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\centerline{\bf Electrodynamics B (PHY 5347) Winter/Spring 2017 
Solutions} \bigskip

{\bf Set 2:} \bigskip

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% \end{figure} \centerline{Potential as function of $z$.}

{\bf 2.\ Magnetic dipoles.}
\bigskip

\noindent The magnetic dipole approximation holds
$$ \vec{B}(\,\vec{r}\,)\,=\,
   \frac{3\hat{r}\,(\,\hat{r}\cdot\vec{m}\,)-\vec{m}}{|\,\vec{r}\,|^3}\,,
   ~~~\hat{r}=\frac{\vec{r}}{r}\,,~~ r=|\,\vec{r}\,|\ .$$
\begin{enumerate}
\item Now we consider $\vec{m}_1$ in the field of $\vec{m}_2$,
$\vec{r}=\vec{r}_1-\vec{r_2}$:
$$ U\,=\,-\vec{m}_1\cdot\vec{B_2}\,=\, - \left(
   \frac{3\,(\vec{m}_1\cdot\hat{r})\,(\,\hat{r}\cdot\vec{m}_2\,)
   -\vec{m}_1\cdot\vec{m}_2}{|\,\vec{r}\,|^3}\right)\,, $$
\item The force acting on either of the magnets follows from
$$ \vec{F}\,=\,-\nabla U\,,\qquad \nabla |\vec{r}|\,=\,\hat{r}\,
    =\, \frac{\vec{r}}{r}\ .$$
We keep the position of $\vec{m}_2$ fixed and calculate the force 
acting on $\vec{m}_1$ at the position $\vec{r}$:
\begin{eqnarray} \nonumber
  \vec{F} &=& + \nabla \frac{3\,(\vec{m}_1\cdot\vec{r})\,
  (\,\vec{r}\cdot\vec{m}_2\,)}{|\,\vec{r}\,|^5} -
   \nabla \frac{\vec{m}_1\cdot\vec{m}_2}{|\,\vec{r}\,|^3}
  \\ \nonumber &=& 
  \frac{3\,\vec{m}_1\,(\,\vec{r}\cdot\vec{m}_2\,)}{|\,\vec{r}\,|^5} +
  \frac{3\,\vec{m}_2\,(\,\vec{r}\cdot\vec{m}_1\,)}{|\,\vec{r}\,|^5} -
  15\,\vec{r}\,\frac{(\vec{m}_1\cdot\vec{r})\,
  (\,\vec{r}\cdot\vec{m}_2\,)}{|\,\vec{r}\,|^7} +
  3\,\vec{r}\,\frac{\vec{m}_1\cdot\vec{m}_2}{|\,\vec{r}\,|^5}\ .
\end{eqnarray}
\item The rest position is at $U=U_{\min}$. Let us choose coordinates,
so that $\hat{r}=\hat{z}$. Then
$$U \,=\, \frac{-2\,m_1^z\,m_2^z+m_1^x\,m_2^x+m_1^y\,m_2^y}
               {|\,\vec{r}\,|^3}\ . $$
We find negative contributions for ${\rm sign}(m_1^z)=
{\rm sign}(m_2^z)$, ${\rm sign}(m_1^x)=-{\rm sign}(m_2^x)$ and 
${\rm sign}(m_1^y)= -{\rm sign}(m_2^y)$, which are therefore assumed 
to be chosen by the system. In particular, this leads to 
$$ \vec{m}_2^{\perp} = -\alpha\,\vec{m}_1^{\perp}\,,~~~~
   \alpha > 0 $$
for the components perpendicular to $\hat{z}$ (i.e. they are vectors in 
the $x$--$y$ plane). For $i=1,2$ let $m_i^{\perp}=|\vec{m}_i^{\perp}|$. 
Then (all terms give negative contributions)
$$ U \,=\, \frac{-2\,m_1^z\,m_2^z-m_1^{\perp},m_2^{\perp} }
               {|\,\vec{r}\,|^3}\ . $$
With $m_i=|\vec{m}_i|$ for $i=1,2$ we can write
$$ m_1^z=m_1\,\cos(\alpha)\,,~~m_1^{\perp}=m_1\,\sin(\alpha)~~{\rm 
   and}~~m_2^z=m_2\,\cos(\beta)\,, ~~m_2^{\perp}=m_2\,\sin(\beta) $$
and obtain
