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\begin{document}
\begin{center}
  {\bf University of Kentucky, Physics 306 \\
    Homework \#10, Rev. A, due Wednesday, 2024-03-27 }
\end{center}

\vspace{.3 in}

% ~/hw/phy416_fa11/ps_03.tex #1a, ~/hw/phy416_fa14/ps_02.tex #1g
{\bf 1.} The {\bf gradient}, {\bf curl}, and {\bf divergence} are all applications of $d$ in different dimensions.  They are defined as\  $df=\bs\nabla f\cdot\bs{dl}$,\ \  $d(\bs A\cdot \bs{dl})=(\bs\nabla\times\bs A)\cdot\bs{da}$,\ \   $d(\bs B\cdot \bs{da})=(\bs\nabla\cdot\bs B)\cdot d\tau$, respectively, in terms of the differential operator $d=dq^{i}\partial_{i}$ and elements\  $\bs{dl} = \bh{e}_ih_idq^i$,\ \  $\bs{da} = \frac12 \bs{dl}\times\bs{dl}$,\ \  $d\tau = \frac13\bs{dl}\cdot\bs{da}$.

{\bf \quad a)} Apply these definitions to scalar $f$, polar vector $\bs A$, and axial vector $\bs B$ fields to obtain

$\bs\nabla f 
= \frac{df}{\bs{dr}}
= \frac{\bh e_i}{h_i}\frac{\partial}{\partial q^i}f$,\quad
$\bs\nabla\times\bs A 
= \frac{d\, \bs{dr}\cdot}{\bs{da}}\bs A
= \frac{\bh e_i}{h_jh_k}\frac{\partial}{\partial q^j} h_k A_k$,\quad $\bs\nabla\cdot\bs B 
= \frac{d\, \bs{da}\cdot}{\bs{d\tau}}\bs B
= \frac{1}{h_1h_2h_3} \frac{\partial}{\partial q^k} h_ih_jB_k$,
where $i,j,k$ are cyclic and $\frac{1}{\bs{dr}}$ is the inverse transformation of $\bs{dr}\cdot$ from $\{\bh x, \bh y, \bh z\}$ to $\{dx,dy,dz\}$, etc.

{\bf \quad b)} Expand a) in Cartesian, cylindrical, and spherical coordinates (compare with \href{https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates}{this table}).

{\bf \quad c)} Calculate the gradient of $f(r,\theta,\phi)= r^\ell P_\ell(\cos(\theta))$, \href{https://en.wikipedia.org/wiki/Legendre_polynomials#Recurrence_relations}{using} $\frac{x^{2}-1}{\ell}P'_{\ell}(x)=xP_{\ell}(x)-P_{\ell-1}(x)$,

{\bf \quad d)} the curl of $\bs A(x,y,z) = (\bh y x - \bh x y)/(x^{2}+y^{2})^n$ in Cartesian and \mbox{cylindrical coordinates, and the}
% \hat\phi/\rho^..

{\bf \quad e)} divergence of $\bs B(x,y,z) = (\bh x x+\bh y y+\bh z z)/(x^2+y^2+z^2)^n$ in Cartesian and \mbox{spherical coordinates.}
% \hat\r/r^..

{\bf \quad f)} Derive the formulas for $d^{2} f$ and $d^{2} \bs A \cdot \bs{dl}$ in terms of their partial derivatives to show that the trivial 2$^{\rm{nd}}$ derivatives $\bs\nabla\times\bs\nabla f$ and $\bs\nabla\cdot\bs\nabla\times \bs A$ are special cases of $d^2=0$.

\newpage

.%Coulomb's law, shell, +Biot-Savart, ring, line

{\bf 2.} The magnetic analog of {\it Coulomb's law} (with a scalar charge element $dq=\lambda dl=\sigma da = \rho d\tau$) is the {\bf Biot-Savart law} (with a vector current element $\bs{v}dq=I\bs{dl}=\bs{K}da=\bs{J}d\tau$):
\begin{eqnarray}
  \label{eq:biot}
  {\bs B} &=& \frac{\mu_0}{4\pi} \oint' \frac{\bs{v}dq' \times \brcurs}{\rcurs^3}
  ~\approx~ \sum_i \left( \Delta\bs B = \frac{\mu_0}{4\pi} \frac{I \Delta \bs \ell \times \brcurs_0}{\rcurs_0^3}\right )_i,
\end{eqnarray}

where $\Delta\bs \ell$ is the displacement vector from the beginning to the end of each current segment, and $\brcurs_0$ is the displacement vector from the middle of each current segment $\bs r'_0$ to the field point $\bs r$.  The approximation is that all of the current is concentrated at $\bs r'_0$ instead of spread out along the length of the segment from $\bs r'_0-\Delta \bs \ell/2$ to $\bs r'_0+\Delta \bs \ell/2$.  In this problem we first calculate a correction term to account for this difference, and then calculate the exact B-field due to each straight segment.
%
\includegraphics[width=\textwidth]{H07_fig1}
%

