
\noindent
{\bf Problem (\ref{ex_rad_Y21a}): Vector potential of two current loops.}
\smallskip

\noindent We define
$$ x_0 = \cos(\theta_0) = \frac{d}{r_0}~~{\rm with}~~
   r_0 = \sqrt{R^2+d^2}~~~{\rm and}~~~x'=\cos(\theta')\ .$$
The currents flow in $\hat{\phi}$ direction and their densities are 
given by
$$ J^{(1,2)}_{\phi} = \pm I\,\sin(\theta')\,
   \delta(\cos\theta_0\mp\cos\theta')\,r_0^{-1}\,\delta(r'-r_0)\ . $$
Note the (correct) normalization
$$ \int_0^{\infty} r'dr' \int_0^{\pi} \sin\theta'\,d\theta'\,
   J^{(1,2)}_{\phi} = \pm I $$
for the currents. Using Cartesian coordinates we have
$$ \vec{J}^{(1,2)} = - J_{\phi}^{1,2}\,\sin(\phi')\,\hat{x}
                     + J_{\phi}^{1,2}\,\cos(\phi')\,\hat{y}\ . $$
(a) For the calculation of the vector potential $\vec{A}$ we exploit
cylindrical symmetry and choose the observation point at $\phi=0$, so
that the $\sin(\phi')\,\hat{x}$ component does not contribute. It 
remains to calculate the $\hat{y}$ component, $\hat{\phi}=\hat{y}$,
$r\gg r'$ now:
$$ \vec{A}_{\phi} = \frac{I}{cr_0} \int_0^{2\pi}d\phi'\,\cos(\phi')
   \int_{-1}^{+1} dx' \int_0^{\infty} r'^2dr'\,J_{\phi}\, \frac{
   e^{i\,k\,|\vec{x}-\vec{x}^{\,'}|}}{|\vec{x}-\vec{x}^{\,'}|}\,, $$
$$ \frac{e^{i\,k\,|\vec{x}-\vec{x}^{\,'}|}}{|\vec{x}-\vec{x}^{\,'}|}
   = 4\pi\,i\,k \sum_{l=0}^{\infty} j_l(kr')\,h_l^+(kr) 
   \sum_{m=-l}^{+l} Y_l^{\ m}(\theta,\phi)\, 
         \overline{Y}_l^{\ m}(\theta',\phi')\,. $$
After the $r'$-integration we have
$$ \vec{A}_{\phi} = 4\pi\,i\,k \frac{I\,r_0}{c} \sum_{l=0}^{\infty} 
   j_l(kr_0)\,h_l^+(kr) \sum_{m=-l}^{+l} Y_l^{\ m}(\theta,\phi)\, 
   \int_0^{2\pi}d\phi'\, \cos(\phi') $$
$$ \int_{-1}^{+1} dx'\,\sqrt{1-x'^2}
   \,\left[\delta(x_0-x')-\delta(x_0+x')\right]\, 
         \overline{Y}_l^{\ m}(\theta',\phi')\,. $$
As 
$$ \int_0^{2\pi}d\phi'\, \cos(\phi') = 0 $$
holds, it follows that $Y_0^{\ 0}$, $Y_1^{\ 0}$ and $Y_2^{\ 0}$ do not 
contribute. For $Y_1^{\ 1}$ we have
$$ \int_{-1}^{+1} dx'\,(1-x'^2)\,
   \left[\delta(x_0-x')-\delta(x_0+x')\right] = 0 $$
and for $Y_2^{\ 2}$
$$  \int_0^{2\pi}d\phi'\,\cos(\phi')\,e^{i\,2,\phi'} = 0 $$
so that $Y_2^{\ 1}$ gives the leading contribution, which is a 
quadrupole. We find:
$$ \overline{Y}_2^{\ 1} = - \sqrt{\frac{15}{8\pi}}\, \sin\theta\,
   \cos\theta\, e^{-i\,\phi} $$
$$ \int_0^{2\pi}d\phi'\,\cos(\phi')\,e^{-i\,\phi'} = \pi\,, $$
$$ \int_{-1}^{+1} dx'\,(1-x'^2)\,x\,
   \left[\delta(x_0-x')-\delta(x_0+x')\right] $$
$$ = 2\,(1-x_0^2)\,x_0 = 2\,\sin^2(\theta_0)\,\cos(\theta_0)\,, $$
and in this approximation the potential is
$$ A_{\phi} = -\frac{I\,r_0}{c}\,\sqrt{\frac{15}{2}}\, 
   \sin^2(\theta_0)\,\cos(\theta_0)\,i\,k\,4\pi\,j_2(kr_0)\,
   h_2^+(kr)\,Y_2^{\ 1}(\theta,\phi=0)  $$
where
$$ Y_2^{\ 1}(\theta,\phi=0) = - \sqrt{\frac{15}{8\pi}}\, 
   \sin\theta\, \cos\theta\, .$$
In the $r\to\infty$ far field approximation we have
$$ h_2^+(kr) = i\,\frac{e^{i\,k\,r}}{r} $$
and, consequently,
$$ A_{\phi}(r,\theta) = A^0_{\phi}\,\sin(\theta)\,\cos(\theta)\,
                        \frac{e^{i\,k\,r}}{r}\ . $$
\bigskip


\noindent
{\bf Problem (\ref{ex_rad_Y21b}): Far field approximation of two 
    current loops. }
\smallskip

\noindent Using the result of the previous problem 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
$$ \vec{S}{\rm av} = \frac{c}{8\pi}\,\vec{E}\times\vec{B}^* =
    \frac{c}{8\pi}\,\hat{r}\,k^2\,\left(A^0_{\phi}\right)^2\,
    \frac{\sin^2(\theta)\,\cos^2(\theta)}{r^2}\ . $$
(d) Therefore, the angular intensity distribution is
$$ {\rm const}\,\frac{\sin^2(\theta)\,\cos^2(\theta)}{r^2} $$
and for fixed intensity we find
$$ r^2 = {\rm const}'\,\sin^2(\theta)\,\cos^2(\theta) $$
or
$$ r = {\rm const}''\,\sin(2\theta) $$
which may be plotted in a polar intensity diagram. This is the
quadrupole case (see the figure of the script or of Jackson).
(e)~The maxima of the angular intensity distribution are at 
$$ \theta_1 = \frac{\pi}{4}~~{\rm and}~~\theta_2=\frac{3\pi}{4} $$
for which we have $\sin(2\,\theta_{1,2})=1$. The gain of the
intensity at these values over the average intensity is given by 
$$ g = \frac{\sin^2(\theta_{1,2})\,\cos^2(\theta_{1,2})}
   {\int_0^{2\pi}\int_{-1}^{+1} \sin^2(\theta)\,\cos^2(\theta)
   \,d\cos(\theta)\,d\phi/(4\pi)}\ . $$  
Using $\sin^2(\theta_{1,2})=\cos^2(\theta_{1,2})=1/2$ we
have
$$ g = \frac{1}{4\int_0^1 (1-x^2)\,x^2\,dx} =
   \frac{1}{4(1/3-1/5)} = \frac{15}{8}\ .  $$

\bigskip

