From: W. Trevor King <wking@drexel.edu>
Date: Tue, 25 Aug 2009 04:26:57 +0000 (-0400)
Subject: Completed rec9 solutions
X-Git-Url: http://git.tremily.us/?a=commitdiff_plain;h=a78e5c604a6f2a37c955a1498970578e373d682c;p=course.git

Completed rec9 solutions
---

diff --git a/latex/problems/Young_and_Freedman_12/problem28.12.tex b/latex/problems/Young_and_Freedman_12/problem28.12.tex
index 4b0ce19..30df69a 100644
--- a/latex/problems/Young_and_Freedman_12/problem28.12.tex
+++ b/latex/problems/Young_and_Freedman_12/problem28.12.tex
@@ -1,25 +1,54 @@
 \newcommand{\dB}{d\vect{B}}
 \newcommand{\dl}{d\vect{l}}
-\newcommand{\rhat}{\hat{r}}
 
 \begin{problem*}{28.12}
-
+Two parallel wires are $5.00\U{cm}$ apart and carry currents in
+opposite directions, as shown in Fig.~28.37.  Find the magnitude and
+direction of the magnetic field at point $P$ due to the two
+$1.50\U{mm}$ segments of wire that are opposite each other and
+$8.00\U{cm}$ from point $P$.
 \begin{center}
 \begin{asy}
 import Mechanics;
 import ElectroMag;
+import Circ;
 
-real u = 0.1cm;
+real u = 0.6cm;
 real Ysep = 5u;
 real hypot = 8u;
 real Xsep = sqrt(hypot**2 - (Ysep/2)**2);
 real Xslush = 1u;
+real Xseg = 1u;
+
+Distance dtop = Distance((-Xsep,Ysep/2), (0,0), "$8.00\U{cm}$");
+Distance dbot = Distance((-Xsep,-Ysep/2), (0,0), "$8.00\U{cm}$");
+
+dtop.draw(labeloffset=8pt);
+dbot.draw(labeloffset=-8pt);
+dot("P", (0,0));
+
+
+Wire wbot_seg = Wire((-Xsep-Xseg/2, -Ysep/2), (-Xsep+Xseg/2, -Ysep/2), red);
+Wire wbot = Wire((-Xsep-Xslush, -Ysep/2), (Xslush,-Ysep/2));
 
-Wire wtop_seg = Wire(
-Wire wtop = Wire((), (Xslush,Ysep/2));
+wbot.draw();
+wbot_seg.draw();
+TwoTerminal Ibot = current((0,-Ysep/2), 180, "$24.0\U{A}$");
+Distance dSegbot = Distance((-Xsep-Xseg/2,-Ysep/2), (-Xsep+Xseg/2,-Ysep/2),
+                            offset=2mm, "$1.50\U{mm}$");
+dSegbot.draw(labeloffset=-8pt);
+
+
+Wire wtop_seg = Wire((-Xsep-Xseg/2, Ysep/2), (-Xsep+Xseg/2, Ysep/2), red);
+Wire wtop = Wire((-Xsep-Xslush, Ysep/2), (Xslush,Ysep/2));
 
 wtop.draw();
 wtop_seg.draw();
+TwoTerminal Itop = current((Ibot.end.x,Ysep/2), "$12.0\U{A}$");
+Distance dSegtop = Distance((-Xsep-Xseg/2,Ysep/2), (-Xsep+Xseg/2,Ysep/2),
+                            offset=-2mm, "$1.50\U{mm}$");
+dSegtop.draw(labeloffset=8pt);
+
 \end{asy}
 \end{center}
 \end{problem*}
@@ -36,6 +65,32 @@ on both segments.  All of the components are given except for
 $\dl\times\rhat$.  Drawing a picture for the top segment
 \begin{center}
 \begin{asy}
+import Mechanics;
+import Circ;
+
+real u = 0.6cm;
+real Ysep = 5u;
+real hypot = 8u;
+real Xsep = sqrt(hypot**2 - (Ysep/2)**2);
+real Xslush = 1u;
+real Xseg = 1u;
+