$$ U \,=\, \frac{-2\,m_1\,m_2\,\cos(\alpha)\,\cos(\beta) - 
   m_1\,m_2\,\sin(\alpha)\,\sin(\beta) } {|\,\vec{r}\,|^3}\ . $$
So, we have to minimize $-2\,\cos(\alpha)\,\cos(\beta)-\sin(\alpha)\,
\sin(\beta)$. Using well-known trigonometric equations, this becomes
$$-\frac{3}{2}\,\cos(\alpha-\beta)-\frac{1}{2}\,\cos(\alpha+\beta)\,.$$
As the maximum of $\cos(\phi)$ is at $\phi=0$, the only solution for
the minimum of $U$ is $\alpha=\beta=0$. Hence, $\vec{m}_1$ is parallel 
to $\vec{m_2}$ along the $\hat{r}$ axis.

In this configuration there is an attractive force of magnitude
$$ F_{\min} = |3+3-15+3|\,\frac{m_1\,m_2}{r^4} 
            = 6\,\frac{m_1\,m_2}{r^4} $$
between the dipoles. The magnetic field of this configuration is
\begin{eqnarray} \nonumber
  \vec{B}(\vec{r}\,)&=& \sum_{i=1}^2\,\frac{3\,(\vec{r}-z_i\,\hat{z})\,
  m_i\,\hat{z}\cdot(\vec{r}-z_i\,\hat{z})-m_i\,\hat{z}\,(\vec{r}-z_i\,
  \hat{z})^2}{|\vec{r}-z_i\,\hat{z}|^5} \\ \nonumber &=&
  \sum_{i=1}^2\,\frac{3\,(\vec{r}-z_i\,\hat{z})\,m_i\,(r\,\cos\theta-
  z_i)-m_i\,\hat{z}\,[r^2-2\,r\,z_i\,\cos\theta+(z_i)^2]} 
  {[r^2-2\,r\,z_i\,\cos\theta+(z_i)^2 ]^{5/2}} \,.
\end{eqnarray}
\end{enumerate}
\bigskip

{\bf 3.\ Magnetic moment of a disk.}

\begin{enumerate}
\item The charge density on the perimeter is $\lambda=q/(2\pi\,R)$, 
      so the current density is (with $r=|\vec{r}|$)
$$ \vec{J}(\vec{r}) = \omega\,R\,\lambda\,\delta(\cos\theta)\,R^{-1}
    \delta(r-R)\,\hat{\phi}\,,$$
and the current becomes
$$ \vec{I} = \int_{-1}^{+1} d\cos\theta\,\int_0^{\infty} r\,dr\, 
   \vec{J}(\vec{r}) = \omega\,R\,\lambda\,\hat{\phi}\,.       $$
The magnetic moment is now 
\begin{eqnarray} \nonumber
  \vec{\mu} &=& \frac{1}{2c} \int \vec{r}\times\vec{J}\,d^3x 
  \\ \nonumber
  &=& \frac{2\pi}{2c}\,\omega\,\lambda\,(-\hat{\theta}) 
  \int_{-1}^{+1}d\cos\theta\,\delta(\cos\theta)\,\int_0^{\infty} 
  r^3 dr\,\delta(r-R) \\ \nonumber
  &=& \frac{\pi}{c}\,\omega\,R^3\,\lambda\,\hat{z}\ =\ 
      \frac{1}{2c}\,\omega\,R^2\,q\,\hat{z} \,.
\end{eqnarray}
\item Let $\sigma_m$ be the mass density of the disk. The angular 
      momentum is
\begin{eqnarray} \nonumber
  \vec{L} = L_z\,,~~~L_z = 2\pi\,\sigma_m\,\omega \int_0^{R} r^3 dr
  = \frac{\pi}{2}\,\sigma_m\,\omega\,R^4\,, 
\end{eqnarray}
and the total mass is given by
$$ M = 2\pi \int_0^r r dr\,\sigma_m = \pi\,R^2\,\sigma_m~~
   \Rightarrow\\ L_z = \frac{M}{2}\,\omega\,R^2\,. $$
\end{enumerate}
Therefore, the magnetic moment can be written 
$$ \mu_z = g\,\frac{q}{2M\,c}\,L_z~~~{\rm with}~~g = 2\, $$
which is the $g$ factor of the electron.
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