% BB Mar 06 2014.
%\documentclass[aps,twocolumn]{revtex4}
%\documentclass[aps,preprint]{revtex4}
\documentclass[aps,12pt]{revtex4}
\usepackage{epsfig}
\newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}}
\newcommand{\ba}{\begin{array}}
\newcommand{\ea}{\end{array}}
\newcommand{\tq}{\tilde{q}}
                                                                                
\begin{document}

\centerline{\bf Electrodynamics B (PHY 5347) Winter/Spring 2014 
Solutions for the Midterm.} \bigskip

\begin{enumerate}

\item {\bf Bar sliding in a magnetic field.} \smallskip

\begin{description}
\item{(a)} Since the $B$ field is constant the magnetic flux is
$$ \Phi = B\,w\,(l-v\,t)~~{\rm and}~~{d\Phi\over dt} = - B\,w\,v\,.$$
Therefore, the induced emf in the loop is
$$ \varepsilon_{\rm emf} = - \frac{1}{c}\,{d\phi\over dt} 
                         = c^{-1}\,B\,w\,v\,. $$
The same result is obtained using the Lorentz force:
$$ \varepsilon_{\rm emf} = - \frac{1}{c}\,\oint_C \left(\vec{v}\times
   \vec{B}\right)\cdot d\vec{l} = - \frac{B\,v}{c}\,\int_0^w dl
   = c^{-1}\,B\,w\,v\,. $$
\item{(b)} Since the resistance of the loop is $R(t)=[2w+2(l-vt)]\,r$,
we have the current
$$I(t) = {\varepsilon_{\rm emf}\over R(t)}
       = {B\,w\,v\over c\,r\,[2w+2(l-vt)]}~~{\rm for}~~vt\le l\,. $$
\end{description}
\bigskip

\item {\bf Four-potential of a moving point particle.} \smallskip

\begin{figure}[th] % From figprob2/.
\centerline{\hbox{ \psfig{figure=figprob2.eps,width=10cm} }}
\caption{Backward light cone and worldline in Minkowski space.
\label{fig_prob2} } \end{figure}

The worldline is given by $y^1=0.5\,y^0=0.5\,ct$. As light travels in 
one $[ps]$ by $3\,[dm]$, the equation for the relevant part of the 
backward light cone is $y^1=3\,[dm]-y^0$. Both meet when times and 
positions agree. Subtracting the first from the second equation gives
$$ 0=3\,[dm]-1.5\,y^0=3\,[dm]-1.5\times 3\times 10^9\,[dm/s]\,t\,,$$
which solves for
$$ t\ =\ \frac{2}{3}\times 10^{-9}\,[s]\ =\ \frac{2}{3}\,[ps]~~
   {\rm and}~~R = \beta\,c\,t = 1~[dm]\,.$$
We find the potentials now from the equations 
\begin{eqnarray} \nonumber
  \Phi\left(t,\vec{x}\right) = \left[\frac{q}{\left(1-
  \vec{\beta}\cdot\hat{n}\right)\,R}\right]_{\rm ret}\!\!,~~~
  \vec{A}\left(t,\vec{x}\right) = \left[\frac{q\,\vec\beta}{\left(
  1-\vec{\beta}\cdot\hat{n}\right)\,R}\right]_{\rm ret}\!\! .
\end{eqnarray}
We have $\vec{\beta}=\beta\,\hat{y}^1=(1/2)\,\hat{y}_1$ and, as the 
point charge moves away from the observer, $\vec{R}=R\,\hat{n}=-R\,
\hat{y}^1$, i.e., $\hat{n}=-\hat{y}_1$. Therefore,
\begin{eqnarray} \nonumber
  \Phi\left(1\,[ps],\vec{0}\right)&=&\frac{q_e}{(1+1/2)\,R}\ =\ 
  \frac{2}{3}\,[q_e/dm]\,, \\ \nonumber 
  \vec{A}\left(1\,[ps],\vec{0}\right)&=&\frac{q_e\,(1/2)\,\hat{y}^1}
  {(1+1/2)\,R}\ =\ \frac{1}{3}\,\hat{y}^1[q_e/dm]\,.
\end{eqnarray}
\medskip


\item {\bf Euler-Lagrange equation.} \smallskip

\noindent First,
$$ \frac{\partial{\cal L}}{\partial h_{\alpha\beta}} =
   \partial_{\gamma}\,\frac{\partial{\cal L}}
   {\partial\,(\partial_{\gamma}h_{\alpha\beta})} = 0\ \Rightarrow\
   h^{\alpha\beta} + \partial^{\beta}\,A^{\alpha}
                   - \partial^{\alpha}\,A^{\beta} = 0\ . $$
So, $h^{\alpha\beta}$ is antisymmetric and agrees with the
electromagnetic field tensor $h^{\alpha\beta}=F^{\alpha\beta}$.
Second,
$$ \partial_{\alpha}\,\frac{\partial{\cal L}}{\partial\,
   (\partial_{\alpha} A_{\beta})} = 
   \frac{\partial{\cal L}}{\partial A_{\beta}}\ \Rightarrow\
   \partial_{\alpha}\,\left( h^{\beta\alpha}-h^{\alpha\beta}
   \right) = - \frac{8\pi}{c}\,J^{\beta}\ . $$
Using the antisymmetry of $h^{\alpha\beta}$, this reads
$$ \partial_{\alpha}\,h^{\alpha\beta}=\frac{4\pi}{c}\,J^{\beta}\ .$$
The Lagrangian has led to the inhomogenous Maxwell equations.
\vfil

\end{enumerate}

\end{document}

