%!TEX root = 497Notes-Temple.tex

\section{MgNet, pre-act ResNet, variants and generalizations}\label{sec:relation}
%\subsection{Some properties of MgNet}
%So, this MgNet is corresponded to the multigrid methods
%with iteration in function space. A natural idea is that there is also a dual version of
%MgNet similar with the multigrid methods with iteration in dual space.
The MgNet model algorithm is one very basic and it can be generalized
in many different ways. It can also be used as a guidance to modify and
extend many existing CNN models.

The following result show how MgNet is related to the pre-act ResNet \cite{he2016identity}.
\begin{theorem}\label{thm:mgnet1}
The MgNet model Algorithm \ref{alg:mgnet},
admits the following identities
\begin{equation}\label{dualmgnet}
r^{\ell, i} = r^{\ell, i-1} -  A^{\ell} \circ \sigma \circ B^{\ell,i}\circ \sigma (r^{\ell,i-1}), \quad i = 1:\nu_\ell, \\
\end{equation}
where
\begin{equation}
  \label{eq:5}
	r^{\ell,i} = f^{\ell} - A^{\ell} \ast u^{\ell,i}.
\end{equation}
Furthermore, \eqref{dualmgnet} represents pre-act ResNet~\cite{he2016identity}
as shown before.
\end{theorem}

\begin{proof}
	Because of the linearity of $A^\ell$ and invariant within the same grid $\ell$,
	we can apply $A^\ell$ on both sides of \eqref{mgnet} and minus with
	$f^\ell$, thus we have
	$$
	f^{\ell} - A^{\ell} \ast u^{\ell,i} = f^{\ell} - A^\ell \ast u^{\ell,i-1} -
	A^{\ell} \ast \sigma \circ B^{\ell,i}\circ \sigma (f^\ell - A^\ell \ast u^{\ell,i-1}).
	$$
This finish the proof with definition in \eqref{eq:5}.
\end{proof}

The above result is very simple but critically important.
In view of Theorem \ref{thm:mgnet1}, it shows how multigrid and
CNN are intimately related. Furthermore, it provides a different version
of iResNet, which can be viewed as the dual version of the original pre-act ResNet.
This relation is quit similar with the dual relation of $u$ and $f$
in multigrid method \cite{xu2017algebraic}.
\begin{lemma}\label{thm:mgnet2}
	The ResNet~\cite{he2016deep} step
 	 as in \eqref{eq:ResNet}
%	\begin{equation}\label{resnet}
%	f^{\ell,i} = \sigma( f^{\ell, i-1} - \xi^{\ell,i} \circ \sigma \circ \eta^{\ell,i} (f^{\ell,i-1}) ).
%	\end{equation}
admits the following relation:
%(which resembles closely with \eqref{dualmgnet})
\begin{equation}\label{tilde-resnet}
\tilde r^{\ell,i+1} =\sigma(\tilde r^{\ell,i}) -
A^{\ell,i} \ast \sigma \circ B^{\ell,i}\ast \sigma( \tilde r^{\ell,i}),
\end{equation}
where
\begin{equation}\label{tilde-f}
\tilde r^{\ell,i} = r^{\ell, i-1} -A^{\ell,i} \ast \sigma \circ B^{\ell,i} \ast r^{\ell,i-1}.
\end{equation}
\end{lemma}
\begin{proof}
%	Now, we will establish the connection between classical ResNet and MgNet.
	First, we apply $ A^{\ell,i+1} \circ \sigma \circ B^{\ell,i+1}$
	on the both sides of \eqref{eq:ResNet} and get
	\begin{equation}\label{resnet1}
	A^{\ell,i+1} \ast \sigma \circ B^{\ell,i+1} \ast r^{\ell,i} =
	A^{\ell,i+1} \ast \sigma \circ B^{\ell,i+1}\ast \sigma( \tilde r^{\ell,i} ).
	\end{equation}
	Minus by $r^{\ell,i}$ on the both sides and recall the definition in \eqref{tilde-f}, we have
	\begin{equation*}
	\tilde r^{\ell,i+1} = r^{\ell,i} - A^{\ell,i+1} \ast \sigma \circ B^{\ell,i+1}\ast \sigma( \tilde r^{\ell,i}).
	\end{equation*}
	By the definition of $r^{\ell,i} = \sigma(\tilde r^{\ell,i})$, we finish this proof.
\end{proof}

We call the above form \eqref{tilde-resnet} as
$\sigma$-ResNet, similar to the MgNet we replace $A^{\ell,i}$ by $A^{\ell}$  and get
the next Mg-ResNet form as:
\begin{equation}\label{mg-resnet}
r^{\ell,i} =\sigma(r^{\ell,i-1}) -
A^{\ell} \ast \sigma \circ B^{\ell,i}\ast \sigma(r^{\ell,i-1}).
\end{equation}

If we take these pooling and prolongation operators
as discussed in the previous sections and focus on
the iterative forms on a certain grid $\ell$, we may
compare them all as:
\begin{table}[!htbp]
	\caption{Comparison for all iterative forms }
	\label{comparison-ALL}
	\begin{center}%\scriptsize
		\resizebox{1.0\textwidth}{!}{
			\begin{tabular}{|c|c|c|}
				\hline
				Primal-Dual & Model & Iterative form \\
				\hline
				\multirow{3}{*}{Feature space} & Abstract-MgNet & Solving $A^\ell(u^\ell) = f^\ell$ \\
				\cline{2-3}
				& General-MgNet & $u^{\ell,i} = u^{\ell, i-1} + B^{\ell,i} (f^\ell - A^{\ell}(u^{\ell,i-1}))$ \\
				\cline{2-3}
				& {MgNet} & $u^{\ell,i} = u^{\ell, i-1} + \sigma \circ B^{\ell,i}\circ \sigma (f^\ell - A^{\ell}(u^{\ell,i-1}))$ \\
				\hline
			\multirow{5}{*}{Data space} & pre-act ResNet & $ r^{\ell,i} = r^{\ell, i-1} -  A^{\ell,i} \ast \sigma \circ B^{\ell,i} \ast \sigma (r^{\ell,i-1})$ \\
				\cline{2-3}
				& Mg pre-act ResNet & $r^{\ell,i} = r^{\ell, i-1} -  A^{\ell} \ast \sigma \circ B^{\ell,i}\ast \sigma (r^{\ell,i-1})$ \\
				\cline{2-3}
				& Mg-ResNet & $r^{\ell,i} = \sigma(r^{\ell,i-1}) - A^{\ell} \ast \sigma \circ B^{\ell,i}\ast \sigma( r^{\ell,i-1})$ \\
				\cline{2-3}
				& $\sigma$-ResNet & $r^{\ell,i} = \sigma(r^{\ell,i-1}) - A^{\ell,i} \ast \sigma \circ B^{\ell,i}\ast \sigma( r^{\ell,i-1})$ \\
				\cline{2-3}
				& ResNet & $r^{\ell,i} = \sigma(r^{\ell, i-1} -  A^{\ell,i} \ast \sigma \circ B^{\ell,i} \ast r^{\ell,i-1})$ \\
				\hline
			\end{tabular}
		}
	\end{center}
\end{table}
We can have these connections for all iterative scheme in data space:
%\begin{equation}
%\text{ ResNet} \xleftrightarrow{\eqref{tilde-f}} \sigma\text{-ResNet } \xleftrightarrow{A^{\ell,i} \leftrightarrow A^{\ell}} \text{Mg-ResNet}
%\xleftrightarrow{\sigma(f^{\ell,i-1}) \leftrightarrow f^{\ell, i-1} } \text{Mg pre-act ResNet} \xleftrightarrow{ A^{\ell} \leftrightarrow A^{\ell,i}} \text{pre-act ResNet}.
%\end{equation}
\vspace{-19pt}
\begin{figure}[H]
	\begin{center}
		\includegraphics[width=1.0\textwidth]{Mgnetrelation}
	\end{center}
	%\caption{Connections}
	\label{fig:mgnet}
\end{figure}
\vspace{-26pt}
%Because of the linearity of $\xi^{\ell,i}$, the above forms of iResNet and ResNet
%are equivalent to the previous as \eqref{dualmgnet} and \eqref{tilde-resnet}.
In this sense, these MgNet related models can be understood as
 models between pre-act ResNet and ResNet. And all these models can be
 understood as iteration in the data space as a dual relationship with
 feature space as MgNet.

The rationality of replacing  $A^{\ell,i}$ by layer independent $A^{\ell}$ may
be justified by the following theorem.
\begin{theorem}\label{thm:CNN}
On each grid $\mathcal T_\ell$,
\begin{enumerate}
	\item Any CNN model with
%	CNN and Mg-ResNet]
	\begin{equation}
	\label{CNN1}
	f^{\ell,i} =   \chi^{\ell,i} \circ \sigma (f^{\ell,i-1}),
	\end{equation}
	can be written as
	\begin{equation}\label{Res-CNN1}
	f^{\ell,i} = \sigma(f^{\ell,i-1}) - \xi^{\ell} \circ \sigma \circ \eta^{\ell,i} \circ\sigma ( f^{\ell,i-1}).
	\end{equation}
	\item Any CNN model with
%	[CNN and ResNet]
	\begin{equation}
	\label{CNN2}
	f^{\ell,i} =   \sigma\circ\chi^{\ell,i} (f^{\ell,i-1}).
	\end{equation}
	can be written as
	\begin{equation}\label{Res-CNN2}
	f^{\ell,i} = \sigma\left(f^{\ell,i-1} - \xi^{\ell} \circ \sigma \circ \eta^{\ell,i}  ( f^{\ell,i-1})\right).
	\end{equation}
\end{enumerate}

%\begin{equation}
% \chi^{\ell,i}: \mathbb{R}^{n_\ell \times n_\ell \times c_\ell}
% \mapsto \mathbb{R}^{n_\ell \times n_\ell \times c_\ell},
%\end{equation}
\end{theorem}
\begin{proof}
%	Without loss of generality, consider the classical CNN with
%	\begin{equation}
%	f^{\ell,i} =   ({\rm id}  + \tilde \eta^{\ell,i} )\circ \sigma (f^{\ell,i-1}).
%	\end{equation}
Let use prove the first case as an example,
the second case can be proven with the same process.

With similar structure in MgNet, we can take
\begin{equation}
\label{xi-cnn1}
\xi^{\ell}= \hat \delta^\ell :=  [\hat \delta_1, \cdots, \hat \delta_{{c_\ell}}],
\end{equation}
and
\begin{equation}
\label{eta-ell}
\eta^{\ell,i} = [{\rm id}_{c_\ell}, -{\rm id}_{c_\ell}] \circ (\chi^{\ell,i} - {\rm id}_{c_\ell}).
\end{equation}
Here
\begin{equation}
{\rm id}_{c_\ell}: \mathbb{R}^{n_\ell \times n_\ell \times c_\ell}
\mapsto \mathbb{R}^{n_\ell \times n_\ell \times c_\ell},
\end{equation}
is the identity map and
\begin{equation}
\hat \delta_k :  \mathbb{R}^{n_\ell \times n_\ell \times 2c_\ell}
\mapsto \mathbb{R}^{n_\ell \times n_\ell},
\end{equation}
with
\begin{equation}\label{eq:hatdelta}
\hat \delta_k([X ,Y]) = -([X]_k + [Y]_k),
\end{equation}
for any $X, Y \in \mathbb{R}^{n_\ell \times n_\ell \times c_\ell}$
and $[X,Y] \in \mathbb{R}^{n_\ell \times n_\ell \times 2c_\ell} $.

