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\begin{center}
\textbf{Phys 742 -- Assignment 6 -- Due 26 March 96}\ 
\end{center}

\noindent \textbf{1. }(Similar to Jackson's 5.2)

\textbf{(a)} Consider a solenoid of cross-sectional area $A$ and length $L,$
wound with $N$ turns per unit length and carrying a current $I.$ In the
limit $L^2>>A,$ find the $\mathbf{B}$ field inside the solenoid (not too
close to the ends) by a simple argument based on Amp\`{e}re's circuitation
law. Give the result in MKSA units as well as in gaussian units.

\textbf{(b) }When the solenoid has a circular cross-section, with $A=\pi
R^2, $ obtain $B_z$ on the axis in terms of $R,$ $L$ and the distance $z$
from the center. Compare with the result given by Jackson in problem 5.2a.

\textbf{(c) }Evaluate the result (b) on an end face of the solenoid (for
instance, at $z=L/2$). Show that for a long, thin solenoid it reduces to one
half of the $B_z$ of part (a). Is this result valid for any cross-sectional
shape?

\smallskip\ 

\noindent \textbf{2. }(Qualifying exam problem 88)

A current $I_1$ flows around in a thin wire that forms a circle of radius $a$
and a current $I_2$ flows in an infinitely long thin wire in the same plane.
The center of the circle is at a distance $b$ from the straight wire. You
may find it convenient to define $\sin \alpha =a/b.$

\textbf{(a) }Is there a net force between the two wires? If there is, what
are its direction and magnitude?

\textbf{(b) }Is there a net torque on the wire loop (with respect to its
center)? If there is, what are its direction and magnitude?

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\noindent \textbf{3. }

In problem 2, replace the circle by a square of side $a.$ The problem is now
very easy. To make it interesting, rotate the square by $\alpha $ around the
axis through its center that is parallel to the straight wire, or
equivalently put the straight wire in the plane obtained by rotating the
plane of the square. For any $\alpha ,$ answer the same questions as in
problem 2.

\smallskip\ 

\noindent \textbf{4.}

In a thin sheet of thickness $d$ and conductivity $\sigma $ there are two
circular current leads of radii $a$ and $b,$ separated by a distance $D$
between their centers ($D>a+b$).

\textbf{(a) }Find the resistance $R$. You have seen this boundary value
problem already, you can just quote mathematical results.

\textbf{(b) }What is the numerical value of $R$ (in ohms) when $\sigma =1\,$(%
$\Omega \cdot $cm)$^{-1}$ and $d=1\,$mm?

\newpage\ 

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

The line element of the circle is $d\mathbf{l}=a(-\sin \phi ,\,\cos \phi
,0)d\phi .$ The field due the straight wire is $(0,0,2I_2/cR),\,$ with $%
R=b-a\cos \phi .$ The elemental force is 
\[
d\mathbf{F=}\frac{I_1}c\,d\mathbf{l}\times \mathbf{B}=\frac{2I_1I_2}{Rc^2}%
\;(\cos \phi ,\;\sin \phi ,\,0)\,d\phi 
\]
In words, $d\mathbf{F}$ is radial, as can also be seen from the figure.

\textbf{(a) }Clearly $F_y=0$ by symmetry and we have 
\[
F_x=\frac{2I_1I_2}{c^2}\,\int \frac{\cos \phi \,d\phi }{b-a\cos \phi }=\frac{%
2I_1I_2}{bc^2}\,\int \frac{\cos \phi \,d\phi }{1-\sin \alpha \cos \phi } 
\]
where we have conveniently put $\sin \alpha =a/b.$

------------------

\textbf{c) }Same as above when the solenoid has a square cross-section, with 
$A=a^2$. Check that the results of (b) and (c) reduce to that of part (a) in
the appropriate limit.

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