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

\textbf{Final exam - 6 May 1996}
\end{center}

\smallskip\ 

\noindent \textbf{1.}

\textbf{(a) }Find the mutual inductance $M$ of two coaxial circular loops of
radii $R_1$ and $R_2$ separated by a vertical distance $h.$ The result can
be left in the form of a single integral to be evaluated numerically or
approximated.

\textbf{(b) }Find the force $F$ between the two loops when currents $I_1$
and $I_2$ flow in them. Again, the result can be left in the form of a
single integral. When the currents are parallel, is there attraction or
repulsion?

\smallskip\ 

\noindent \textbf{2.}

\textbf{(a) }Consider two planar loops of areas\textbf{\ }$A_1$ and $A_2$
not too different in shape from circles and stacked above each other at a
distance $h$ such that $h^2>>$\textbf{\ }$A_1$ and $h^2>>$\textbf{\ }$A_2$.
Find the force $F$ between the two loops when currents $I_1$ and $I_2$ flow
in them. Also find the mutual inductance of the two loops. Explicit formulas
are required.

\textbf{(b)} Check that for two circular loops the results of part (a) agree
with those of problem 1 when $h>>$\textbf{\ }$R_1$ and $h>>$\textbf{\ }$R_2.$

\textbf{(c)} Find the force between a circular loop of radius $R$ and a much
smaller loop of area $A$ located on its axis, when currents $I_1$ and $I_2$
flow in the two loops. An explicit formula is required, and it must be valid
for all values of the distance $z$ (or $h)$ along the axis. Sketch the
result.

\smallskip\ \ 

\noindent \textbf{3.}

Recall that the Stokes parameters are defined as 
\begin{eqnarray*}
s_0 &=&a_1^2+a_2^2 \\
s_1 &=&a_1^2-a_2^2 \\
s_2 &=&2a_1a_2\cos (\delta _2-\delta _1) \\
s_3 &=&2a_1a_2\sin (\delta _2-\delta _1)
\end{eqnarray*}
where you are expected to supply the definitions of $a_1,\,a_2,\,\delta
_1,\,\delta _2$ (and also of $\mathbf{\epsilon }_1$ here below). Taking $%
s_0=1$ for convenience, what are the values of the other Stokes parameters
for radiation that is

\textbf{(a)} linearly polarized at 45$^{\circ }$ to the $\mathbf{\epsilon }%
_1 $ vector

\textbf{(b)} circularly polarized, counterclockwise (left handed in the
usual optics language)

\textbf{(c)} totally unpolarized

\textbf{(d) }elliptically polarized, counterclockwise, with semiaxes $a$ and 
$b$ in the ratio $a/b=2$ and the major axis at 30$^{\circ }$ from $\mathbf{%
\epsilon }_1.$

\newpage\ 

\smallskip\ \ 

\noindent \textbf{4.}

It is possible to put on a surface an optical coating of thickness $w$ and
refractive index $n,$ such that no light will be reflected, at normal
incidence. Solve the problem of such a coating on a slab having refractive
index $n^{\prime },$ taking the thickness of the slab to be effectively
infinite. What must the thickness of the coating be if $n^{\prime }=2n=4$
and the incident wavelength (in air) is 5000\AA ? Why is this called a
quarter\thinspace -\thinspace wavelength optical coating?

(Air can be taken to have refractive index $1$ for this problem)

\smallskip\ \ 

\noindent \textbf{5.}

Solve the boundary-value problem for a superconducting sphere of radius $a$
in a weak external magnetic field $\mathbf{B}_0.$ Assume that the magnetic
flux is totally expelled from the volume occupied by the sphere. (You may
wish to describe this by saying that in the sphere $\mathbf{M}=-\mathbf{H}%
/4\pi ,$ or that a surface current $\mathbf{K}$ flows on the sphere).

\textbf{(a)} Find the total $\mathbf{B}$ outside the sphere

\textbf{(b)} Choose polar coordinates in such a way that $\mathbf{K}$ has
only the component $K_\phi $ and give the explicit expression for $K_\phi .$

\textbf{(c) }Find a vector potential $\mathbf{A}$ that describes $\mathbf{B}$
everywhere.

\smallskip\ \ 

\noindent \textbf{6.}

True or false? If the assertion is false, explain briefly why, or give a
counterexample.