	First, we see that $\eta^{\ell,i}$ with the above
	form is a convolution from $\mathbb{R}^{n_\ell \times n_\ell \times c_\ell}$
	to  $\mathbb{R}^{n_\ell \times n_\ell \times 2c_\ell}$.
	Following the identity
	\begin{equation}
	ReLU(x) + ReLU(-x) = x,
	\end{equation}
	and the definition of $\xi^{\ell}$ i.e.
	\begin{equation}
	\xi^{\ell} = \hat \delta^\ell,
	\end{equation}
	as a special case in MgNet.
	For more details, we can give a exact form of
	$\hat \delta_k$ as in \eqref{eq:hatdelta} with
	\begin{equation}
	\hat \delta_k = [0, \cdots,0, -\delta, \cdots 0;  0, \cdots,0, -\delta, \cdots 0],  \quad k = 1:{c_\ell},
	\end{equation}
	where $\delta$ is the identity kernel during one channel.

	At last, we have
	\begin{align}
	\left[\xi^{\ell} \circ \sigma \circ [{\rm id}_{c_\ell}, -{\rm id}_{c_\ell}] (x) \right]_k &=  \left[\xi^{\ell} \circ \sigma \circ [x, -x]  \right]_k, \\
	&= \hat \delta_k ( [\sigma(x), \sigma(-x)]),  \\
	&= -\delta([\sigma(x)]_k) - \delta([\sigma(-x)]_k),\\
	&=-( \sigma([x]_k)+ \sigma(-[x]_k)) , \\
	&=  -[x]_k
	\end{align}
	Thus to say,
	\begin{equation}
	\xi^{\ell} \circ \sigma \circ [{\rm id}_{c_\ell}, -{\rm id}_{c_\ell}]  = -{\rm id}_{c_\ell}.
	\end{equation}
	Then the modified dual form of MgNet in \eqref{tilde-resnet} becomes
	\begin{align}
	f^{\ell,i} &= \sigma(f^{\ell,i-1}) - \xi^{\ell,i} \circ \sigma \circ \eta^{\ell,i} \circ\sigma ( f^{\ell,i-1}) , \\
	&=  \sigma(f^{\ell,i-1}) - \left( \xi^{\ell} \circ \sigma \circ [{\rm id}_{c_\ell}, -{\rm id}_{c_\ell}] \right)
	\circ (\chi^{\ell,i} - {\rm id}_{c_\ell})\circ \sigma(f^{\ell,i-1})\\
	&=\sigma(f^{\ell,i-1}) + (\chi^{\ell,i} -{\rm id}_{c_\ell})\circ \sigma(f^{\ell,i-1}),  \\
	&=\chi^{\ell,i} \circ  \sigma (f^{\ell,i-1}).
	\end{align}
	This covers \eqref{Res-CNN1}.
\end{proof}

\begin{remark}
Theorems~\ref{thm:CNN} shows that general CNN in
the forms of either \eqref{CNN1} or \eqref{CNN2} can be written recast
as \eqref{Res-CNN1} or \eqref{Res-CNN2} with the data-feature mapping
$A^\ell=\xi^\ell$ that is not only independent of the layers, but is
actually given a priori as in \eqref{xi-cnn1}.  In
view of Theorems~\ref{thm:mgnet1} and \ref{thm:mgnet2}, the classic
CNN models can be essentially recovered from MgNet by choosing
$\xi^\ell$ a priori as in  \eqref{xi-cnn1}.  Since
the classic CNN models have been extensively tested to be successful,
the more general MgNet with more general $\xi^\ell$ (to be trained)
are expected to be more efficient than the classic CNN models.
\end{remark}

%At last we have the next relation:
%\input{MgNet-relation.tex}


% !TEX root = hott_intro.tex

\section{The Blakers-Massey theorem}
The Blakers-Massey theorem is a connectivity theorem which can be used to prove the Freudenthal suspension theorem, giving rise to the field of \emph{stable homotopy theory}. It was proven in the setting of homotopy type theory by Lumsdaine et al, and their proof was the first that was given entirely in an elementary way, using only constructions that are invariant under homotopy equivalence.

\subsection{The Blakers-Massey theorem}
Consider a span $A \leftarrow S \rightarrow B$, consisting of an $m$-connected map $f:S\to A$ and an $n$-connected map $g:S\to B$. We take the pushout of this span, and subsequently the pullback of the resulting cospan, as indicated in the diagram
\begin{equation}\label{eq:BM}
\begin{tikzcd}
S \arrow[drr,bend left=15,"g"] \arrow[ddr,bend right=15,swap,"f"] \arrow[dr,densely dotted,"u" near end] \\
& A \times_{(A \sqcup^S B)} B \arrow[r,"\pi_2"] \arrow[d,"\pi_1"] & B \arrow[d,"\inr"] \\
& A \arrow[r,swap,"\inl"] & A \sqcup^S B.
\end{tikzcd}
\end{equation}
The universal property of the pullback determines a unique map $u:S\to A \times_{(A\sqcup^S B)} B$ as indicated.

\begin{thm}[Blakers-Massey]
The map $u:S\to A \times_{(A\sqcup^S B)} B$ of \cref{eq:BM} is $(n+m)$-connected.
\end{thm}

\subsection{The Freudenthal suspension theorem}

\begin{thm}
  If $X$ is a $k$-connected pointed type, then the canonical map
  \begin{equation*}
    X \to \loopspace{\susp{X}}
  \end{equation*}
  is $2k$-connected.
\end{thm}

\begin{thm}
  $\pi_n(\sphere{n})=\Z$ for $n\geq 1$.
\end{thm}

\subsection{Higher groups}
\label{sec:higher-groups}

Recall that types in HoTT may be viewed as $\infty$-groupoids:
elements are objects, paths are morphisms, higher paths are higher
morphisms, etc.

It follows that \emph{pointed connected} types $B$ may be viewed as higher
groups, with \define{carrier} $\loopspacesym B$.
The neutral element is the identity path,
the group operation is given by path composition,
and higher paths witness the unit and associativity laws.
Of course, these higher paths are themselves subject to further laws,
etc., but the beauty of the type-theoretic definition is
that we don't have to worry about that:
all the (higher) laws follow from the rules of the identity types.
Writing $G$ for the carrier $\loopspacesym B$, it is common to write $BG$ for the pointed
connected type $B$, which comes equipped with an identification $G = \loopspacesym BG$.
We call $BG$ the \define{delooping} of $G$.

The type of pointed types is
$\UU_\pt \defeq  \sm{A:\UU} A$. The type of $n$-truncated types is
$\UU^{\le n} \defeq  \sm{A:\UU}\istrunc{n}{A}$ and for $n$-connected types it is
$\UU^{>n} \defeq  \sm{A:\UU}\mathsf{is\usc{}conn}_n(A)$. We will combine these notations as needed.

\begin{defn}
We define the type of \define{higher groups}, or \define{$\infty$-groups}, to be
\begin{equation*}
\infty\mathsf{Grp}\defeq \sm{G:\UU}{BG:\UU_\pt^{>0}} \eqv{G}{\loopspacesym BG}.
\end{equation*}
When $G$ is an $\infty$-group, we also write $G$ for its first projection, called the \define{carrier} of $G$.
\end{defn}

\begin{rmk}
Note that we have equivalences
\begin{align*}
  \infty\mathsf{Grp}
  &\jdeq   \sm{G:\UU}{BG:\UU_\pt^{>0}} \eqv{G}{\loopspacesym BG} \\
  &\eqvsym \sm{G:\UU_\pt}{BG:\UU_\pt^{>0}} G \eqvsym_\pt \loopspacesym BG \\
  &\eqvsym \UU_\pt^{>0}
\end{align*}
for the type of higher groups.
\end{rmk}

Automorphism groups form a major class of examples of $\infty$-groups.
Given \emph{any} type $A$ and any object $a : A$, the automorphism group at $a$ is defined as
\define{automorphism group} $\Aut a\defeq (a=a)$.
This is indeed an $\infty$-group, because it is the loop space of the connected component of $A$ at $a$, i.e. we define~$\BAut a \defeq  \im(a : 1 \to A) = (x : A) \times \trunc{-1}{a=x}$.
From this definition it is immediate that $\Aut a=\loopspacesym\BAut a$, so we see that $\Aut a$ is indeed an example of an $\infty$-group.

If we take $A = \mathsf{Set}$, we get the usual symmetric groups
$\Sym_n \defeq  \Aut(\Fin(n))$, where $\Fin(n)$ is a set with $n$
elements. (Note that $\BS_n = \BAut(\Fin (n))$ is the type of all
$n$-element sets.)

We recover the ordinary set-level groups by requiring that $G$ is a $0$-type, or equivalently, that $BG$
is a $1$-type. This leads us to introduce:

\begin{defn}
We define the type of \define{groupal $(n-1)$-groupoids}, or \define{$n$-groups}, to be
\begin{equation*}
n\mathsf{Grp} \defeq \sm{G:\UU_\pt^{<n}}{BG :\UU_\pt^{>0}} G \eqvsym_\pt \loopspacesym BG.
\end{equation*}
We write $\mathsf{Grp}$ for the type of $1$-groups.
\end{defn}

The type of $n$-groups is therefore equivalent to the type of pointed connected $(n+1)$-types. Note that if $A$ is an $(n+1)$-type, then $\Aut a$ is an $(n+1)$-group because $\Aut a$ is $n$-truncated.

For example, the integers $\mathbb{Z}$ as an additive group are from this
perspective represented by their delooping $\mathop{\mathrm{B}\mathbb{Z}}=\bS^1$, i.e., the circle.
Indeed, any set-level group $G$ is represented as its delooping $BG\defeq K(G,1)$.

Moving across the homotopy hypothesis, for every pointed type $(X,x)$
we have the \define{fundamental $\infty$-group of $X$},
$\Pi_\infty(X,x)\defeq \Aut x$. Its $(n-1)$-truncation (an instance of
decategorification, see \cref{sec:stabilization}) is the
\define{fundamental $n$-group of $X$}, $\Pi_n(X,x)$,
with corresponding delooping $\mathrm{B}\Pi_n(X,x) = \trunc{n}{\BAut x}$.

Double loop spaces are more well-behaved than mere loop
spaces. For example, they are commutative up to homotopy
by the Eckmann-Hilton argument~\cite[Theorem~2.1.6]{hottbook}.
Triple loop spaces are even better behaved than double loop spaces, and so on.

\begin{defn}
A type $G$ is said to be \define{$k$-tuply groupal} if it comes equipped with a \define{$k$-fold delooping}, i.e.~ a pointed $k$-connected
$B^kG : \UU_\pt^{\ge k}$ and an equivalence $G \eqvsym \loopspacesym^kB^kG$.

Mixing the two directions, we also define
\begin{align*}
  (n,k)\GType
  &\defeq  \sm{G : \UU_\pt^{\le n}}{B^kG : \UU_\pt^{\ge k}}
    G \eqvsym_\pt \loopspacesym^kB^kG \\
  & \phantom{:}\eqvsym \UU_\pt^{\ge k,\le n+k}
\end{align*}
for the type of \define{$k$-tuply groupal $n$-groupoids}\footnote{This
  is called $n\UU_k$ in \cite{BaezDolan1998}, but here we give equal
  billing to $n$ and $k$,
  and we add the ``G'' to indicate group-structure.}.
We allow taking $n=\infty$, in which case the truncation requirement
is simply dropped.
\end{defn}

Note that $n\mathsf{Grp} = (n-1,1)\GType$. This shift in indexing is slightly
annoying, but we keep it to stay consistent with the literature.

