In a small timer interval $\delta t$,
a charge $\delta Q = I \delta t$ flows around $C$
and the work done on it by the Lorentz force is
\[ \delta W = \mathcal{E} \delta Q = -LI
\dfrac{I}{t} \delta t \]
So the rate at which work is done \emph{by} the
current \emph{on} the EM field is
\[ -\dfrac{W}{t} = LI \dfrac{I}{t} = \dfrac{}{t}
(\half LI^2) \]
Consider reaching a magnetostatic state by
building up the current from $0$ to $I$. The
energy stored is
\begin{align*}
    U
    &= \half LI^2 \\
    &= \half I\mathcal{F} \\
    &= I \int_C \bf{A} \cdot \dd \bf{x} \\
    &= \half \int \bf{J} \cdot \bf{A} \dd V
\end{align*}
analogous to $U = \half \int \rho \Phi \dd V$ in
electrostatics.

\myskip
Now, using $(M3')$ we have
\[ U = \frac{1}{2\mu_0} \int (\nabla \times
\bf{B}) \cdot \bf{A}\ dd V \]
and $(\nabla \times \bf{B}) \cdot \bf{A} = \nabla
\cdot (\bf{B} \times \bf{A}) + \bf{b} \cdot
(\nabla \times \bf{A})$. If we take the integral
over all space then the first term gives zero by
the divergence theorem, since
\[ |\bf{B}| = O \left( \frac{1}{|\bf{x}|^3}
\right) \qquad \text{and} \qquad |\bf{A}| = O
\left( \frac{1}{|\bf{x}|^2} \right) \]
as $|\bf{x}| \to \infty$ for a finite current
distribution, leaving
\[ \boxed{U = \int \frac{|\bf{B}|^2}{2\mu_0} \dd V} \]
as the energy stored in the magnetic field.

\subsection{Ohm's law}

In a stationary conductor
\[ \boxed{\bf{J} = \sigma\bf{E}} \]
where $\sigma$ is the \emph{electrical
conductivity}. This is not a fundamental physical
law but a constitutive relation, a macroscopic
property of a material.

\myskip
Inverse relation
\[ \bf{E} = \sigma^{-1} \bf{J} \]
where $\sigma^{-1}$ is the \emph{resistivity}
(usually denoted $\rho$ -- both $\sigma$ and
$\rho$ conflict with notation for charge
densities).

\myskip
A \emph{perfect conductor} corresponds to the
limit $\sigma \to \infty$ (so $\bf{E} = \bf{0}$)
and a \emph{perfect insulator} to $\sigma \to 0$
(so $\bf{J} = \bf{0}$).

\myskip
Consider a straight wire of length $L$ in the
direction of the unit vector $\bf{n}$ and with
uniform cross-sectional area $A$ and conductivity
$\sigma$. If the electric field is $\bf{E} = E
\bf{n}$ where $E = \text{const}$ then $\bf{J} =
\sigma E \bf{n}$ and the total current is $I =
\sigma EA$. The potential difference (voltage)
along the wire is
\[ V = \int \bf{E} \cdot \dd \bf{x} = EL =
\frac{IL}{\sigma A} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/f5f60574adde11ed.png}
\end{center}
\[ \boxed{V = IR} \]
where $R = \frac{L}{\sigma A}$ is the
\emph{resistance} of the wire.

\myskip
Accompanying the resistance of a wire is
\emph{Joule heating} (or \emph{Ohmic heating}):
conversion of EM energy into heat at the rate
$I^2R$. If the voltage $V$ is maintained by a
batter, then $VI = I^2R$ is the rate at which the
emf of the battery ($\mathcal{E} = V$) does work
to maintain the $I$.

\subsection{Time-dependent electric fields}

\subsubsection*{Scalar and vector potentials}

In electrodynamics we can no longer write $\bf{E}
= -\nabla \Phi$. But (M2) still allows us to write
\[ \boxed{\bf{B} = \nabla \times \bf{A}} \]
and (M3) then gives
\[ \nabla \times \left( \bf{E} + \pfrac{\bf{A}}{t}
\right) = \bf{0} \]
allowing us to write
\[ \boxed{\bf{E} = -\nabla \Phi - \pfrac{\bf{A}}{t}} \]
generalising the electrostatic expression. Under a
time-dependent gauge transformation
\[ \bf{A}' = \bf{A} + \nabla \chi, \qquad \Phi' =
\Phi - \pfrac{\chi}{t} \]
where $\chi(\bf{x}, t)$ is any scalar field, then
both $\bf{E}$ and $\bf{B}$ are unchanged.

\subsubsection*{The displacement current}

In magnetostatics we used Amp\`ere's law
\[ \int_C \bf{B} \cdot \dd \bf{x} = \mu_0 \int_S
\bf{J} \cdot \dd \bf{S} = \mu_0 I \]
or its differential form (M4')
\[ \nabla \times \bf{B} = \mu_0 \bf{J} \]
For $t$-dependent situations (M4)
\[ \nabla \times \bf{B} = \mu_0 \left( \bf{J} +
\eps_0 \pfrac{\bf{E}}{t} \right) \]
contains an extra term, the \emph{displacement
current}.

\myskip
Why is it needed? Without it, we would have
$\nabla \cdot \bf{J} = 0$, which describes charge
conservation in a situation where $\rho$ is
constrained to remain constant.

\myskip
But suppose we place free particles of positive
charge in some locallised region. Repulsice
Coulomb forces cause the particles to separate,
implying $\nabla \cdot \bf{J} > 0$.
\begin{center}
    \includegraphics[width=0.2\linewidth]
    {images/c225a0b8ade011ed.png}
\end{center}
We have seen that charge conservation in the
correct form
\[ \pfrac{\rho}{t} + \nabla \cdot \bf{J} = 0 \]
follows from Maxwell's equations including the
displacement current.