Completed rec9 solutions

[course.git] / latex / problems / Young_and_Freedman_12 / problem28.62.tex

diff --git a/latex/problems/Young_and_Freedman_12/problem28.62.tex b/latex/problems/Young_and_Freedman_12/problem28.62.tex
index cda5006b0b823b63bab676b0f424a6c490c1d5b6..4ad478996c88d13713ddd4d4001b9ebd93c4ab84 100644 (file)
--- 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}