Note that for each $k\geq 0$ there is a forgetful map
\begin{equation*}
(n,k+1)\GType \to (n,k)\GType,
\end{equation*}
given by $B^{k+1}G\mapsto \loopspacesym B^{k+1}G$, defining a sequence
\begin{equation*}
\begin{tikzcd}
\cdots \arrow[r] & (n,2)\GType \arrow[r] & (n,1)\GType \arrow[r] & (n,0)\GType.
\end{tikzcd}
\end{equation*}
Thus we define $(n,\infty)\GType$ as the limit of this sequence:
\begin{align*}
(n,\infty)\GType & \defeq  \lim_k{}(n,k)\GType \\
&\phantom{:}\eqvsym \sm{B^{\blank}G : \prd{k : \bN}\UU_\pt^{\ge k,\le n+k}}\prd{k : \bN} B^kG \eqvsym_\pt \loopspacesym B^{k+1}G.
\end{align*}
In \cref{sec:stabilization} we prove the stabilization theorem
(\cref{thm:stabilization}), from which it follows that
$(n,\infty)\GType=(n,k)\GType$ for $k\geq n+2$.

The type $(\infty,\infty)\GType$ is the type of \define{stably groupal $\infty$-groups},
also known as \define{connective spectra}. If we also relax the
connectivity requirement, we get the type of all spectra, and we can
think of a spectrum as a kind of $\infty$-groupoid with $k$-morphisms
for all $k\in\mathbb{Z}$.

The double hierarchy of higher groups is summarized in~\cref{tab:periodic}.
We shall prove the correctness of the $n=0$ column in~\cref{sec:n=0}.
\begin{table}
  \caption{\label{tab:periodic}Periodic table of $k$-tuply groupal $n$-groupoids.}
  \centering
  \begin{tabular}{clllll} \toprule
    $k\setminus n$ & $0$ & $1$ & $2$ & $\cdots$ & $\infty$ \\
    \midrule
    $0$ & pointed set & pointed groupoid & pointed $2$-groupoid & $\cdots$ & pointed $\infty$-groupoid \\
    $1$ & group & $2$-group & $3$-group & $\cdots$ & $\infty$-group \\
    $2$ & abelian group & braided $2$-group & braided $3$-group & $\cdots$ & braided $\infty$-group \\
    $3$ & \ditto & symmetric $2$-group & sylleptic $3$-group & $\cdots$ & sylleptic $\infty$-group \\
    $4$ & \ditto & \ditto & symmetric $3$-group & $\cdots$ & ?? $\infty$-group \\
    $\vdots$ & \mbox{}\quad$\vdots$ & \mbox{}\quad$\vdots$ & \mbox{}\quad$\vdots$ & $\ddots$ & \mbox{}\quad$\vdots$ \\
    $\loopspacesym$ & \ditto & \ditto & \ditto & $\cdots$ & connective spectrum \\
    \bottomrule
  \end{tabular}
\end{table}

A homomorphism between higher groups is any
function that can be suitably delooped.

\begin{defn}
For $G,H : (n,k)\GType$, we define
\begin{align*}
\hom_{(n,k)}(G,H) & \defeq
\sm{h: G \to_\pt H}{B^k h: B^kG \to_\pt B^kH} \loopspacesym^k(B^k h) \sim_\pt h \\
&\phantom{:}\eqvsym (B^k h: B^kG \to_\pt B^kH).
\end{align*}
For (connective) spectra we need
pointed maps between all the deloopings and pointed homotopies showing
they cohere.
\end{defn}

Note that if $h,k : G \to H$ are homomorphisms between set-level
groups, then $h$ and $k$ are \define{conjugate} if $Bh, Bk : BG \to_\pt BH$ are
\define{freely} homotopic (i.e., equal as maps $BG \to BH$).

Also observe that
\begin{align*}
\pi_j(B^kG \to_\pt B^kH) & \eqvsym \trunc{0}{B^kG \to_\pt \loopspacesym^jB^kH} \\
& \eqvsym \trunc{0}{\Sigma^jB^kG \to_\pt B^kH} \\
& \eqvsym 0
\end{align*}
for $j>n$, which suggests that $\hom_{(n,k)}(G,H)$ is $n$-truncated. To prove this, we deviate slightly from the approach in \cite{BuchholtzDoornRijke} and use the following intermediate result.

\subsection{The stabilization theorem for higher groups}
\label{sec:stabilization}

\begin{defn}
The \define{decategorification} $\Decat G$ of a $k$-tuply groupal $(n+1)$-group is defined to be the $k$-tuply groupal $n$-group $\trunc{n-1}{G}$, which has delooping $\trunc{n+k-1}{B^kG}$. Thus, decategorification is an operation
\begin{equation*}
\Decat : (n,k)\GType \to (n-1,k)\GType.
\end{equation*}
The functorial action of $\Decat$ is defined in the expected way. We also define the \define{$\infty$-decategorification} $\iDecat G$ of a $k$-tuply groupal $\infty$-group as the $k$-tuply groupal $n$-group $\trunc{n}{G}$, which has delooping $\trunc{n+k}{B^k G}$.
\end{defn}

\begin{defn}
The \define{discrete categorification} $\Disc G$ of a $k$-tuply-groupal $(n+1)$-group is defined to be the same $\infty$-group $G$, now considered as a $k$-tuply groupal $(n+2)$-group. Thus, the discrete categorification is an operation
\begin{equation*}
\Disc : (n,k)\GType \to (n+1,k)\GType.
\end{equation*}
Similarly, the \define{discrete $\infty$-decategorification} $\iDisc G$ of a $k$-tuply groupal $(n+1)$-group is defined to be the same group, now considered as a $k$-tuply groupal $\infty$-group.
\end{defn}

\begin{rmk}
The decategorification and discrete categorification functors make the $(n+1)$-category $(n,k)\GType$ a reflective sub-$(\infty,1)$-category of $(n+1,k)\GType$. That is, there is an adjunction ${\Decat} \dashv {\Disc}$. These properties are straightforward consequences of the universal property of truncation.
Similarly, we have ${\iDecat} \dashv {\iDisc}$ such that the counit induces an isomorphism ${\iDecat} \circ {\iDisc} = \idfunc$.
\end{rmk}

For the next constructions, we need the following properties.
\begin{defn}
  For $A : \UU_\pt$ we define the \define{$n$-connected cover} of $A$ to be
  $A{\angled n} \defeq  \fibf{A \to \trunc{n}{A}}$. We have the projection $p_1: A{\angled n} \to_\pt A$.
\end{defn}

\begin{lem} \label{lem:connected-cover-univ}
  The universal property of the $n$-connected cover states the following. For any $n$-connected pointed type $B$, the pointed map
  $$(B \to_\pt A{\angled n}) \to_\pt (B \to_\pt A),$$
  given by postcomposition with $p_1$, is an equivalence.\\
\end{lem}

\begin{proof}
  Given a map $f:B\to_\pt A$, we can form a map $\widetilde f: B \to A{\angled n}$. First note that for $b:B$ the type $\truncunit{fb}_n=_{\trunc{n}{A}}\truncunit{\pt}_n$ is $(n-1)$-truncated and inhabited for $b=\pt$. Since $B$ is $n$-connected, the universal property for connected types shows that we can construct a $qb:\truncunit{fb}_n=\truncunit{\pt}_n$ for all $b$ such that $q_0:qb_0\cdot\mathsf{ap}_{\truncunit{\blank}_n}(f_0)=1$. Then we can define the map $\widetilde f(b)\defeq (fb, qb)$. Now $\widetilde f$ is pointed, because $(f_0,q_0):(fb_0,qb_0)=(a_0,1)$.

  Now we show that this is indeed an inverse to the given map. On the one hand, we need to show that if $f: B \to_\pt A$, then $\proj 1 \circ \widetilde f=f$. The underlying functions are equal because they both send $b$ to $f(b)$. They respect points in the same way, because
  $\mathsf{ap}{p_1}(\widetilde f_0)=f_0$. The proof that the other composite is the identity follows from a computation using fibers and connectivity, which we omit here, but can be found in the formalization.
\end{proof}

The next reflective sub-$(\infty,1)$-category is formed by looping and delooping.
\begin{description}
\item[looping] $\loopspacesym : (n,k)\GType \to (n-1,k+1)\GType$ \\
  $\angled{G,B^kG} \mapsto \angled{\loopspacesym G,B^kG{\angled k}}$
\item[delooping] $\B : (n,k)\GType \to (n+1,k-1)\GType$ \\
  $\angled{G,B^kG} \mapsto \angled{\loopspacesym^{k-1}B^kG,B^kG}$
\end{description}
We have ${\B} \dashv {\loopspacesym}$, which follows from Lemma \ref{lem:connected-cover-univ} %note: autoref writes "Theorem"
and $\loopspacesym\circ{\B} = \idfunc$, which follows from the fact that $A{\angled n}=A$ if $A$ is $n$-connected.

The last adjoint pair of functors is given by stabilization and forgetting. This does not form a reflective sub-$(\infty,1)$-category.
\begin{description}
\item[forgetting] $F : (n,k)\GType \to (n,k-1)\GType$ \\
  $\angled{G,B^kG} \mapsto \angled{G,\loopspacesym B^kG}$
\item[stabilization] $S : (n,k)\GType \to (n,k+1)\GType$ \\
  $\angled{G,B^kG} \mapsto \angled{SG,\trunc{n+k+1}{\susp B^kG}}$,\\
  where $SG = \trunc{n}{\loopspacesym^{k+1}\susp B^kG}$
\end{description}
We have the adjunction ${S} \dashv {F}$ which follows from the suspension-loop adjunction $\Sigma\dashv\loopspacesym$ on pointed types.

The next main goal in this section is the stabilization theorem,
stating that the ditto marks in~\cref{tab:periodic} are justified.

The following corollary is almost \cite[Lemma~8.6.2]{hottbook}, but
proving this in Book HoTT is a bit tricky. See the
formalization for details.
\begin{lem}[Wedge connectivity]
  \label{lem:wedge-connectivity}
  If $A : \UU_\pt$ is $n$-connected and $B: \UU_\pt$ is
  $m$-connected, then the map $A \vee B \to A \times B$ is
  $(n+m)$-connected.
\end{lem}

Let us mention that there is an alternative way to prove the wedge
connectivity lemma: Recall that if $A$ is $n$-connected and $B$ is
$m$-connected, then $A \ast B$ is
$(n+m+2)$-connected~\cite[Theorem~6.8]{joinconstruction}. Hence the
wedge connectivity lemma is also a direct consequence of the following lemma.
\begin{lem}
Let $A$ and $B$ be pointed types.
The fiber of the wedge inclusion $A\vee B\to A\times B$ is equivalent to
$\loopspacesym{A}\ast\loopspacesym{B}$.
\end{lem}
\begin{proof}
Note that the fiber of $A\to A\times B$ is $\loopspacesym B$, the fiber of $B\to A\times B$ is $\loopspacesym A$, and of course the fiber of $1\to A\times B$ is $\loopspacesym A\times \loopspacesym B$. We get a commuting cube
\begin{equation*}
\begin{tikzcd}
& \loopspacesym A\times \loopspacesym B \arrow[dl] \arrow[d] \arrow[dr] \\
\loopspacesym B \arrow[d] & 1 \arrow[dl] \arrow[dr] & \loopspacesym A \arrow[dl,crossing over] \arrow[d] \\
A \arrow[dr] & 1 \arrow[d] \arrow[from=ul,crossing over] & B \arrow[dl] \\
& A\times B
\end{tikzcd}
\end{equation*}
in which the vertical squares are pullback squares.

