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\begin{document}

\centerline{\bf Electrodynamics B (PHY 5347): Spring 2017 
Solutions for the Final.} \bigskip


\begin{enumerate}
\item {\bf Magnetic moment of a sphere.}

The charge density on the surface is $\sigma=q/(4\pi\,R^2)$, 
so the current on the sphere becomes (with $r=|\vec{r}|$)
$$ \vec{J}(\vec{r}) = 
   \omega\,r\,\sin(\theta)\,\sigma\,\delta(r-R)\,\hat{\phi}\ .$$
The magnetic moment is then
$$ \vec{\mu} = \frac{1}{2c} \int \vec{r}\times\vec{J}\,d^3x =
   \frac{1}{2c} \int \omega^2\,r^2\,\sin(\theta)\,\sigma\,
   \delta(r-R)\,(-\hat{\theta})\,d^3x\ . $$
By symmetry reasons only $\mu_z\ne 0$. The projection on the
$z$-axis is found using $\hat{z}\cdot\hat{\theta}=-\sin(\theta)$. So,
$$ \mu_z = \frac{2\pi}{2c}\,\omega\,R^4\,\sigma 
   \int_0^{\pi} \sin^3(\theta)\,d\theta = \frac{\pi}{c}\,
   \omega\,R^4\,\sigma \int_{-1}^{+1} \left(1-x^2\right)\,dx\,.    $$
The integral has the value $4/3$ and our final result is
$$ \mu_z = \frac{4\pi}{3c}\,\omega\,R^4\,\sigma = 
           \frac{\omega\,R^2\,q}{3c}\,.$$
\smallskip % \bigskip


\item{\bf Principal value integrals and Green functions
      for the wave equation.} \smallskip

For $\tau>0$ we have to close in the lower half-plane, where we 
get no contribution from closing ($e^{-i\,i\,v\,\tau}=e^{v\,\tau}
\Rightarrow v<0$ required). The integration path of Fig.~\ref{fig1} 

\begin{figure}[th] % From ../Fortran/4Potential Spring 2015 PHY5347.
\centerline{\hbox{ \psfig{figure=IPp1,width=10cm} }}
\caption{A possible integration path $I_p(\eta)$, where $\eta>0$
is the radius of the half-circles, and we close in the lower plane. 
\label{fig1} } \end{figure}

\begin{figure}[th] % From ../Fortran/4Potential Spring 2015 PHY5347.
\centerline{\hbox{ \psfig{figure=IPp2,width=10cm} }}
\caption{Another integration path $I_p(\eta)$, where $\eta>0$
is the radius of the half-circles. \label{fig2} } \end{figure}

leads to the retarded Green function and we have
$$ \lim_{\eta\to 0} I_p(\eta)\ =\ I_p^r\ =\ -\frac{i}{2\omega_p}\,
   \left(e^{+i\omega_p\tau}-e^{-i\omega\tau}\right)\ = $$
$$ \frac{1}{2\pi} \lim_{\eta\to 0} \left[ 
   \int_{-\infty}^{-\omega_p-\eta} du\,\frac{e^{-i\,u\,\tau}}
   {\omega_p^2-u^2} + \int_{-\omega_p+\eta}^{\omega_p-\eta} 
   du\,\frac{e^{-i\,u\,\tau}}{\omega_p^2-u^2} + 
   \int_{\omega_p+\eta}^{\infty}
   du\,\frac{e^{-i\,u\,\tau}}{\omega_p^2-u^2} \right] $$
$$ + \frac{1}{2\pi} \lim_{\eta\to 0} \left[ 
  \int_{C_+(-\omega_p,\eta)}dw\,\frac{e^{-i\,w\,\tau}}{\omega_p^2-w^2}+ 
  \int_{C_+(+\omega_p,\eta)}dw\,\frac{e^{-i\,w\,\tau}}{\omega_p^2-w^2}
  \right]\,, $$
where $C_+(-\omega_p,\eta)$ denotes the clockwise half-circle of radius 
$\eta$ about $-\omega_p$ and $C_+(+\omega_p,\eta)$ the clockwise 
half-circle of radius $\eta$ about $+\omega_p$. In the limit $\eta\to 
0$ integration over the half-circles gives half the contribution from 
the residue theorem for the entire integration path, and the integrals 
on the real axis add up to the principal value definition. So we have
$$ I_p^r\ =\ \frac{1}{2\pi}\,P \int_{-\infty}^{+\infty} 
   du\,\frac{e^{-i\,u\,\tau}}{\omega_p^2-u^2} + \frac{1}{2}\,I_p^r\
   \Rightarrow\ \frac{1}{2\pi}\,P \int_{-\infty}^{+\infty} 
   du\,\frac{e^{-i\,u\,\tau}}{\omega_p^2-u^2} = \frac{1}{2}\,I_p^r\,.$$
For $\tau<0$ we have to close in the upper half-plane and the 
integration path of Fig.~\ref{fig1} gives
$$ 0\ =\ \frac{1}{2\pi}\,P \int_{-\infty}^{+\infty} 
   du\,\frac{e^{-i\,u\,\tau}}{\omega_p^2-u^2} - \frac{i}{4\omega_p}\,
   \left(e^{+i\omega_p\tau}-e^{-i\omega_p\tau}\right)~~\Rightarrow $$
$$ \frac{1}{2\pi}\,P\int_{-\infty}^{+\infty}du\,\frac{e^{-i\,u\,\tau}}
   {\omega_p^2-u^2} = \frac{i}{4\omega_p}\,\left(e^{+i\omega_p\tau}
   -e^{-i\omega_p\tau}\right)\ =\ \frac{1}{2}\,I^a_p\,. $$