+draw((-Xsep,Ysep/2)--(0,0)--(-Xsep,0)--cycle, dashed);
+label("$8.00$", (-Xsep/2,Ysep/4), NE);
+label("$2.50$", (-Xsep,Ysep/4), W);
+
+dot("P", (0,0));
+
+Angle theta = Angle((0,Ysep/2), (-Xsep,Ysep/2), (0,0), 10mm, "$\theta$");
+theta.draw();
+Angle theta2 = Angle((-Xsep,Ysep/2), (0,0), (-Xsep,0), 10mm, "$\theta$");
+theta2.draw();
+
+Vector dL = Vector((-Xsep, Ysep/2), mag=3u, "$\dl$");
+Vector rhat = Vector((-Xsep, Ysep/2), mag=3u, dir=degrees((Xsep,-Ysep/2)),
+                     "$\rhat$");
+dL.draw();
+rhat.draw();
 \end{asy}
 \end{center}
 Using our knowledge of cross products and trigonometry
@@ -43,7 +98,6 @@ Using our knowledge of cross products and trigonometry
   |\dl\times\rhat|=|\dl|\cdot|\rhat|\cdot\sin\theta
     = 1.50\U{mm}\cdot1\cdot\sin\theta
     = 1.50\U{mm}\cdot\frac{2.50\U{cm}}{8.00\U{cm}}
-    = 1.50\U{mm}\cdot\frac{2.50\U{cm}}{8.00\U{cm}}
     = 4.6875\E{-4}\U{m} \;.
 \end{equation}
 It's also pretty clear that this cross product will have the same
@@ -53,7 +107,7 @@ work that out in detail if you didn't notice right off.
 The net magnetic field $B_p$ is then
 \begin{equation}
   B_p = B_{pt} + B_{pb}
-    = \frac{$\mu_0$}{4\pi}\cdot\frac{4.6875\E{-4}\U{m}}{(8.00\E{-2}\U{m})^2}
+    = \frac{\mu_0}{4\pi}\cdot\frac{4.6875\E{-4}\U{m}}{(8.00\E{-2}\U{m})^2}
       \cdot(12.0\U{A}+24.0\U{A})
     = \ans{264\U{nT}} \;.
 \end{equation}
diff --git a/latex/problems/Young_and_Freedman_12/problem28.18.tex b/latex/problems/Young_and_Freedman_12/problem28.18.tex
index b585771..0fc786a 100644
--- a/latex/problems/Young_and_Freedman_12/problem28.18.tex
+++ b/latex/problems/Young_and_Freedman_12/problem28.18.tex
@@ -1,5 +1,40 @@
 \begin{problem*}{28.18}
+Two long, straight wires, one above the other, are seperated by a
+distance $2a$ and are parallel to the $x$-axis.  Let the $+y$-axis be
+in the plane of the wires in the direction from the lower wire to the
+upper wire.  Each wire carries current $I$ in the $+x$-direction.
+What are the magnitude and direction of the net magnetif field of the
+two wires at a point in the plane of the wires \Part{a} midway between
+them; \Part{b} at a distance $a$ above the upper wire; \Part{c} at a
+distance $a$ below the lower wire?
 \end{problem*}
 
 \begin{solution}
+\Part{a}
+Using the right hand rules for magnetic field from a wire, we see that
+the upper wire will create a magnetic field into the page while the
+lower wire will create a magnetic field out of the page.  The
+magnitude of the field from a single wire is given by
+\begin{equation}
+  B = \frac{\mu_0 I}{2\pi r} \;,
+\end{equation}
+with the same current $I$ and distance $r=a$ for both wires.
+Therefore the net magnetic field is \ans{zero}.