By the descent theorem for pushouts it now follows that $\loopspacesym A\ast \loopspacesym B$ is the fiber of the wedge inclusion.
\end{proof}

The second main tool we need for the stabilization theorem is:
\begin{thm}[Freudenthal]
  If $A : \UU_\pt^{>n}$ with $n\ge 0$, then the map
  $A \to \loopspacesym\susp A$ is $2n$-connected.
\end{thm}
This is \cite[Theorem~8.6.4]{hottbook}.

The final building block we need is:
\begin{lem}
  There is a pullback square
  \[
    \begin{tikzcd}
      \susp\loopspacesym A \ar[d,"\varepsilon_A"']\ar[r] & A \vee A \ar[d] \\
      A \ar[r,"\Delta"'] & A \times A
    \end{tikzcd}
  \]
  for any $A : \UU_\pt$.
\end{lem}

\begin{proof}
Note that the pullback of $\Delta:A\to A\times A$ along either inclusion $A\to A\times A$ is contractible. So we have a cube
\begin{equation*}
\begin{tikzcd}
& \loopspacesym A \arrow[dl] \arrow[d] \arrow[dr] \\
1 \arrow[d] & 1 \arrow[dl] \arrow[dr] & 1 \arrow[dl,crossing over] \arrow[d] \\
A \arrow[dr] & A \arrow[d,"\Delta"] \arrow[from=ul,crossing over] & A \arrow[dl] \\
& A\times A
\end{tikzcd}
\end{equation*}
in which the vertical squares are all pullback squares. Therefore, if we pull back along the wedge inclusion, we obtain by the descent theorem for pushouts that the square in the statement is indeed a pullback square.
\end{proof}

\begin{thm}[Stabilization]
  \label{thm:stabilization}
  If $k\ge n+2$, then $S : (n,k)\GType \to (n,k+1)\GType$ is an
  equivalence, and any $G : (n,k)\GType$ is an infinite loop space.
\end{thm}
\begin{proof}
  We show that $F\circ S=\idfunc=S\circ F : (n,k)\GType \to (n,k)\GType$
  whenever $k\ge n+2$.

  For the first, the unit map of the adjunction factors as
  \[
    B^kG \to \loopspacesym\susp B^kG \to \loopspacesym\trunc{n+k+1}{\susp B^kG}
  \]
  where the first map is $2k-2$-connected by Freudenthal, and the
  second map is $n+k$-connected. Since the domain is $n+k$-truncated,
  the composite is an equivalence whenever $2k-2 \ge n+k$.

  For the second, the counit map of the adjunction factors as
  \[
    \trunc{n+k}{\susp\loopspacesym B^kG} \to \trunc{n+k}{B^kG} \to B^kG,
  \]
  where the second map is an equivalence. By the two lemmas above, the
  first map is $2k-2$-connected.
\end{proof}
For example, for $G : (0,2)\GType$ an abelian group, we have
$B^nG = K(G,n)$, an Eilenberg-MacLane space.

The adjunction ${S} \dashv {F}$ implies that the free group on a
pointed set $X$ is $\loopspacesym\trunc{1}{\susp X}=\pi_1(\susp X)$.  If $X$
has decidable equality, $\susp X$ is already $1$-truncated. It is an
open problem whether this is true in general.

Also, the abelianization of a set-level group $G : 1\mathsf{Grp}$ is
$\pi_2(\susp BG)$. If $G : (n,k)\GType$ is in the stable range ($k \ge
n+2$), then $SFG=G$.

\subsection{Eilenberg-Mac Lane spaces}

\begin{exercises}
\exercise Show that if $X$ is $m$-connected and $f:X\to Y$ is $n$-connected, then the map
\begin{equation*}
X \to \fib{m_f}{\ast}
\end{equation*}
where $m_f:Y\to M_f$ is the inclusion of $Y$ into the cofiber of $f$, is $(m+n)$-connected.
\exercise Suppose that $X$ is a connected type, and let $f:X\to Y$ be a map.
Show that the following are equivalent:
\begin{enumerate}
\item $f$ is $n$-connected.
\item The mapping cone of $f$ is $(n+1)$-connected.
\end{enumerate}
\exercise Apply the Blakers-Massey theorem to the defining pushout square of the smash product to show that if $A$ and $B$ are $m$- and $n$-connected respectively, then there is a $(m+n+\min(m,n)+2)$-connected map
\begin{equation*}
\join{\loopspace{A}}{\loopspace{B}}\to \loopspace{A \wedge B}.
\end{equation*}
\exercise Show that the square
\begin{equation*}
\begin{tikzcd}
\unit \arrow[r] \arrow[d] & \bool \arrow[d] \\
X \arrow[r] & X+\unit
\end{tikzcd}
\end{equation*}
is both a pullback and a pushout. Conclude that the result of the Blakers-Massey theorem is not always sharp.
\item Show that for every pointed type $X$, and any $n:\N$, there is a fiber sequence
  \begin{equation*}
    K(\pi_{n+1}(X),n+1)\hookrightarrow \trunc{n+1}{X}\twoheadrightarrow \trunc{n}{X}.
  \end{equation*}
\end{exercises}


\chapter{Maxwell's Equations }
\section{Electrodynamics Before Maxwell}
Till now in electromagnetic theory we found out four equations which contribute to the foundation of Electromagnetic theory, and they are,
\begin{align*}
\vec{\nabla} \cdot \vec{E}&=\frac{\rho}{\epsilon_{0}} \quad && \Rightarrow \text { Gauss' Law }  \\
\vec{\nabla} \cdot \vec{B}&=0 \quad && \Rightarrow \text { Gauss' Law for magnetism } \\
\vec{\nabla} \times \vec{E}&=-\frac{\partial \vec{B}}{\partial t} && \Rightarrow \text { Faraday's Law } \\
\vec{\nabla} \times \vec{B}&=\mu_{0}\left(\vec{J}\right) && \Rightarrow \text { Ampere's Law }
\end{align*}
There is a fatal inconsistency in these formulas. It has something to do with the old rule that divergence of curl is always zero $ (\nabla .(\nabla \times \mathbf{E})=0)$. If you apply the divergence to Faraday's law, everything works out:
$$\nabla \cdot(\nabla \times \mathbf{E})=\nabla \cdot\left(-\frac{\partial \mathbf{B}}{\partial t}\right)=-\frac{\partial}{\partial t}(\nabla \cdot \mathbf{B})$$
The left side is zero because divergence of curl is zero; the right side is zero by virtue of Gauss law for magnetism. But when you do the same thing to Ampere's law, you get into trouble:$$\boldsymbol{\nabla} \cdot(\boldsymbol{\nabla} \times \mathbf{B})=\mu_{0}(\boldsymbol{\nabla} \cdot \mathbf{J})$$ The left side must be zero, but the right side, in general, is not. For steady currents, the divergence of $\mathbf{J}$ is zero, but evidently when we go beyond magnetostatics Ampère's law cannot be right.
\subsection{How Maxwell Fixed Ampere's Law}
Applying the continuity equation  and Gauss's law, in  Ampere's law the offending term can be rewritten as:
\begin{align*}
\nabla \cdot \mathbf{J}&=-\frac{\partial \rho}{\partial t}\\
&=-\frac{\partial}{\partial t}\left(\epsilon_{0} \nabla \cdot \mathbf{E}\right)\\&=-\nabla \cdot\left(\epsilon_{0} \frac{\partial \mathbf{E}}{\partial t}\right)
\end{align*}
If we were to combine $\epsilon_{0}(\partial \mathbf{E} / \partial t)$ with $\mathbf{J}$, in Ampère's law, it would be just right to kill off the extra divergence:$$\nabla \times \mathbf{B}=\mu_{0} \mathbf{J}+\mu_{0} \epsilon_{0} \frac{\partial \mathbf{E}}{\partial t}$$ Such a modification changes nothing, as far as magnetostatics is concerned: when $\mathbf{E}$ isconstant, we still have $\boldsymbol{\nabla} \times \mathbf{B}=\mu_{0} \mathbf{J}$.\\
Just as a changing magnetic field induces an electric field (Faraday's law), so \textbf{A changing electric field induces a magnetic field}.  Maxwell called his extra term the displacement current:

\hspace{5.10cm}\framebox{

	\parbox[t][2cm]{4cm}{

		\addvspace{0.2cm} \centering
		Displacement Current
		$$\mathbf{J}_{d} \equiv \epsilon_{0} \frac{\partial \mathbf{E}}{\partial t}$$}
}
\\\\ It's a misleading name, since $\epsilon_{0}(\partial \mathbf{E} / \partial t)$ has nothing to do with current, except that it adds to $\mathbf{J}$ in Ampère's law.
\subsection{Maxwell's Equations}
\begin{center}
	\framebox{
		\parbox[t][4.2cm]{3.5cm}{

			\addvspace{0.2cm} \centering

			\begin{align*}
			\begin{array}{lll}
			\textbf{(i)} & \nabla \cdot \mathbf{E}=\frac{1}{\epsilon_{0}} \rho&\text{(Gauss's law)} \\\\ \textbf{(ii)} & \nabla \cdot \mathbf{B}=0&\text{(No name.)}\\\\ \textbf{(iii)}& \boldsymbol{\nabla} \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}&\text{(Faraday's law)}\\\\\textbf{(iv)}&\boldsymbol{\nabla} \times \mathbf{B}=\mu_{0} \mathbf{J}+\mu_{0} \epsilon_{0} \frac{\partial \mathbf{E}}{\partial t}&\text{(Ampère's law with
				Maxwell's correction)}
			\end{array}
			\end{align*}} }
\end{center}
Every phenomenon in electricity and magnetism can be derived from these equations. Many of our most important tools for various analyses come from the integral version of these equations, which are.
\begin{center}
	\framebox{
		\parbox[t][4.2cm]{3.5cm}{

			\addvspace{0.2cm} \centering

			\begin{align*}
			\begin{array}{lll}
			\textbf{(i)} & \iint_{S} \overrightarrow{\mathbf{E}} \cdot d \overrightarrow{\mathbf{A}}=\frac{Q}{\varepsilon_{0}} &\text{(Gauss's law)} \\\\ \textbf{(ii)} & \iint_{S} \overrightarrow{\mathbf{B}} \cdot d \overrightarrow{\mathbf{A}}=0&\text{(No name.)}\\\\ \textbf{(iii)}& \oint \overrightarrow{\mathbf{E}} \cdot d \overrightarrow{\mathbf{s}}=-\frac{d \mathbf{\phi_B}}{d t}&\text{(Faraday's law)}\\\\\textbf{(iv)}&\oint \overrightarrow{\mathbf{B}} \cdot d \overrightarrow{\mathbf{s}}=\mu_{0} I+\mu_{0} \varepsilon_{0} \frac{d \Phi_{E}}{d t}&\text{(Ampère's law with
				Maxwell's correction)}
			\end{array}
			\end{align*}} }
\end{center}
Together with the force law,\ $\mathbf{F}=q(\mathbf{E}+\mathbf{v} \times \mathbf{B})$,\ they summarize the entire theoretical content of classical electrodynamics. Even the continuity equation,\ $\boldsymbol{\nabla} \cdot \mathbf{J}=-\frac{\partial \rho}{\partial t}$ \ which is the mathematical expression of conservation of charge, can be derived from Maxwell's equations by applying the divergence to Ampere's law.
\subsection{Maxwell's Equations In Free Space}
An extremely important limit of Maxwell's equations is found when there are no sources: $\rho=0, \vec{J}=0 .$ The equations become,
\begin{center}
	\framebox{
		\parbox[t][4.2cm]{3.5cm}{