Alternatively, one could have used the integration path of 
Fig.~\ref{fig2} and obtains the same results.
\bigskip 


\item{\bf Far field approximation of two current loops. }
\smallskip

\noindent Using the given potential $\vec{A}$, we have (a)
$$ \vec{B} = i\,k\,\hat{r}\times\vec{A} = -i\,k\,\hat{\theta}\,
   A^0_{\phi}\,\sin(\theta)\,\cos(\theta)\,
   \frac{e^{i\,k\,r-i\,\omega\,t}}{r}  $$
and (b)
$$ \vec{E} = -\hat{r}\times\vec{B} = i\,k\,\hat{\phi}\,
   A^0_{\phi}\,\sin(\theta)\,\cos(\theta)\,
   \frac{e^{i\,k\,r-i\,\omega\,t}}{r} .  $$
(c) The time-averaged Poynting vector becomes 
$(\hat{\phi}\times\hat{\theta}=-\hat{r})$
$$ \vec{S}_{\rm av} = \frac{c}{8\pi}\,\vec{E}\times\vec{B}^* =
    \frac{c}{8\pi}\,\hat{r}\,k^2\,\left|A^0_{\phi}\right|^2\,
    \frac{\sin^2(\theta)\,\cos^2(\theta)}{r^2}\ . $$
(d) Therefore, the angular intensity distribution is
$$ \left(\frac{dP}{d\Omega}\right) = r^2\,\hat{r}\cdot\vec{S}_{\rm av}
   = \frac{c}{8\pi}\,k^2\,\left|A^0_{\phi}\right|\,\sin^2(\theta)
   \,cos^2(\theta)\,, $$
which may be plotted in a polar intensity diagram. This is the
quadrupole case (see the figure of the lecture notes).

(e)~The maxima of the angular intensity distribution are at 
$$ \theta_1 = \pm\frac{\pi}{4}~~{\rm and}~~\theta_2=
   \pm\frac{3\pi}{4} $$
for which we have $\sin^2(\theta_{1,2})=\cos^2(\theta_{1,2})=1/2$, 
and, therefore,
$$ \left(\frac{dP}{d\Omega}\right)_{\max} = 
   \frac{c}{32\pi}\,k^2\,\left|A^0_{\phi}\right|\,.  $$
(d)~The intensity at these values over the average intensity is 
given by 
$$ g = \frac{\sin^2(\theta_{1,2})\,\cos^2(\theta_{1,2})}
   {\int_0^{2\pi}\int_{-1}^{+1} \sin^2(\theta)\,\cos^2(\theta)
   \,d\cos(\theta)\,d\phi/(4\pi)}\ . $$  
$$ = \frac{1}{4\int_0^1 (1-x^2)\,x^2\,dx} =
   \frac{1}{4(1/3-1/5)} = \frac{15}{8}\ .  $$


\end{enumerate}
\end{document}