+
+\Part{b}
+Both wires create a magnetic field out of the page.  The magnitude of
+the toal field will be
+\begin{equation}
+  B = \frac{\mu_0 I}{2\pi a} + \frac{\mu_0 I}{2\pi (3a)}
+    = \ans{\frac{2 \mu_0 I}{3\pi a}} \;.
+\end{equation}
+
+\Part{c}
+Both wires create a magnetic field into the page.  The magnitude of
+the toal field will be
+\begin{equation}
+  B = \frac{\mu_0 I}{2\pi (3a)} + \frac{\mu_0 I}{2\pi a}
+    = \ans{\frac{2 \mu_0 I}{3\pi a}} \;,
+\end{equation}
+the same as the magnitude for \Part{b}.
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem28.23.tex b/latex/problems/Young_and_Freedman_12/problem28.23.tex
index e844ac4..cb59801 100644
--- a/latex/problems/Young_and_Freedman_12/problem28.23.tex
+++ b/latex/problems/Young_and_Freedman_12/problem28.23.tex
@@ -1,5 +1,138 @@
 \begin{problem*}{28.23}
+Four long, parallel power lines each carry $100\U{A}$ currents.  A
+cross-sectional diagram of these lines if a square, $20.0\U{cm}$ on
+each side.  For each of the three cases shown in Fig.~28.41, calculate
+the magnetic field at the center of the square.
 \end{problem*}
 
+\begin{nosolution}
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+real a = 1u;
+real dx = 3u;
+pen ipen = red+blue;
+
+real x = 0;
+real d = a/2;
+
+Vector Vs[];
+
+label("(a)", (x,-d), S);
+Vs.push(Vector((x-d,-d), phi=-90, ipen));
+Vs.push(Vector((x+d,-d), phi=-90, ipen));
+Vs.push(Vector((x+d, d), phi=-90, ipen));
+Vs.push(Vector((x-d, d), phi=-90, ipen));
+x += dx;
+
+label("(b)", (x,-d), S);
+Vs.push(Vector((x-d,-d), phi= 90, ipen));
+Vs.push(Vector((x+d,-d), phi=-90, ipen));
+Vs.push(Vector((x+d, d), phi= 90, ipen));
+Vs.push(Vector((x-d, d), phi=-90, ipen));
+x += dx;
+
+label("(c)", (x,-d), S);
+Vs.push(Vector((x-d,-d), phi= 90, ipen));
+Vs.push(Vector((x+d,-d), phi= 90, ipen));
+Vs.push(Vector((x+d, d), phi=-90, ipen));
+Vs.push(Vector((x-d, d), phi=-90, ipen));
+
+for (int i=0; i<Vs.length; i+=1)
+  Vs[i].draw();
+\end{asy}
+\end{center}
+\end{nosolution}
+
 \begin{solution}
+Note that the magnitude of magnetic field at the center of the square
+from any corner wire will be
+\begin{equation}
+  B_w = \frac{\mu_0 I}{2\pi \frac{a}{\sqrt{2}}} \;,
+\end{equation}
+where $a=20.0\U{cm}$ is the side length of the square, and
+$r=a/\sqrt{2}=a\cos(45\dg)$ is the distance from the corner of the
+square to the center.