			\addvspace{0.2cm} \centering

			\begin{align*}
			\begin{array}{lll}
			\textbf{(i)} & \nabla \cdot \mathbf{E}=0&\text{(Gauss's law)} \\\\ \textbf{(ii)} & \nabla \cdot \mathbf{B}=0&\text{(No name.)}\\\\ \textbf{(iii)}& \boldsymbol{\nabla} \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}&\text{(Faraday's law)}\\\\\textbf{(iv)}&\boldsymbol{\nabla} \times \mathbf{B}=\mu_{0} \epsilon_{0} \frac{\partial \mathbf{E}}{\partial t}&\text{(Ampère's law with
				Maxwell's correction)}
			\end{array}
			\end{align*}} }
\end{center}
\subsection{Maxwell's Equations in Matter}
For inside polarized matter there will be accumulations of "bound" charge and current over which you exert no direct control. It would be nice to reformulate Maxwell's equations in such a way as to make explicit reference only to those sources we control directly: the "free" charges and currents. \\
We have already learned, from the static case, that an electric polarization $\mathbf{P}$ produces a bound charge density $$\rho_{b}=-\nabla \cdot \mathbf{P}$$ Likewise, a magnetic polarization (or "magnetization") $\mathbf{M}$ results in a bound current $$\mathbf{J}_{b}=\nabla \times \mathbf{\mathbf { M }}$$ There's just one new feature to consider in the nonstatic case: Any change in the electric polarization involves a flow of (bound) charge (call it $\mathbf{J}_{p}$), which must be included in the total current. For suppose we examine a tiny chunk of polarized material The polarization introduces a charge density $\sigma_{b}=P$ at one end and $-\sigma_{b}$ at the other. If $P$ now increases a bit, the charge on each end increases accordingly. \\giving a net current $$d I=\frac{\partial \sigma_{b}}{\partial t} d a_{\perp}=\frac{\partial P}{\partial t} d a_{\perp}$$
\begin{figure}[H]
	\begin{center}
		\includegraphics[width=4cm,height=2cm]{02-crop}
	\end{center}
\end{figure}
The current density, therefore, is $$\mathbf{J}_{p}=\frac{\partial \mathbf{P}}{\partial t}$$
This polarization current has nothing whatever to do with the bound current $\mathbf{J}_{b}$. The latter is associated with magnetization of the material and involves the spin and orbital motion of electrons; $\mathbf{J}_{p}$, by contrast, is the result of the linear motion of charge when the electric polarization changes. If $\mathbf{P}$ points to the right and is increasing, then each plus charge moves a bit to the right and each minus charge to the left; the cumulative effect is the polarization current $\mathbf{J}_{p}$\\\\
In fact, $\mathbf{J}_{p}$ is essential to account for the conservation of bound charge.\\
In view of all this, the total charge density can be separated into two parts: $$\rho=\rho_{f}+\rho_{b}=\rho_{f}-\boldsymbol{\nabla} \cdot \mathbf{P}$$ and the current density into three parts:
$$\mathbf{J}=\mathbf{J}_{f}+\mathbf{J}_{b}+\mathbf{J}_{p}=\mathbf{J}_{f}+\mathbf{\nabla} \times \mathbf{M}+\frac{\partial \mathbf{P}}{\partial t}$$ Gauss's law can now be written as $$\nabla \cdot \mathbf{E}=\frac{1}{\epsilon_{0}}\left(\rho_{f}-\nabla \cdot \mathbf{P}\right)$$ \text{or}$$\boldsymbol{\nabla} \cdot \mathbf{D}=\rho_{f}$$ where $\mathbf{D}$, as in the static case, is given by $$\mathbf{D} \equiv \epsilon_{0} \mathbf{E}+\mathbf{P}$$ Meanwhile, Ampère's law (with Maxwell's term) becomes $$\nabla \times \mathbf{B}=\mu_{0}\left(\mathbf{J}_{f}+\nabla \times \mathbf{M}+\frac{\partial \mathbf{P}}{\partial t}\right)+\mu_{0} \epsilon_{0} \frac{\partial \mathbf{E}}{\partial t}$$\text{or}$$\boldsymbol{\nabla} \times \mathbf{H}=\mathbf{J}_{f}+\frac{\partial \mathbf{D}}{\partial t}$$ where, as before,$$\mathbf{H} \equiv \frac{1}{\mu_{0}} \mathbf{B}-\mathbf{M}$$ Faraday's law and $\nabla \cdot \mathbf{B}=0$ are not affected by our separation of charge and current into free and bound parts, since they do not involve $\rho$ or $\mathbf{J}$. \\
In terms of free charges and currents, then, Maxwell's equations read\
\begin{center}
	\framebox{
		\parbox[t][2.5cm]{4cm}{

			\addvspace{0cm} \centering

			\begin{align*}
			\begin{array}{llll}
			\textbf{(i)} & \nabla \cdot \mathbf{D}=\rho_{f} &\textbf{(iii)} & \boldsymbol{\nabla} \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t},\\\\
			\textbf{(ii)}&\boldsymbol{\nabla} \cdot \mathbf{B}=0&\textbf{(iv)}& \boldsymbol{\nabla} \times \mathbf{H}=\mathbf{J}_{f}+\frac{\partial \mathbf{D}}{\partial t}
			\end{array}
			\end{align*}} }
\end{center}
for linear media
\begin{align*}
\begin{array}{lll}
\mathbf{P}=\epsilon_{0} \chi_{e} \mathbf{E} & and& \mathbf{M}=\chi_{m} \mathbf{H}\\\\
\mathbf{D}=\epsilon \mathbf{E}&and&\mathbf{H}=\frac{1}{\mu} \mathbf{B}
\end{array}
\end{align*}
where $\epsilon \equiv \epsilon_{0}\left(1+\chi_{e}\right)$ and $\mu \equiv \mu_{0}\left(1+\chi_{m}\right)$ D is called the electric "displacement"; that's why the second term in the Ampère/Maxwelf equation (iv) is called the displacement current,\\
In integral form Maxwell's equations in matter can be written as\\
\begin{align*}
\text{(i)}\hspace{0.5cm}\oint_{s}D\cdot da&=\theta_{fenc}\\
\text{(ii)}\hspace{0.5cm}\oint_{s}B\cdot da&=0\\
\text{(iii)}\hspace{0.5cm}\oint_{p}E\cdot dl&=\frac{-d}{dt}\int_{s} B\cdot da\\
\text{(iv)}\hspace{0.5cm}\oint_{p}H\cdot dl&=I_{end}+\frac{d}{dt}\int_{s} D\cdot da\\
\end{align*}
$P$ is the closed loop enclosing the surface $S$
\section{Boundary conditions}
In general the fields \textbf{E,B,D,} and \textbf{H} will be discontineous at a boundary between two different media or at surface that carries charge density $\sigma$ or current density \textbf{K}. The explicit form of the discontinities can be deduced from Maxwell's equations,in their integral form\\\\
$\left. \right. $\hspace{0.28cm} (i) $\oint_s \mathbf{D} \cdot d \mathbf{a}=Q_{f_{\mathrm{enc}}}$\\\\
$\left. \right. $\hspace{0.3cm}(ii) $\oint_s \mathbf{B} \cdot d \mathbf{a}=0$\\\\
$\left.\begin{array}{ll}\text { (iii) } \oint_l \mathbf{E} \cdot d \mathbf{l} & =-\frac{d}{d t} \int_s \mathbf{B} \cdot d \mathbf{a} \\ \\
\text { (iv) } \oint_l \mathbf{H} \cdot d \mathbf{l} & =I_{f_{\text {enc }}}+\frac{d}{d t} \int_s \mathbf{D} \cdot d \mathbf{a}\end{array}\right\} \begin{aligned}&\text { for any surface } s \\&\text { bounded by the } \\&\text { closed loop } l .\end{aligned}$\\\\
Applying (i) to a tiny, wafer-thin Gaussian pillbox extending just slightly into the material on either side of the boundary, we obtain \\
\begin{figure}[H]
	\centering
	\includegraphics[height=4cm,width=7cm]{diagram-20220103(8)-crop}
	\caption{}
	\label{}
\end{figure}
$$\mathbf{D}_{1} \cdot \mathbf{a}-\mathbf{D}_{2} \cdot \mathbf{a}=\sigma_{f} a$$
Thus ,the component of D that is perperdicular to the interface is discontineous in the amount \\
$$D_{1}^{\perp}-D_{2}^{\perp}=\sigma_{f}$$
Identical reasoning,applied to equation(ii) yields\\
$$B_1^{\perp}-B_2^{\perp}=0$$
consider equation(iii) a very thin Amperian loop straddling the surface gives\\
\begin{figure}[H]
	\centering
	\includegraphics[height=4cm,width=7cm]{diagram-20220103(9)-crop}
	\caption{}
	\label{}
\end{figure}
$$E_1\cdot L-E_2\cdot L=-\frac{d}{dt} \int_{s} B\cdot da$$
But in the limit as the width of the loop goes to zero ,the flux vanishes.Therefore\\
$$E_1^{\parallel}-E_2^{\parallel}=0$$
That is the comonents of E parallel to the interface are contineous across the boundary.\\
equation (iv)  implies \\
$$H_1 \cdot L-H_2\cdot L=I_{f_{enc}}$$
Where $I_{f_{enc}}$ is the free current passing through the Amperian loop.No volume current density will contribute but a surface current can . In fact if $\hat{n}$ is a unit vector perpendicular to the interface so that $(\hat{n}\times L)$ is normal to the Amperian loop then,
$$I_{f_{enc}}=K_f \cdot (\hat{n}\times L)=(K_f\times \hat{n})\cdot L$$
And hence $$H_1^{\parallel}-H_2^{\parallel}=K_f\times \hat{n}$$
So the parallel components  H are discontinuous by an amount proportional to the free surface charge density.\\
In the case of linear media they can be expressed in terms of E and B alone

\begin{enumerate}[label=(\roman*)]
	\item $\epsilon_{1} \mathbf{E}_{1}^{\perp}-\epsilon_{2} \mathbf{E}_{2}^{\perp}=\sigma_{f}$
	\item  $\mathbf{E}_{1}^{\|}-\mathbf{E}_{2}^{\|}=0$
	\item $\mathbf{B}_{1}^{\perp}-\mathbf{B}_{2}^{\perp}=0$
	\item $\frac{1}{\mu_{1}} \mathbf{B}_{1}^{\|}-\frac{1}{\mu_{2}} \mathbf{B}_{2}^{\|}=\mathbf{K}_{f} \times \hat{\mathbf{n}}$
\end{enumerate}

