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\begin{center}
\textbf{Phys 742 - EM I}

\textbf{In-class test 2 - 4 April 1996}
\end{center}

\smallskip\ 

\textbf{1.}

Solve the boundary-value problem for a spontaneously magnetized sphere of
radius $a.$ Assuming that $\mathbf{M}$ is constant in the sphere, find the $%
\mathbf{H}$ and $\mathbf{B}$ fields everywhere.

The magnetization is equivalent to a surface current $\mathbf{K}$. Choose
polar coordinates in such a way that $\mathbf{K}$ has only the component $%
K_\phi $ and give the explicit expression for $K_\phi .$

Find a vector potential $\mathbf{A}$ that describes $\mathbf{B}$ everywhere.
Give explicitly the components of $\mathbf{A}$ in the coordinates of your
choice.

\smallskip\ 

\textbf{2.}

A dielectric slab of thickness $a$ and width $b$ is partially inserted
between the plates of a capacitor (see sketch on blackboard). Is the slab
attracted or ejected? Assuming a linear dielectric constant $\epsilon ,$
find the force on the slab when the potential difference between the plates
is $V$. What is a typical value of $\epsilon $ and what is the corresponding
numerical value of the force when $a=1$ \negthinspace cm, $b=10$
\negthinspace cm and $V=1000$ Volts?

\smallskip\ \ 

\textbf{3.}

\textbf{(a) }A solenoid consists of $N$ turns of area $A$ and carries a
current $I.$ Find the field $\mathbf{B}$ at large distances, far outside the
solenoid.

\textbf{(b) }Find the torque on the solenoid when its center is at a
distance $d$ from the surface of a medium of linear permeability $\mu $ and
its axis makes an angle $\theta $ with the normal to the surface. Assume
that $d$ is much larger than the dimensions of the solenoid.

\smallskip\ 

\noindent \hrulefill 

\smallskip\ 
\[
Y_{00}=\frac 1{\sqrt{4\pi }}\quad \quad \quad Y_{10}=\sqrt{\frac 3{4\pi }\,}%
\,\cos \theta \quad \quad \quad Y_{11}=-\sqrt{\frac 3{8\pi }\,}\,\sin \theta
\,e^{i\phi } 
\]

\newpage\ 

\begin{center}
\textsf{Answer 1}
\end{center}

The dipole moment is $m=(4\pi /3)Ma^3.$ The field for $r>a$ is 
\[
\mathbf{B}^{out}\mathbf{=}-m\mathbf{\nabla }(z/r^3) 
\]
\[
B_r^{out}=2m\cos \theta \,/r^3\quad \quad \quad B_\theta ^{out}=m\sin \theta
/r^3 
\]
From the continuity conditions, the fields for $r<a$ are $B^{in}=(8\pi /3)M$
and $H^{in}=-(4\pi /3)M$, in agreement with $B^{in}=H^{in}+4\pi M.$

From $\mathbf{K}=c\mathbf{M}\times \mathbf{n}$ get $K_\phi =cM\sin \theta .$
The vector potential can be computed from 
\[
\mathbf{A}(\mathbf{r})=\frac 1c\int \frac{\mathbf{K}(\mathbf{r}^{\prime })}{%
\left| \mathbf{r-r}^{\prime }\right| }\,d^2r^{\prime }
\]
as shown in Jackson, page. The result is 
\[
A_\phi ^{out}=m\sin \theta \,/r^2\quad \quad \quad A_\phi ^{in}=mr\sin
\theta \,/a^3=m\rho /a^3
\]
One can check that $\mathbf{B}=\mathbf{\nabla \times A}.$ For instance, 
\[
B_z^{in}=\frac 1\rho \,\frac \partial {\partial \rho }\,(\rho A_\phi ^{in})=%
\frac{2m}{a^3}
\]
It is also possible to use this equation backwards, to find $A_\phi ^{in}$
given that 

\begin{center}
\textsf{Answer 2}
\end{center}

As the slab moves in an amount $\delta x,$ the enegy increases by $\frac
12(\epsilon -1)(b\delta x/a)V^2.$ The force is in the direction of motion
(rmember that at constant $V$ one has $F_x=+\partial U/\partial x)$ and has
magnitude 
\[
F=(\epsilon -1)\frac{bV^2}{2a}\simeq 50(\epsilon -1)\,\text{dynes} 
\]

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