+
+Using our right-hand rules to determine the direction of the magnetic
+field from each wire
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+real a = 1u;
+real dx = 3u;
+pen iUL = red;
+pen iUR = blue;
+pen iLL = green;
+pen iLR = yellow;
+
+real x = 0;
+real d = a/2;
+real dy = 1mm;
+
+Vector Vs[];
+
+label("(a)", (x,-d), S);
+Vs.push(Vector((x-d,-d), phi=-90, iLL));
+Vs.push(Vector((x,0), dir=-45, iLL));
+Vs.push(Vector((x+d,-d), phi=-90, iLR));
+Vs.push(Vector((x,0), dir=45, iLR));
+Vs.push(Vector((x+d, d), phi=-90, iUR));
+Vs.push(Vector((x,0), dir=135, iUR));
+Vs.push(Vector((x-d, d), phi=-90, iUL));
+Vs.push(Vector((x,0), dir=-135, iUL));
+x += dx;
+
+label("(b)", (x,-d), S);
+Vs.push(Vector((x-d,-d), phi= 90, iLL));
+Vs.push(Vector((x,0), dir=135, iLL));
+Vs.push(Vector((x+d,-d), phi=-90, iLR));
+Vs.push(Vector((x,0), dir=45, iLR));
+Vs.push(Vector((x+d, d), phi= 90, iUR));
+Vs.push(Vector((x,0), dir=-45, iUR));
+Vs.push(Vector((x-d, d), phi=-90, iUL));
+Vs.push(Vector((x,0), dir=-135, iUL));
+x += dx;
+
+label("(c)", (x,-d), S);
+Vs.push(Vector((x-d,-d), phi= 90, iLL));
+Vs.push(Vector((x,+dy), dir=135, iLL));
+Vs.push(Vector((x+d,-d), phi= 90, iLR));
+Vs.push(Vector((x,-dy), dir=-135, iLR));
+Vs.push(Vector((x+d, d), phi=-90, iUR));
+Vs.push(Vector((x,0), dir=135, iUR));
+Vs.push(Vector((x-d, d), phi=-90, iUL));
+Vs.push(Vector((x,0), dir=-135, iUL));
+Vs.push(Vector((x,0), dir=-180)); // Net B field.
+Vs[Vs.length-1].mag *= 4*Cos(45);
+
+for (int i=0; i<Vs.length; i+=1)
+  Vs[i].draw();
+\end{asy}
+\end{center}
+
+For both \Part{a} and \Part{b}, the magnetic fields cancel out and the
+net magnetic field at the center of the square is \ans{zero}.  We can
+also see this by symmetry, without using the right-hand rule.
+The \Part{a} configuration is symmetric to $90\dg$ rotations about the
+square center, so the magnetic field at the center must also be
+symmetric to $90\dg$ rotations about the square center.  A vector with
+non-zero length is only symmetric to $360\dg$ rotations, so there
+cannot be any magnetic field in the center of \Part{a}.  Similarly,
+the \Part{b} configuration is symmetric to $180\dg$ rotations, so it
+cannot have magnetic field at the center.  \Part{c} is only symmetric
+to $360\dg$ rotations, so it can have a magnetic field at the center.
+
+The magnitude of the magnetic field for \Part{c} is given by the sum
+of the horizontal components of the corner magnetic fields, since the
+vertical components cancel out.
+\begin{equation}
+  B_c = 4 B_w \cos(45\dg)
+    = 4 \cdot \frac{\mu_0 I}{2\pi \frac{a}{\sqrt{2}}} \cdot \frac{1}{\sqrt{2}}
+    = \ans{\frac{2\mu_0 I}{\pi a}} \;.
+\end{equation}
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem28.30.tex b/latex/problems/Young_and_Freedman_12/problem28.30.tex
index 778235f..b7177fa 100644
--- a/latex/problems/Young_and_Freedman_12/problem28.30.tex
+++ b/latex/problems/Young_and_Freedman_12/problem28.30.tex
@@ -1,5 +1,52 @@
+% Commented out because these are defined in problem28.12.tex
+%\newcommand{\dB}{d\vect{B}}
+%\newcommand{\dl}{d\vect{l}}
+
 \begin{problem*}{28.30}
+Calculate the magnitude and direction of the magnetic field at point
+$P$ due to the current in the semicircular section of wire shown in
+Fig.~28.46.  (\emph{Hint:} Does the current in the long, straight
+section of the wire produce any field at $P$?)