In particular, if there is no free charge or free current at the interface, then
\begin{enumerate}[label=(\roman*)]
	\item $\epsilon_{1} \mathbf{E}_{1}^{\perp}-\epsilon_{2} \mathbf{E}_{2}^{\perp}=0$
	\item $\mathbf{E}_{1}^{\|}-\mathbf{E}_{2}^{\|}=0$
	\item $\mathbf{B}_{1}^{\perp}-\mathbf{B}_{2}^{\perp}=0$
	\item $\frac{1}{\mu_{1}} \mathbf{B}_{1}^{\|}-\frac{1}{\mu_{2}} \mathbf{B}_{2}^{\|}=0$
\end{enumerate}

\newpage
\begin{abox}
	Practise Set-1
\end{abox}
\begin{enumerate}
	\item The $x$ - and $z$-components of a static magnetic field in a region are $B_{x}=B_{0}\left(x^{2}-y^{2}\right)$ and $B_{z}=0$, respectively. Which of the following solutions for its $y$-component is consistent with the Maxwell equations?
	{\exyear{ NET/JRF-(JUNE-2016)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$B_{y}=B_{0} x y$
		\task[\textbf{b.}]$B_{y}=-2 B_{0} x y$
		\task[\textbf{c.}] $B_{y}=-B_{0}\left(x^{2}-y^{2}\right)$
		\task[\textbf{d.}]  $B_{y}=B_{0}\left(\frac{1}{3} x^{3}-x y^{2}\right)$
	\end{tasks}
	\item A current $i_{p}$ flows through the primary coil of a transformer. The graph of $i_{p}(t)$ as a function of time $t$ is shown in the figure below.\\
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=4cm]{diagram-20211011(27)-crop}
	\end{figure}
	Which of the following graphs represents the current $i_{S}$ in the secondary coil?
	{\exyear{NET/JRF(JUNE-2014)}}
	\begin{tasks}(2)
		\task[\textbf{A.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=4cm]{diagram-20211011(28)-crop}
		\end{figure}
		\task[\textbf{B.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=4cm]{diagram-20211011(29)-crop}
		\end{figure}
		\task[\textbf{C.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=4cm]{diagram-20211011(30)-crop}
		\end{figure}
		\task[\textbf{D.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=4cm]{diagram-20211011(31)-crop}
		\end{figure}
	\end{tasks}
	\item A circular conducting wire loop is placed close to a solenoid as shown in the figure bellow. Also shown is the current through the solenoid as a function of solenoid as a function of time.\\
	\begin{figure}[H]
		\centering
		\includegraphics[height=3.5cm,width=9cm]{diagram-20211028(5)-crop}
	\end{figure}
	The magnitude $|i(t)|$ of the induced current in the wire loop, as a function of time $t$, is best represented as.
	{\exyear{NET/JRF(DEC-2019)}}
	\begin{tasks}(2)
		\task[\textbf{A.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=5cm]{diagram-20211028(6)-crop}
		\end{figure}
		\task[\textbf{B.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=5cm]{diagram-20211028(7)-crop}
		\end{figure}
		\task[\textbf{C.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=5cm]{diagram-20211028(8)-crop}
		\end{figure}
		\task[\textbf{D.}] \begin{figure}[H]
			\centering
			\includegraphics[height=3cm,width=5cm]{diagram-20211028(9)-crop}
		\end{figure}
	\end{tasks}
	\item A horizontal metal disc rotates about the vertical axis in a uniform magnetic field pointing up as shown in the figure. A circuit is made by connecting one end A of a resistor to the centre of the disc and the other end $B$ to its edge through a sliding contact. The current that flows through the resistor is
	{\exyear{NET/JRF(DEC-2013)}}
	\begin{figure}[H]
		\centering
		\includegraphics[height=4cm,width=5cm]{diagram-20211011(22)-crop}
	\end{figure}
	\begin{tasks}(4)
		\task[\textbf{A.}] Zero
		\task[\textbf{B.}] $D C$ from $A$ to $B$
		\task[\textbf{C.}] $D C$ from $B$ to $A$
		\task[\textbf{D.}] $A C$,
	\end{tasks}
	\item  A conducting circular disc of radius $r$ and resistivity $\rho$ rotates with an angular velocity $\omega$ in a magnetic field $B$ perpendicular to it. A voltmeter is connected as shown in the figure below. Assuming its internal resistance to be infinite, the reading on the voltmeter
	{\exyear{NET/JRF(DEC-2016)}}
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{diagram-20211011(46)-crop}
	\end{figure}
	\begin{tasks}(1)
		\task[\textbf{A.}] Depends on $\omega, B, r$ and $\rho$
		\task[\textbf{B.}] Depends on $\omega, B$ and $r$ but not on $\rho$
		\task[\textbf{C.}]  Is zero because the flux through the loop is not changing
		\task[\textbf{D.}] Is zero because a current the flows in the direction of $B$
	\end{tasks}
	\item A uniform magnetic field in the positive $z$-direction passes through a circular wire loop of radius $1 \mathrm{~cm}$ and resistance $1 \Omega$ lying in the $x y$-plane. The field strength is reduced from 10 tesla to 9 tesla in $1 s$. The charge transferred across any point in the wire is approximately
	{\exyear{ NET/JRF-(JUNE-2015)	}}
	\begin{tasks}(2)
		\task[\textbf{a.}] $3.1 \times 10^{-4}$ coulomb
		\task[\textbf{b.}]$3.4 \times 10^{-4}$ coulomb
		\task[\textbf{c.}]$4.2 \times 10^{-4}$ coulomb
		\task[\textbf{d.}] $5.2 \times 10^{-4}$ coulomb
	\end{tasks}
	\item A magnetic field $B$ is $B \hat{z}$ in the region $x>0$ and zero elsewhere. A rectangular loop, in the $x y$-plane, of sides $l$ (along the $x$-direction) and $h$ (along the $y$-direction) is inserted into the $x>0$ region from the $x<0$ region at constant velocity $v=v \hat{x}$. Which of the following values of $l$ and $h$ will generate the largest EMF?
	{\exyear{ NET/JRF-(JUNE-2016)	}}
	\begin{tasks}(2)
		\task[\textbf{a.}] $l=8, h=3$
		\task[\textbf{b.}]$l=4, h=6$
		\task[\textbf{c.}]$l=6, h=4$
		\task[\textbf{d.}]  $l=12, h=2$
	\end{tasks}
	\item Consider a solenoid of radius $R$ with $n$ turns per unit length, in which a time dependent current $I=I_{0} \sin \omega t$ (where $\omega R / c<<1$ ) flows. The magnitude of the electric field at a perpendicular distance $r<R$ from the axis of symmetry of the solenoid, is
	{\exyear{NET/JRF-(DEC-2011)	}}
	\begin{tasks}(2)
		\task[\textbf{a.}]0
		\task[\textbf{b.}]$\frac{1}{2 r} \omega \mu_{0} n I_{0} R^{2} \cos \omega t$
		\task[\textbf{c.}]$\frac{1}{2} \omega \mu_{0} n I_{0} r \sin \omega t$
		\task[\textbf{d.}]  $\frac{1}{2} \omega \mu_{0} n I_{0} r \cos \omega t$
	\end{tasks}
	\item A parallel plate capacitor is formed by two circular conducting plates of radius a separated by a distance $d$, where $d \ll a$. It is being slowly charged by a current that is nearly constant. At an instant when the current is $I$, the magnetic induction between the plates at a distance $\frac{a}{2}$ from the centre of the plate, is
	{\exyear{ NET/JRF-(DEC-2016)}}
	\begin{tasks}(2)
		\task[\textbf{a.}]$\frac{\mu_{0} I}{\pi a}$
		\task[\textbf{b.}]$\frac{\mu_{0} I}{2 \pi a}$
		\task[\textbf{c.}]$\frac{\mu_{0} I}{a}$
		\task[\textbf{d.}]  $\frac{\mu_{0} I}{4 \pi a}$
	\end{tasks}
	\item Suppose the $y z$-plane forms a chargeless boundary between two media of permittivities $\epsilon_{\text {left }}$ and $\epsilon_{\text {right }}$ where $\epsilon_{\text {left }}: \epsilon_{\text {right }}=1: 2$, if the uniform electric field on the left is $\vec{E}_{\text {left }}=c(\hat{i}+\hat{j}+\hat{k})$ (where $c$ is a constant), then the electric field on the right $\vec{E}_{\text {right }}$ is
	{\exyear{NET/JRF(JUNE-2015)}}
	\begin{tasks}(4)
		\task[\textbf{A.}]  $c(2 \hat{i}+\hat{j}+\hat{k})$
		\task[\textbf{B.}] $c(\hat{i}+2 \hat{j}+2 \hat{k})$
		\task[\textbf{C.}] $c\left(\frac{1}{2} \hat{i}+\hat{j}+\hat{k}\right)$
		\task[\textbf{D.}] $c\left(\hat{i}+\frac{1}{2} \hat{j}+\frac{1}{2} \hat{k}\right)$
	\end{tasks}
	\item  The half space region $x>0$ and $x<0$ are filled with dielectric media of dielectric constants $\varepsilon_{1}$ and $\varepsilon_{2}$ respectively. There is a uniform electric field in each part. In the right half, the electric field makes an angle $\theta_{1}$ to the interface. The corresponding angle $\theta_{2}$ in the left half satisfies
	{{\exyear{NET/JRF(JUNE-2016)}}}
	\begin{figure}[H]
		\centering
		\includegraphics[height=4cm,width=5cm]{diagram-20211011(41)-crop}
	\end{figure}
	\begin{tasks}(2)
		\task[\textbf{A.}] $\varepsilon_{1} \sin \theta_{2}=\varepsilon_{2} \sin \theta_{1}$
		\task[\textbf{B.}] $\varepsilon_{1} \tan \theta_{2}=\varepsilon_{2} \tan \theta_{1}$
		\task[\textbf{C.}] $\varepsilon_{1} \tan \theta_{1}=\varepsilon_{2} \tan \theta_{2}$
		\task[\textbf{D.}] $\varepsilon_{1} \sin \theta_{1}=\varepsilon_{2} \sin \theta_{2}$
	\end{tasks}
	\item Which of the following is not a correct boundary condition at an interface between two homogeneous dielectric media? (In the following $\hat{n}$is a unit vector normal to the  interface, $\sigma$ and $\vec{j}_s$, are the surface charge and current densities, respectively.)
	{\exyear{NET/JRF(JUNE-2019)}}
	\begin{tasks}(2)
		\task[\textbf{A.}] $\hat{n} \times\left(\vec{D}_{1}-\vec{D}_{2}\right)=0$
		\task[\textbf{B.}] $\hat{n} \times\left(\vec{H}_{1}-\vec{H}_{2}\right)=\vec{j}_{s}$
		\task[\textbf{C.}] $\hat{n} \cdot\left(\vec{D}_{1}-\vec{D}_{2}\right)=\sigma$
		\task[\textbf{D.}] $\hat{n} \cdot\left(\vec{B}_{1}-\vec{B}_{2}\right)=0$
	\end{tasks}
\end{enumerate}
 \colorlet{ocre1}{ocre!70!}
\colorlet{ocrel}{ocre!30!}
\setlength\arrayrulewidth{1pt}
\begin{table}[H]
	\centering
	\arrayrulecolor{ocre}
	\begin{tabular}{|p{1.5cm}|p{1.5cm}||p{1.5cm}|p{1.5cm}|}
		\hline
		\multicolumn{4}{|c|}{\textbf{Answer key}}\\\hline\hline
		\rowcolor{ocrel}Q.No.&Answer&Q.No.&Answer\\\hline
		1&\textbf{b} &2&\textbf{c}\\\hline
		3&\textbf{d} &4&\textbf{c} \\\hline
		5&\textbf{b} &6&\textbf{a} \\\hline
		7&\textbf{b}&8&\textbf{d}\\\hline
		9&\textbf{d}&10&\textbf{c}\\\hline
		11&\textbf{c} &12&\textbf{a}\\\hline