{\bf \quad a)} To analytically integrate the Biot-Savart law along a single straight
segment of the path, parametrize the segment $\bs r'(s)$ with the parameter
$s$, ranging from $s=-\frac12$ at the beginning to $s=+\frac12$ at the end of
the segment.  The parameterization involves the constant vectors $\bs r'_0$
(the center of the segment) and $\Delta \bs \ell$ (displacement along the
segment).  Calculate the line element $\bs{dl}= \frac{d\bs r'}{ds} ds$.
Calculate $\brcurs$ as a function of $\brcurs_0$, $\Delta \bs \ell$, and $s$.
Substitute these into the Biot-Savart formula and factor out the constant approximation of Eq.~\ref{eq:biot}] to obtain
\begin{equation}
  \Delta\bs B(\bs r) = \frac{\mu0}{4\pi}
  \frac{I\Delta \bs \ell {\times} \brcurs_0}{\rcurs_0^3} T(\alpha,\beta),
\end{equation}
where the integral $T(\alpha,\beta)$ along $s$ depends on $\alpha = {\brcurs_0\cdot\Delta \bs \ell}/{\rcurs_0^2}$ and $\beta = {\Delta \ell^2}/{\rcurs_0^2}$.

{\bf \quad b)} [bonus: Approximate the integrand of $T(\alpha,\beta)$ to order $s^2$ and integrate to obtain the correction term $T(\alpha,\beta) \approx 1 + \frac18(5 \alpha^2 - \beta)$ for the case where all the current is at the center of the segment.]

{\bf \quad c)} [bonus: Calculate the exact integral $T(\alpha,\beta)$ and show that
%\begin{equation}
$ {\Delta \bs B} = \frac{\mu_0I}{4\pi} \frac{{\Delta \bs \ell}\times \brcurs_0} {\left({\Delta \bs \ell}\times \brcurs_0\right)^2}\;\Delta\bs \ell \cdot\left(\hrcurs_--\hrcurs_+\right), $
%\end{equation}
where $\brcurs_\pm = \bs r - \bs r'(\pm \frac12)$ is the displacement vector from each end of the segment to the field point and $\hrcurs_\pm=\brcurs_\pm/\rcurs_\pm$. ]

{\bf \quad d)} [bonus: show that this is equivalent to
%\begin{equation}
$ {\Delta \bs B} = \frac{\mu_0I}{4\pi}\frac{(\brcurs_-{\times}\brcurs_+) (\rcurs_-{+}\rcurs_+)} {\rcurs_-\rcurs_+\;(\rcurs_-\rcurs_+{+}\brcurs_-\cdot\brcurs_+)}$. ]
%\end{equation}

%[See \href{http://web.ornl.gov/sci/fed/Theory/publication/pub2002/Biot-Savart_BC_reprint.pdf?origin=publication_detail}{``Compact expressions for the Biot-Savart fields of a filamentary segment''}, Eq.~(9).]

\newpage
[{\it bonus:} {\bf 3. Current sheet}---surface currents can be approximated numerically by a tiling of quadrilaterals like the one shown below, with current $I$ flowing parallel to the top and bottom edges, from left to right.  Let the vector $\bs \ell = \bs \ell_+ = \bs \ell_-$ run along either the top or bottom edge, parallel to the current; $\bs w = \bs w_- = \bs w_+$ from bottom to top along the left or right edge; and $\bs r_0$ be the point at the center of the parallelogram as shown in the diagram.
\begin{center}
  \includegraphics[width=.6\textwidth]{H07_fig2}
\end{center}

{\bf \quad a)} Parametrize the surface of the parallelogram as $\bs r'(s,t)$, with the top and bottom edges are at $t=+\frac12$ and $-\frac12$, and the left and right edges are at $s=+\frac12$ and $-\frac12$ respectively.

{\bf \quad b)} Write down the Biot-Savart integral for the magnetic field in terms of integration parameters $s,t$ and constants $\brcurs_0 \equiv \bs r - \bs r'_0$, $\bs \ell$, and $\bs w$.

{\bf \quad c)} Expand in powers of $s$ and $t$, to calculate the integral up to second order.]

{\bf \quad d)} It is not possible to tile arbitrary surfaces with parallelograms---we need all four points on the quadrilateral to be arbitrary.  To generalize this solution, let $\bs \ell_0$, $\bs w_0$ be the corresponding vectors through the center of the quadrilateral.  We need one more vector $\bs u_0=\bs\ell_+-\bs\ell_-=\bs w_+-\bs w_-$, where $\bs\ell_\pm$ run across the top and bottom, and $\bs w_\pm$ run along the right and left sides of the diagram.  Generalize steps (a)-(c) to calculate $\bs B(\bs r)$.]


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