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\newpage
\section{Electrodynamics}

\subsection{Faraday's law of induction}

Maxwell's equation
\[ \nabla \times \bf{E} = -\pfrac{\bf{B}}{t} \tag{M3} \]
implies that a time-dependent magnetic field must
be accompanied by an electric field. This can
induce a current to flow in a conductor -- a
process known as \emph{electromagnetic induction}.

\subsubsection*{Faraday's law for a static circuit}

Consider a closed curve $C$ that is the boundary
of a time-independent open surface $S$. Integrate
(M3) over $S$ and use Stoke's theorem:
\[ \int_C \bf{E} \cdot \dd \bf{x} = -\int_S
\pfrac{\bf{B}}{t} \cdot \dd \bf{S} = -\dfrac{}{t}
\int_S \bf{B} \cdot \dd \bf{S} \]
This is \emph{Faraday's law of induction} for a
static circuit:
\[ \boxed{\mathcal{E} = -\dfrac{\mathcal{F}}{t}} \]
where
\[ \boxed{\mathcal{E} = \int_C \bf{E} \cdot \dd \bf{x}} \]
is the \emph{electromotive force} (emf) around $C$
and
\[ \boxed{\mathcal{F} = \int_S \bf{B} \cdot \dd
\bf{S}} \]
is the \emph{magnetic flux} through $S$.

\myskip
Since $\nabla \cdot \bf{B} = 0$, the flux
$\mathcal{F}$ is the same for any $S$ such that
$\partial S = C$, so it can be regarded as the
magnetic flux through $C$.

\myskip
Using $\bf{B} = \nabla \times \bf{A}$ and Stoke's
theorem, we can write
\[ \int_C \bf{A} \cdot \dd \bf{x} ,\]
which is invariant under a gauge transformation
$\bf{A}' = \bf{A} + \nabla \chi$.

\myskip
The emf is not actuallly a force. It is the line
integral of the Lorentz force on a particle of
unit charge confined to $C$:
\[ \mathcal{E} = \frac{1}{q} \int_C \bf{F} \cdot
\dd \bf{x} = \int_C (\bf{E} + \dot{\bf{x}} \times
\bf{B}) \cdot \dd \bf{x} = \int_C \bf{E} \cdot \dd
\bf{x} ,\]
since $\dot{\bf{x}}$ is tangent to $C$ for a
particle confined to a $t$-independent curve $C$.

\myskip
We will see later that if $C$ coincides with a
thin wire of resistance $R$, then the current
induced in the wire is $I = \mathcal{E} / R$.

\myskip
There are several ways in which the magnetic flux
through $C$ could change in time:
\begin{itemize}
    \item A magnet is moved near $C$.
    \item A current-carrying circuit is moved near
        $C$.
    \item The current in a nearby circuit is
        changed.
\end{itemize}
All these will induce an emg around $C$ and cause
a current to flow.

\subsubsection*{Faraday's law for a moving circuit}

Now let $C(t)$ be a \emph{time-dependent} closed
curve that is the boundary of an open surface
$S(t)$. How does the magnetic flux through $S$,
\[ \mathcal{F} = \int_S \bf{B} \cdot \dd \bf{S} \]
change in time? We have
\begin{align*}
    \mathcal{F}(t + \delta t) - \mathcal{F}(t)
    &= \int_{S(t + \delta t)} \bf{B}(\bf{x}, t +
    \delta t) \cdot \dd \bf{S} - \int_{S(t)}
    \bf{B}(\bf{x}, t) \cdot \dd \bf{S} \\
    &= \int_{S(t + \delta t)} \left(
    \bf{B}(\bf{x}, t) + \pfrac{\bf{B}}{t} \delta t
    + O(\delta t^2) \right) \cdot \dd \bf{S} -
    \int_{S(t)} \bf{B}(\bf{x}, t) \cdot \dd \bf{S}
    \\
    &= \int_{S(t + \delta t) - S(t)} \bf{B}(\bf{x},
    t) \cdot \dd \bf{S} + \int_{S(t)}
    \pfrac{\bf{B}}{t} \cdot \dd \bf{S} \delta t +
    O(\delta t^2)
\end{align*}
Let $\partial V$ be the volume swept out by
$S(t)$ in the time interval $\delta t$. Its
boundary is the closed surface $S(t + \delta t) -
S(t) + \Sigma$, where $\Sigma$ is the surface
swept out by $C(t)$ in time $\delta t$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/a66d2ec2ac4b11ed.png}
\end{center}
By (M2) and the divergence theorem,
\begin{align*}
    0
    &= \int_{\partial V} (\nabla \cdot \bf{B}) \dd
    V \\
    &= \int_{S(t + \delta t) - S(t)} \bf{B} \cdot
    \dd \bf{S} + \int_\Sigma \bf{B} \cdot \dd
    \bf{S}
\end{align*}
To evaluate the last term, parametrise $C$ as
$\bf{x} = \bf{x}(\lambda, t)$, where $\lambda$ is
a parameter around $C$. An element of $C$ is
\[ \dd \bf{x} = \pfrac{\bf{x}}{\lambda} \dd
\lambda \]
and has velocity $\bf{v} = \pfrac{\bf{x}}{t}$. In
time $\delta t$ it sweeps out the vector area
element
\[ \dd \bf{S} = \dd \bf{x} \times (\bf{v} \delta
t) \]
(points out of $\delta V$, as required). Thus
\begin{align*}
    \int_\Sigma \bf{B} \cdot \dd \bf{S}
    &= \int_C \bf{B} \cdot (\dd \bf{x} \times
    \bf{v}) \delta t + O(\delta t^2) \\
    &= \int_C (\bf{v} \times \bf{B}) \cdot \dd
    \bf{x} \delta t + O(\delta t^2)
\end{align*}
We then have
\begin{align*}
    \mathcal{F}(t + \delta t) - \mathcal{F}(t)
    &= - \int_C (\bf{v} \times \bf{B}) \cdot \dd
    \bf{x} \delta t + \int_S \pfrac{\bf{B}}{t}
    \cdot \dd \bf{S} \delta t + O(\delta t^2)
\end{align*}
from which
\begin{align*}
    \dfrac{\mathcal{F}}{t}
    &= -\int_C (\bf{v} \times \bf{B}) \cdot \dd
    \bf{x} + \int_S \pfrac{\bf{B}}{t} \cdot
    \dd \bf{S} \\
    &= -\int_C (\bf{v} \times \bf{V}) \cdot \dd
    \bf{x} - \int_S (\nabla \times \bf{E}) \cdot
    \dd \bf{S} \\
    &= -\int_C (\bf{e} + \bf{v} \times \bf{B})
    \cdot \dd \bf{x}
\end{align*}
We recover Faraday's law,
\[ \mathcal{E} = -\dfrac{\mathcal{F}}{t} ,\]
with the redefined emf
\[ \mathcal{E} = \int_C (\bf{E} + \bf{v} \times
\bf{B}) \cdot \dd \bf{x} \]
This $\mathcal{E}$ is again the line integral
adound $C$ of the Lorentz force on particle of
unit charge confined to $C$ (for which the
perpendicular components of $\dot{\bf{x}}$ must
agree with those of the curve velocity
$\bf{v}$).