	\end{tabular}
\end{table}
\newpage
\begin{abox}
	Practise Set-2
\end{abox}
\begin{enumerate}
	\item Two rails of a railroad track are insulated from each other and from the ground, and are connected by a millivoltmeter. What is the reading of the millivoltmeter when a train travels at the speed $90 \mathrm{~km} / \mathrm{hr}$ down the track? Assume that the vertical component of the earth's magnetic field is $0.2$ gauss and that the tracks are separated by two meters. Use 1 gauss $=10^{-4}$ Tesla $=10^{-4} \mathrm{~V} \cdot \mathrm{sec} / \mathrm{m}^{2}$
	{\exyear{ JEST-2020}}
	 \begin{tasks}(2)
		\task[\textbf{a.}]10
		\task[\textbf{b.}]1
		\task[\textbf{c.}]$0.2$
		\task[\textbf{d.}]180
	\end{tasks}
\item 	Which of the following expressions represents an electric field due to a time varying magnetic field?
	{\exyear{ JEST-2015}}
 \begin{tasks}(2)
	\task[\textbf{a.}]$K(x \hat{x}+y \hat{y}+z \hat{z})$
	\task[\textbf{b.}] $K(x \hat{x}+y \hat{y}-z \hat{z})$
	\task[\textbf{c.}]$K(x \hat{x}-y \hat{y})$
	\task[\textbf{d.}]$K(y \hat{y}-x \hat{y}+2 z \hat{z})$
\end{tasks}
\item Two parallel rails of a railroad track are insulated from each other and from the ground. The distance between the rails is 1 meter. A voltmeter is electrically connected between the rails. Assume the vertical component of the earth's magnetic field to the $0.2$ gauss. What is the voltage developed between the rails when a train travels at a speed of $180 \mathrm{~km} / \mathrm{h}$ along the track? Give the answer in milli-volts.
	{\exyear{ JEST-2018}}
\item 	A very long solenoid (axis along $z$ direction) of $n$ turns per unit length carries a current which increases linearly with time, $i=K t$. What is the magnetic field inside the solenoid at a given time $t$ ?
	{\exyear{ JEST-2019}}
 \begin{tasks}(2)
	\task[\textbf{a.}]$\vec{B}=\mu_{0} n K t \hat{z}$
	\task[\textbf{b.}]$\vec{B}=\mu_{0} n K \hat{z}$
	\task[\textbf{c.}]$\vec{B}=\mu_{0} n K t(\hat{x}+\hat{y})$
	\task[\textbf{d.}] $\vec{B}=\mu_{0} c n K t \hat{z}$
\end{tasks}
\item 	A circular metal loop of radius $a=1 \mathrm{~m}$ spins with a constant angular velocity $\omega=20 \pi \mathrm{rad} / \mathrm{s}$ in a magnetic field $B=3$ Tesla, as shown in the figure. The resistance of the loop is 10 ohms. Let $P$ be the power dissipated in one complete cycle. What is the value of $\frac{P}{\pi^{4}}$ in Watts?
		{\exyear{ JEST-2019}}
	\begin{figure}[H]
		\centering
		\includegraphics[height=4cm,width=2.8cm]{ED-20}
	\end{figure}
\item Self inductance per unit length of a long solenoid of radius $R$ with $n$ turns per unit length is:
	{\exyear{ JEST-2016}}
 \begin{tasks}(2)
	\task[\textbf{a.}]$\mu_{0} \pi R^{2} n^{2}$
	\task[\textbf{b.}]$2 \mu_{0} \pi R^{2} n$
	\task[\textbf{c.}]$2 \mu_{0} \pi R^{2} n^{2}$
	\task[\textbf{d.}] $\mu_{0} \pi R^{2} n$
\end{tasks}
\item The $x-y$ plane is the boundary between free space and a magnetic material with relative permeability $\mu_{r}$. The magnetic field in the free space is $B_{x} \hat{i}+B_{z} \hat{k}$. The magnetic field in the magnetic material is
	{\exyear{ GATE- 2016}}
 \begin{tasks}(2)
	\task[\textbf{a.}]$B_{x} \hat{i}+B_{z} \hat{k}$
	\task[\textbf{b.}]$B_{x} \hat{i}+\mu_{r} B_{z} \hat{k}$
	\task[\textbf{c.}] $\frac{1}{\mu_{r}} B_{x} \hat{i}+B_{z} \hat{k}$
	\task[\textbf{d.}] $\mu_{r} B_{x} \hat{i}+B_{z} \hat{k}$
\end{tasks}
\item 	At a surface current, which one of the magnetostatic boundary condition is $\underline{\text { NOT }}$ CORRECT?
{\exyear{ GATE- 2013}}
	 \begin{tasks}(2)
		\task[\textbf{a.}]Normal component of the magnetic field is continuous.
		\task[\textbf{b.}]Normal component of the magnetic vector potential is continuous.
		\task[\textbf{c.}] Tangential component of the magnetic vector potential is continuous.
		\task[\textbf{d.}] Tangential component of the magnetic vector potential is not continuous.
	\end{tasks}
\item A circular loop made of a thin wire has radius $2 \mathrm{~cm}$ and resistance $2 \Omega$. It is placed perpendicular to a uniform magnetic field of magnitude $\left|\vec{B}_{0}\right|=0.01$ Tesla. At time $t=0$ the field starts decaying as $\vec{B}=\vec{B}_{0} e^{-t / t_{0}}$, where $t_{0}=1 s$. The total charge that passes through a cross section of the wire during the decay is $Q$. The value of $Q$ in $\mu C$ (rounded off to two decimal places) is
{\exyear{ GATE- 2019}}
\item A long solenoid is embedded in a conducting medium and is insulated from the medium. If the current through the solenoid is increased at a constant rate, the induced current in the medium as a function of the radial distance $r$ from the axis of the solenoid is proportional to
{\exyear{GATE- 2015}}
 \begin{tasks}(2)
	\task[\textbf{a.}] $r^{2}$ inside the solenoid and $\frac{1}{r}$ outside
	\task[\textbf{b.}]$r$ inside the solenoid and $\frac{1}{r^{2}}$ outside
	\task[\textbf{c.}]$r^{2}$ inside the solenoid and $\frac{1}{r^{2}}$ outside
	\task[\textbf{d.}] $r$ inside the solenoid and $\frac{1}{r}$ outside
\end{tasks}
\item Consider an infinitely long solenoid with $N$ turns per unit length, radius $R$ and carrying a current $I(t)=\alpha \cos \omega t$, where $\alpha$ is a constant and $\omega$ is the angular frequency. The magnitude of electric field at the surface of the solenoid is
 \begin{tasks}(2)
	\task[\textbf{a.}]$\frac{1}{2} \mu_{0} N R \omega \alpha \sin \omega t$
	\task[\textbf{b.}]$\frac{1}{2} \mu_{0} \omega N R \cos \omega t$
	\task[\textbf{c.}]$\mu_{0} N R \omega \alpha \sin \omega t$
	\task[\textbf{d.}] $\mu_{0} \omega N R \cos \omega t$
\end{tasks}
\item A medium $\left(\varepsilon_{r}>1, \mu_{r}=1, \sigma>0\right)$ is semi-transparent to an electromagnetic wave when
{\exyear{ GATE- 2020}}
 \begin{tasks}(2)
	\task[\textbf{a.}]Conduction current $\gg>$ Displacement current
	\task[\textbf{b.}]Conduction current $<<$ Displacement current
	\task[\textbf{c.}]Conduction current $=$ Displacement current
	\task[\textbf{d.}] Both Conduction current and Displacement current are zero
\end{tasks}
\item A sinusoidal voltage of the form $V(t)=V_{0} \cos (\omega t)$ is applied across a parallel plate capacitor placed in vacuum. Ignoring the edge effects, the induced emf within the region between the capacitor plates can be expressed as a power series in $\omega$. The lowest nonvanishing exponent in $\omega$ is -----------
{\exyear{ GATE- 2020}}
 \colorlet{ocre1}{ocre!70!}
\colorlet{ocrel}{ocre!30!}
\setlength\arrayrulewidth{1pt}
\begin{table}[H]
	\centering
	\arrayrulecolor{ocre}
	\begin{tabular}{|p{1.5cm}|p{1.5cm}||p{1.5cm}|p{1.5cm}|}
		\hline
		\multicolumn{4}{|c|}{\textbf{Answer key}}\\\hline\hline
		\rowcolor{ocrel}Q.No.&Answer&Q.No.&Answer\\\hline
		1&\textbf{b} &2&\textbf{d}\\\hline
		3&\textbf{1.0} &4&\textbf{a} \\\hline
		5&\textbf{18} &6&\textbf{a} \\\hline
		7&\textbf{d}&8&\textbf{d}\\\hline
		9&\textbf{6.28}&10&\textbf{d}\\\hline
		11&\textbf{a} &12&\textbf{b}\\\hline
		13&\textbf{2}&14&\textbf{}\\\hline