+\begin{center}
+\begin{asy}
+import Mechanics;
+import Circ; // TODO: wire(path);
+
+real u = 1cm;
+real r = u;
+
+Distance R = Distance((0,0), (r*Cos(45),r*Sin(45)), "$r$");
+R.draw();
+draw((-2r,0)--(-r,0){N}..(0,r){E}..{S}(r,0)--(2r,0));
+TwoTerminal Il = current((-1.5r,0), "I");
+TwoTerminal Ir = current((1.5r,0), "I");
+dot("$P$", (0,0), S);
+\end{asy}
+\end{center}
 \end{problem*}
 
 \begin{solution}
+The Biot-Savart law
+\begin{equation}
+  \dB = \frac{\mu_0}{4\pi}\cdot\frac{I\dl\times\rhat}{r^2}
+\end{equation}
+gives the magnetic field from an infinitesimal chunk of current.
+The straight sections of wire do not contribute any magnetic field at $P$, because
+\begin{equation}
+  |\dl\times\rhat| = |\dl|\cdot|\rhat|\sin\theta = 0 \;,
+\end{equation}
+since $\theta=0\dg$ to the left and $180\dg$ to the right.
+
+For all portions of wire along the semicircle, $\theta=90\dg$, and
+$\dl\times\rhat$ points into the page.  The net magnetic field at $P$
+is therefore
+\begin{equation}
+  B = \int_s \frac{\mu_0}{4\pi}\cdot\frac{I\dl\times\rhat}{r^2}  
+    = \frac{\mu_0}{4\pi}\cdot\frac{I}{r^2} \int_s |\dl|\cdot|\rhat|\sin(90\dg)  
+    = \frac{\mu_0 I}{4\pi r^2} \int_s |\dl|
+    = \frac{\mu_0 I}{4\pi r^2} \cdot \pi r
+    = \ans{\frac{\mu_0I}{4r}} \;.
+\end{equation}
+where $\int_s |\dl|$ is just the length of the semicircle, which is
+half the $2\pi r$ circumference of the entire circle.
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem28.60.tex b/latex/problems/Young_and_Freedman_12/problem28.60.tex
index 570f2a7..bbda9f7 100644
--- a/latex/problems/Young_and_Freedman_12/problem28.60.tex
+++ b/latex/problems/Young_and_Freedman_12/problem28.60.tex
@@ -1,5 +1,141 @@
 \begin{problem*}{28.60}
+Figure~28.54 shows an end view of two long, parallel wires
+perpendicular to the $xy$-plane, each carrying a current $I$ but in
+opposite directions.  \Part{a} Copy the diagram, and draw vectors to
+show the \vect{B} field of each wire and the net \vect{B} field at a
+point $P$.  \Part{b} Derive the expression for the magnitude of
+\vect{B} at any point on the $x$-axis in terms of the $x$-coordinate
+of the point.  What is the direction of \vect{B}?  \Part{c} Graph the
+magnitude of \vect{B} at points on the $x$-axis.  \Part{d} At what
+value of $x$ is the magnitude of \vect{B} a maximum?  \Part{e} What is
+the magnitude of \vect{B} when $x\gg a$?