\smallskip\ 

\textbf{(a)} The dielectric constant $\varepsilon $ of a ``passive'' medium
is always $\geq 1$ at zero frequency

\textbf{(b)} The dielectric constant $\varepsilon $ of a ``passive'' medium
is always $\geq 1$ at all frequencies

\textbf{(c) }The magnetic permeability\textbf{\ }$\mu $ of any medium is
always $\geq 1$ at zero frequency

\textbf{(d) }An isolated, uncharged body at rest is always attracted to a
region of stronger $E$

\textbf{(e) }An isolated, uncharged body at rest is always attracted to a
region of stronger $B$

\textbf{(f) }The Lorentz gauge condition is\textbf{\ }$\nabla \cdot \mathbf{A%
}=0$

\textbf{(g)} Any analytic function $f(z),$ where $z=x+iy,$ satisfies $\dfrac{%
\partial ^2f}{\partial x^2}+\dfrac{\partial ^2f}{\partial y^2}=0$

\newpage\ 

\textbf{Answer 1}

\smallskip\ 

\noindent \textbf{(a) }Use $MI_1I_2=(1/c)\int \mathbf{A}_1\mathbf{\cdot J}%
_2\,d^3x=(2\pi /c)R_2I_2A_1$ with 
\[
A_1=\frac{I_1}c\int_0^{2\pi }\frac{R_1\cos \phi ^{\prime }\,d\phi ^{\prime }%
}{\sqrt{h^2+R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime }}} 
\]
as in Jackson's eq. (5.36). Obtain 
\[
M=\frac{2\pi }{c^2}R_1R_2\int_0^{2\pi }\frac{\cos \phi ^{\prime }\,d\phi
^{\prime }}{\sqrt{h^2+R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime }}} 
\]

\noindent \textbf{(b)} $F=I_1I_2\,\partial M/\partial h$ (note the sign!)
gives 
\[
F=-\,\frac{2\pi }{c^2}I_1I_2R_1R_2\int_0^{2\pi }\frac{h\cos \phi ^{\prime
}\,d\phi ^{\prime }}{\left( h^2+R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime
}\right) ^{3/2}}
\]

$F$ is attractive when the currents are parallel, i.e., when $I_1I_2>0.$

\hrulefill 

\smallskip\ 

\textbf{Answer 2}

\smallskip\ 

\noindent \textbf{(a) }The magnetic moment of a loop is $m=I\mathcal{A}/c.$
Then 
\[
F=-6m_1m_2/h^4=-6I_1I_2\mathcal{A}_1\mathcal{A}_2/c^2h^4 
\]
\[
M=2\mathcal{A}_1\mathcal{A}_2/c^2h^3 
\]

\noindent \textbf{(b)} Use the expansion 
\[
\frac 1{\sqrt{h^2+R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime }}}\simeq \frac
1h\left( 1-\frac 12\frac{R_1^2+R_2^2-2R_1R_2\cos \phi ^{\prime }}{h^2}%
\right) 
\]
in the answer 1(a) to obtain 
\[
M\simeq \frac{2\pi R_1^2R_2^2}{c^2h^3}\int_0^{2\pi }\cos ^2\phi ^{\prime
}\,d\phi ^{\prime }=\frac{2\pi ^2R_1^2R_2^2}{c^2h^3} 
\]

\smallskip\ 

\noindent \textbf{(c)} On-axis, the field of the large loop is 
\[
B_1=\frac{2\pi R^2I_1}{c\left( R^2+z^2\right) ^{3/2}} 
\]
The magnetic moment of the small loop is $m_2=I_2\mathcal{A}/c.$ Then 
\[
F=m_2\frac{\partial B_1}{\partial z}=-\frac{6\pi R^2\mathcal{A}}{c^2}\frac
z{\left( R^2+z^2\right) ^{5/2}} 
\]
This agrees with the result of (a) when $z>>R.$ Here is a plot of $-z/\left(
1+z^2\right) ^{5/2}:$