	\end{tabular}
\end{table}
\end{enumerate}

\newpage
\begin{abox}
	Practise Set-3
\end{abox}
\begin{enumerate}
	\item 	The $x$ and $z$-components of a static magnetic field in a region are $B_{x}=B_{0}\left(x^{2}-y^{2}\right)$ and $B_{z}=0$, respectively. Find one of the possible solution for its $y$-component which is consistent with the Maxwell equations?
	\begin{answer}
		\begin{align*}
		 B_{x}&=B_{0}\left(x^{2}-y^{2}\right), B_{z}=0\\
		\because \vec{\nabla} \cdot \vec{B}&=0 \Rightarrow \frac{\partial B_{x}}{\partial x}+\frac{\partial B_{y}}{\partial y}+\frac{\partial B_{z}}{\partial z}=0 \Rightarrow \frac{\partial B_{y}}{\partial y}\\&=-\frac{\partial B_{x}}{\partial x}=-2 B_{0} x \Rightarrow B_{y}=-2 B_{0} x y
		\end{align*}
	\end{answer}
	\item 	Which of the following expressions represent an electric field due to a time varying magnetic field?
	\begin{tasks}(2)
		\task[\textbf{a.}]$K(x \hat{x}+\hat{y} \hat{y}+z \hat{z})$
		\task[\textbf{b.}]$K(x \hat{x}+y \hat{y}-z \hat{z})$
		\task[\textbf{c.}] $K(x \hat{x}-y \hat{y})$
		\task[\textbf{d.}] $K(y \hat{y}-x \hat{y}+2 z \hat{z})$
	\end{tasks}
\begin{answer}
	\begin{align*}
	\text { For time varying fields } &\vec{\nabla} \times \vec{E} \neq 0\\
	\text { (a) } \vec{\nabla} \times \vec{E}=K\left|\begin{array}{ccc}
	\hat{x} & \hat{y} & \hat{z} \\
	x & \partial / \partial y & \partial / \partial z \\
	y & z
	\end{array}\right|&=\hat{x}\left(\frac{\partial z}{\partial y}-\frac{\partial y}{\partial z}\right)+\hat{y}\left(\frac{\partial x}{\partial z}-\frac{\partial z}{\partial x}\right)+\hat{z}\left(\frac{\partial y}{\partial x}-\frac{\partial x}{\partial y}\right)=0\\
	\text { (b) } \vec{\nabla} \times \vec{E}=K\left|\begin{array}{ccc}
	\hat{x} & \hat{y} & \hat{z} \\
	\partial / \partial x & \partial / \partial y & \partial / \partial z \\
	x & y & -z
	\end{array}\right|&=\hat{x}\left(-\frac{\partial z}{\partial y}-\frac{\partial y}{\partial z}\right)+\hat{y}\left(\frac{\partial x}{\partial z}+\frac{\partial z}{\partial x}\right)+\hat{z}\left(\frac{\partial y}{\partial x}-\frac{\partial x}{\partial y}\right)=0\\
	\text { (c) } \vec{\nabla} \times \vec{E}=K\left|\begin{array}{ccc}
	\hat{x} & \hat{y} & \hat{z} \\
	\partial / \partial x & \partial / \partial y & \partial / \partial z \\
	x & -y & 0
	\end{array}\right|&=\hat{x}\left(0+\frac{\partial y}{\partial z}\right)+\hat{y}\left(\frac{\partial x}{\partial z}-0\right)+\hat{z}\left(-\frac{\partial y}{\partial x}-\frac{\partial x}{\partial y}\right)=0\\
	\text { (d) } \vec{\nabla} \times \vec{E}=K\left|\begin{array}{ccc}
	\hat{x} & \hat{y} & \hat{z} \\
	\partial / \partial x & \partial / \partial y & \partial / \partial z \\
	y & -x & 2 z
	\end{array}\right|&=\hat{x}\left(\frac{\partial(2 z)}{\partial y}+\frac{\partial x}{\partial z}\right)+\hat{y}\left(-\frac{\partial x}{\partial z}-\frac{\partial(2 z)}{\partial x}\right)+\hat{z}\left(\frac{\partial y}{\partial x}-\frac{\partial y}{\partial y}\right)\\
	\Rightarrow \vec{\nabla} \times \vec{E}&=-\hat{z} \neq 0
	\end{align*}
\end{answer}
	\item A uniform magnetic field in the positive $z$-direction passes through a circular wire loop of radius $1 \mathrm{~cm}$ and resistance $3.14 \Omega$ lying in the $x y$-plane. The field strength is reduced from 10 tesla to 9 tesla in $1 s$. Find the charge transferred across any point in the wire.
	\begin{answer}
		\begin{align*}
		\varepsilon&=-\frac{d \phi}{d t} \Rightarrow I=\frac{d q}{d t}=\frac{\varepsilon}{R}=-\frac{1}{R} \frac{d \phi}{d t}\\
		\Rightarrow d q&=-\frac{A}{R} d B=\frac{-\pi r^{2}}{R} d B\\
		\Rightarrow d q&=\frac{-3.14 \times\left(10^{-2}\right)^{2}}{3.14} \times 1=10^{-4} \text { coulomb }
		\end{align*}
	\end{answer}
	\item A small loop of wire of area $A=0.01 \mathrm{~m}^{2}, N=40$ turns and resistance $R=10 \Omega$ is initially kept in a uniform magnetic field $B$ in such a way that the field is normal to the loop. When it is pulled out of the magnetic field, a total charge of $Q=2 \times 10^{-5} C$ flows through the coil. Find the magnitude of the field $B$.
	\begin{answer}
		\begin{align*}
		\text { Magnetic flux through the loop }& \phi=N B A\\
		\text { Induced e.m.f } \varepsilon&=-\frac{d \phi}{d t} \text { and induced current } i\\&=-\frac{1}{R} \frac{d \phi}{d t}=\frac{d Q}{d t} \Rightarrow-\frac{1}{R} d \phi=d Q \text {. }\\
		\text { Thus } \frac{1}{10} \times(40 \times B \times 0.01)&=2 \times 10^{-5} \Rightarrow B=5 \times 10^{-4} T
		\end{align*}
	\end{answer}
	\item A rectangular loop of dimension $L$ and width $w$ moves with a constant velocity $v$ away from an infinitely long straight wire carrying a current $I$ in the plane of the loop as shown in the figure below. Let $R$ be the resistance of the loop. Show that the current in the Toop at the instant the near side is at a distance $r$ from the wire is $\frac{\mu_{0} I L}{2 \pi R} \frac{w v}{r[r+w]}$.
	\begin{figure}[H]
		\centering
		\includegraphics[height=3.5cm,width=4cm]{Ass-05}
	\end{figure}
\begin{answer}
	\begin{align*}
	\phi_{B}&=\int_{S} \vec{B} \cdot d \vec{a}=\int_{r}^{w} \frac{\mu_{0} I}{2 \pi r} L d r=\frac{\mu_{0} I L}{2 \pi R} \ln \left(\frac{r+w}{r}\right)\\
	\Rightarrow I&=-\frac{1}{R} \frac{d \phi_{B}}{d t}=\frac{\mu_{0} I L}{2 \pi R}\left[\frac{1}{r+w}-\frac{1}{r}\right] \frac{d r}{d t}=\frac{\mu_{0} I L w v}{2 \pi R r(r+w)}
	\end{align*}
\end{answer}
	\item A square loop of side $L$ and mass $M$ is made of a wire of cross-sectional area $A$ and resistance $R$ The loop. moving with a constant velocity $v_{v} \hat{i}$ in the horizontal xy-plane, enters a region $0 \leq x \leq 2 L$ having constant magnetic field $B \hat{k}$
	\begin{figure}[H]
		\centering
		\includegraphics[height=4cm,width=8cm]{Ass-01}
	\end{figure}
	Find an expression for the $x$-component of the force $\vec{F}$ acting on the loop in terms of its velocity $\vec{v}(t), B, L$ and $R$.
	\begin{answer}
		\begin{align*}
		\text { Initial flux } \phi_{0}&=B L x\\
		\text { Flux after time } d t, \phi&=B L(x+d x)\\
		\text { Change in flux } d \phi&=\phi-\phi_{0}=B L d x \text {. }\\
		\text { Induced e.m.f. } \varepsilon&=-\frac{d \phi}{d t}=-B L \frac{d x}{d t}=-B L v_{0} \text {. }\\
		\text { Induced current } I_{i n d}&=-\frac{B L v_{0}}{R} \text { (in clockwise direction } a b c d a \text { ) }\\
		\text { Force on element } b c \vec{F}&=I_{\text {ind }} \int(d \vec{l} \times \vec{B})=-\frac{B^{2} L^{2} v_{0}}{R} \hat{x}
		\end{align*}
	\end{answer}
	\item A coil of 15 turns, each of radius 1 centimeter, is rotating at a constant angular velocity $\omega=300$ radians per second in a uniform magnetic field of $0.5$ tesla, as shown in the figure. Assume at time $t=0$ that the normal $\hat{n}$ to the coil plane is along the $y$-direction and that the self-inductance of the coil can be neglected. If the coil resistance is 9 ohms, what will be the magnitude of the induced current in milliamperes?
	\begin{figure}[H]
		\centering
		\includegraphics[height=3.8cm,width=4.3cm]{Ass-02}
	\end{figure}
\begin{answer}
	The voltage induced is equal to the change in magnetic flux
	\begin{align*}
	\varepsilon=-\frac{\partial \Phi}{\partial t}, \text { where } \Phi=\int \vec{B} \cdot d \vec{A}
	\intertext{Noting the initial condition $(\Phi(t=0)=0)$, since the field and area normal are perpendicular). One finds that $\Phi=\int \vec{B} \cdot d \vec{A}=B \cos (90+\omega t) \pi r^{2}=(-B \sin \omega t) \pi r^{2}$.
		Thus, $|d \Phi / d t|=\omega B \cos (\omega t) \pi r^{2}$.}
	\intertext{Now, to find the current, one uses Ohm's Law in Faraday's Law to get}
	I=\frac{N \dot{\Phi}}{R}&=\frac{N B \omega}{R} \cos (\omega t) \pi r^{2} \text {, where } N \text { is the number of turns. }\\
	\text { Thus, } I=\frac{15 \times 0.5 \times 300}{9} \cos (\omega t) \pi(1 / 100)^{2}&=250 \times 10^{-4} \pi \cos (\omega t) \text { Ampere }I=25 \pi \cos (\omega t) \mathrm{mA}
	\end{align*}
\end{answer}
	\item The circuit shown below is in a uniform magnetic lield that is into the page and is decreasing in magnitude at the rate of 150 Tesla/sec. Then tind the ammeter reading.
	\begin{figure}[H]
		\centering
		\includegraphics[height=4.5cm,width=5cm]{Ass-03}
	\end{figure}
\begin{answer}
 From Ohm's Law $V-\varepsilon=I R$, one can obtain the current. (Note that $V=5.0 \mathrm{~V}$ is the voltage of the battery. The voltage induced acts to oppose this emf from the battery).
	\begin{align*}
	\text { The problem gives } \frac{d B}{d t}&=150 T / \mathrm{s} \text {. The area is just } 0.01 \mathrm{~m}^{2} \text {. }\\
	\text { Thus, the induced emf is, } \varepsilon&=\frac{d B}{d t} \quad A=150 \times 0.01=1.5\\
	\text { Thus, } V-\varepsilon&=3.5=I R \Rightarrow I=0.35 A \text {, since } R=10 \Omega \text {. }
	\end{align*}
\end{answer}
	\item A parallel plate air-gap capacitor is made up of two plates of area $10 \mathrm{~cm}^{2}$ each kept at a distance of $0.88 \mathrm{~mm}$. A sine wave of amplitude $10 \mathrm{~V}$ and frequency $50 \mathrm{~Hz}$ is applied across the capacitor as shown in the figure.
	\begin{tasks}(1)
		\task[\textbf{a.}]Find the amplitude of the displacement current density between the plates.
		\task[\textbf{b.}]
		 Find the r.m.s value of the displacement current density between the plates.
		\task[\textbf{c.}]Find the average value of the displacement current density (in $\mathrm{mA} / \mathrm{m}^{2}$ ) between the plates.
	\end{tasks}
	\begin{figure}[H]
		\centering
		\includegraphics[height=2.5cm,width=4cm]{Ass-04}
	\end{figure}
\begin{answer}
	\begin{align*}
	\text { Displacement current density } J_{d}&=\varepsilon_{0} \frac{\partial E}{\partial t}=\frac{\varepsilon_{0}}{d} \frac{\partial V(t)}{\partial t}=-\frac{\varepsilon_{0} \omega V_{0} \sin \omega t}{d}\\
	\text { (a) Amplitude of the displacement current density } J_{0 d}&=\frac{\varepsilon_{0} \omega V_{0}}{d}=\frac{2 \pi \varepsilon_{0} f V_{0}}{d}\\
	\Rightarrow J_{0 d}=4 \pi \varepsilon_{0} \frac{f V_{0}}{2 d}&=\frac{1}{9 \times 10^{9}} \frac{50 \times 10}{2 \times 88 \times 10^{-5}}=0.03 \mathrm{~mA} / \mathrm{m}^{2}
	\intertext{(b) The r.m.s value of the displacement current density is }
	J_{d, r m s}=\frac{\varepsilon_{0} \omega V_{0}}{d \sqrt{2}}=\frac{2 \pi \varepsilon_{0} f V_{0}}{d \sqrt{2}}&=\frac{0.03}{\sqrt{2}}=0.022 \mathrm{~mA} / \mathrm{m}^{2}
	\intertext{(c) The average value of the displacement current density is zero. }
	\end{align*}
\end{answer}
\end{enumerate}