 \end{problem*}
 
+\begin{nosolution}
+\begin{center}
+\begin{asy}
+import graph;
+import Mechanics;
+
+xaxis("$x$");
+yaxis("$y$");
+
+real u = 1cm;
+real a = 1u;
+real x = 3u;
+pen ipen = red+blue;
+
+Vector Vt = Vector(Scale((0,a)), phi=90, ipen);
+Vector Vb = Vector(Scale((0,-a)), phi=-90, ipen);
+Vt.draw();
+Vb.draw();
+
+dot("P", Scale((x,0)), N);
+
+label("$a$", align=W, Scale((0,a/2)));
+label("$a$", align=W, Scale((0,-a/2)));
+label("$x$", align=S, Scale((x/2,0)));
+\end{asy}
+\end{center}
+\end{nosolution}
+
 \begin{solution}
+\Part{a}
+\begin{center}
+\begin{asy}
+import graph;
+import Mechanics;
+import ElectroMag;
+
+xaxis("$x$");
+yaxis("$y$");
+
+real u = 1cm;
+real a = 1u;
+real x = 3u;
+pen ipen = red+blue;
+
+Vector Vs[];
+Vs.push(Vector(Scale((0,a)), phi=90, ipen));
+Vs.push(Vector(Scale((0,-a)), phi=-90, ipen));
+Vs.push(BField((x,0), dir=degrees((x,-a))+90)); // From top wire
+Vs.push(BField((x,0), dir=degrees((x,a))-90)); // From bottom wire
+Vs.push(BField((x,0)));  // Net
+
+for (int i=0; i<Vs.length; i+=1) {
+  Vs[i].draw();
+}
+
+dot("P", Scale((x,0)), N);
+
+label("$a$", align=W, Scale((0,a/2)));
+label("$a$", align=W, Scale((0,-a/2)));
+label("$x$", align=S, Scale((x/2,0)));
+\end{asy}
+\end{center}
+
+\Part{b}
+The magnitude of magnetic field from each wire at $P$ is
+\begin{equation}
+  B_w = \frac{\mu_0 I}{2\pi r}
+    = \frac{\mu_0 I}{2\pi\sqrt{x^2+a^2}} \;.
+\end{equation}
+The net magnetic field is the sum of the horizontal components, since
+the vertical components cancel.
+\begin{equation}
+  B = 2 B_w \cos\theta
+    = 2\cdot\frac{\mu_0 I}{2\pi\sqrt{x^2+a^2}}\cdot\frac{a}{\sqrt{x^2+a^2}}
+    = \ans{\frac{\mu_0 I a}{\pi(x^2+a^2)}} \;,
+\end{equation}
+where $\cos\theta=a/r$ comes from
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 1cm;
+real a = 1u;
+real x = 3u;
+
+Vector Bt = BField((x,0), dir=degrees((x,-a))+90); // From top wire
+Angle theta1 = Angle((x,0)+dir(0), (x,0), (x,0)+dir(Bt.dir), "$\theta$");
+Angle theta2 = Angle((0,0), (0,a), (x,0), "$\theta$");
+
+theta1.draw();
+theta2.draw();
+draw((0,0)--(0,a)--(x,0)--cycle);
+Bt.draw();
+label("$r$", (x/2,a/2), NE);
+label("$a$", (0,a/2), W);
+\end{asy}
+\end{center}
+
+\Part{c}
+\begin{center}
+\begin{asy}
+import graph;
+size(5cm, 2cm, IgnoreAspect);
+
+real a=1;
+real B(real x){
+  return 1.0/(x**2 + 1);
+}
+
+xaxis("$x$");
+yaxis("$B$");
+draw(graph(B, -2*a, 2*a, n=333), red);
+label("$a$", align=S, Scale((a,0)));
+label("$-a$", align=S, Scale((-a,0)));
+\end{asy}
+\end{center}
+
+\Part{d}
+The magnitude of \vect{B} has a maximum at $\ans{x=0}$.
+
+\Part{e}
+For $x\gg a$, $x^2+a^2\approx x^2$, so
+\begin{equation}
+  B = \frac{\mu_0 I a}{\pi(x^2+a^2)}
+    \approx \ans{\frac{\mu_0 I a}{\pi x^2}} \;.
+\end{equation}
 \end{solution}
diff --git a/latex/problems/Young_and_Freedman_12/problem28.62.tex b/latex/problems/Young_and_Freedman_12/problem28.62.tex
index cda5006..4ad4789 100644
--- a/latex/problems/Young_and_Freedman_12/problem28.62.tex
+++ b/latex/problems/Young_and_Freedman_12/problem28.62.tex
@@ -1,5 +1,91 @@
 \begin{problem*}{28.62}
+A pair of long, rigid metal rods, each of length $L$, lie parallel to
+each other on a perfectly smooth table.  Their ends are connected by
+identical, very light conducting springs of force constant $k$
+(Fig.~28.55) and negligable unstretched length.  If a current $I$ runs
+through this circuit, the springs will stretch.  At what seperation
+will the rods remain at rest?  Assume that $k$ is large enough so that
+the separation of the rods will be much less than $L$.