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\hrulefill

\smallskip\ 

\textbf{Answer 3}

\smallskip\ 

\noindent \textbf{(a)} $s_1=0,\,s_2=1,\,s_3=0.$

\smallskip\ 

\noindent \textbf{(b) }$s_1=0,\,s_2=0,\,s_3=1$

\smallskip\ 

\noindent \textbf{(c) }$s_1=0,\,s_2=0,\,s_3=0$

\smallskip\ 

\noindent \textbf{(d) }$s_1=3/10,\,s_2=3\sqrt{3}/10\simeq
.5196152,\,s_3=4/5. $

One way to solve part (d) is to go to circularly polarized components, where
the Stokes parameters are given by 
\begin{eqnarray*}
s_0 &=&a_{+}^2+a_{-}^2 \\
s_1 &=&2a_{+}a_{-}\cos (\delta _{-}-\delta _{+}) \\
s_2 &=&2a_{+}a_{-}\sin (\delta _{-}-\delta _{+}) \\
s_3 &=&a_{+}^2-a_{-}^2
\end{eqnarray*}
For counterclockwise polarization, $a_{+}=\left( a+b\right) /\sqrt{2}$ and $%
a_{-}=\left( a-b\right) /\sqrt{2}.$ Taking for convenience $a=2$ and $b=1$
we find $a_{+}=3/\sqrt{2}$ and $a_{-}=1/\sqrt{2}.$ Further, $\delta
_{-}-\delta _{+}=2\alpha =60^{\circ }.$ Then $s_0=5,\,s_1=3/2,\,s_2=3\sqrt{3}%
/2,\,s_3=4,$ and we get the answer by normalizing $s_0$ to $1.$

$.$\hrulefill 

\smallskip\ 

\textbf{Answer 4}

\smallskip\ 

Let the electric fields be 
\[
\begin{array}{lll}
E_{0+}\,e^{ik_0z}+E_{0-}\,e^{-ik_0z} & \text{for} & z<0 \\ 
E_{+}\,e^{ikz}+E_{-}\,e^{-ikz} & \text{for} & 0<z<w \\ 
E_{+}^{\prime }\,e^{ik^{\prime }z} & \text{for} & z>w
\end{array}
\]
with $k_0=\omega /c,\,k=nk_{0,}\,k^{\prime }=n^{\prime }k_0.$ The boundary
conditions are 
\[
\left\{ 
\begin{array}{l}
E_{0+}\,+E_{0-}\,=E_{+}\,+E_{-}\, \\ 
E_{0+}\,-E_{0-}\,=n\left( E_{+}\,+E_{-}\right) \,
\end{array}
\right. 
\]
\[
\left\{ 
\begin{array}{l}
\,E_{+}\,e^{ikw}+E_{-}\,e^{-ikw}=E_{+}^{\prime }\,e^{ik^{\prime }w} \\ 
\,n\left( E_{+}\,e^{ikw}-E_{-}\,e^{-ikw}\right) =n^{\prime }E_{+}^{\prime
}\,e^{ik^{\prime }w}
\end{array}
\right. 
\]
The first two equations give 
\[
\left( 
\begin{array}{l}
E_{0+} \\ 
E_{0-}
\end{array}
\right) =\frac 12\left( 
\begin{array}{ll}
1+n & 1-n \\ 
1-n & 1+n
\end{array}
\right) \left( 
\begin{array}{l}
E_{+} \\ 
E_{-}
\end{array}
\right) 
\]
and the last two give similarly 
\[
\left( 
\begin{array}{l}
E_{+} \\ 
E_{-}\,
\end{array}
\right) =\frac 1{2n}\left( 
\begin{array}{l}
\left( n+n^{\prime }\right) e^{-ikw} \\ 
\left( n-n^{\prime }\right) e^{ikw}
\end{array}
\right) E_{+}^{\prime }\,e^{ik^{\prime }w} 
\]
From this we get directly 
\[
\left( 
\begin{array}{l}
E_{0+} \\ 
E_{0-}
\end{array}
\right) =\frac 1{4n}\,\left( 
\begin{array}{l}
\left( 1+n\right) \left( n+n^{\prime }\right) e^{-ikw}+\left( 1-n\right)
\left( n-n^{\prime }\right) e^{ikw} \\ 
\left( 1-n\right) \left( n+n^{\prime }\right) e^{-ikw}+\left( 1+n\right)
\left( n-n^{\prime }\right) e^{ikw}
\end{array}
\right) E_{+}^{\prime }\,e^{ik^{\prime }w} 
\]
The reflection coefficient vanishes when $E_{0-}/E_{0+}=0.$ For $n>1$ and $%
n^{\prime }>n,$ this can happen only if 
\[
e^{2ikw}=-1 
\]
i.e. for $w=(2j+1)(\lambda /4),$ with $j$ integer. Even then, we must have
the condition 
\[
\left( 1-n\right) \left( n+n^{\prime }\right) +\left( 1+n\right) \left(
n^{\prime }-n\right) =0 
\]
which simplifies to give $n^{\prime }=n^2.$ This is satisfied by the
problem's data, $n=2,\,n^{\prime }=4.$

A quarter-wavelength coating corresponds to $j=0.$ Note that $\lambda $ is
the wavelength in the coating, $\lambda =\lambda _0/n.$ In our case, $%
\lambda =2500$ \negthinspace \AA . No reflection occurs for $w[$\AA $%
]=2500/4=625$.

\end{document}