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+real L = 4u;
+real d = 1u;
+real Isep = 6pt;
+real hBar = 1mm;
+pen ipen = red+blue;
+
+Block Bt = Block((0,0), width=L, height=hBar, fill=yellow);
+Block Bb = Block((0,-d), width=L, height=hBar, fill=yellow);
+Spring Sl = Spring((-L/2,-d), (-L/2,0),
+                   deadLength=1mm, unstretchedLength=d, "k");
+Spring Sr = Spring((L/2,0), (L/2,-d),
+                   deadLength=1mm, unstretchedLength=d, "k");
+Vector It = Vector((0,Isep), dir=0, ipen);
+Vector Ib = Vector((0,-d-Isep), dir=180, ipen);
+
+Distance DL = Distance((-L/2,0), (L/2,0), offset=-2Isep, "$L$");
+Distance Dd = Distance((L/2,0), (L/2,-d), offset=-4Isep, "$d$");
+
+Sl.draw(rotateLabel=false);
+Sr.draw(rotateLabel=false);
+Bt.draw();
+Bb.draw();
+It.draw();
+Ib.draw();
+DL.draw();
+Dd.draw(rotateLabel=false);
+\end{asy}
+\end{center}
 \end{problem*}
 
 \begin{solution}
+In order for the rods to remain at rest, the net force on each rod
+must be zero.  The only forces we need to consider are the spring
+forces $F_s$ and magnetic forces $F_B$.  The only other possible force
+for this problem would be a gravitational force $F_g=mg$, but no mass
+$m$ is given for the rods, so we must assume $F_g$ is negligable.
+
+Because of Newton's third law, we know that if the forces of the
+bottom rod on the top rod cancel, then the forces from the top rod on
+the bottom rod must also cancel, and we can restrict our force
+balancing to only the top rod.
+
+The spring force on the top rod is
+\begin{equation}
+  F_s = 2kd
+\end{equation}
+downward, due to a $kx$ force for each spring (left and right), with
+$x=d$ because the unstetched length of the springs is negligable.
+
+The magnetic force on the top rod is
+\begin{equation}
+  \vect{F}_B = I\vect{L}\times\vect{B} \;,
+\end{equation}
+where \vect{B} is the magnetic field along the top rod created by the
+current in the bottom rod.  This magnetic field can be approximated as
+as that due to an infinitely long, straight wire a distance $d$ below
+the top wire
+\begin{equation}
+  B = \frac{\mu_0 I}{2\pi d}
+\end{equation}
+directed into the board.  The bottom wire is not infinitely long, but
+because $d\ll L$, fringe effects due to the wire's finite length are
+small.  Because the magnetic field is into the board and the current
+in the top wire is to the right, the vector $\vect{F}_B$ is
+\begin{equation}
+  F_B = |I\vect{L}|\cdot|\vect{B}|\sin\theta = ILB
+    = \frac{\mu_0 I^2L}{2\pi d}
+\end{equation}
+directed upward.
+
+Putting this all together to balance the forces on the top bar
+\begin{align}
+  0 &= \sum{F_y} = F_B - F_s = \frac{\mu_0 I^2L}{2\pi d} - 2kd \\
+  2kd &= \frac{\mu_0 I^2L}{2\pi d} \\
+  d^2 &= \frac{\mu_0 I^2L}{4\pi k} \\
+  d &= \ans{\sqrt{\frac{\mu_0 L}{4\pi k}}\;I}\;.
+\end{align}
 \end{solution}