Completed rec9 solutions

[course.git] / latex / problems / Young_and_Freedman_12 / problem28.23.tex

diff --git a/latex/problems/Young_and_Freedman_12/problem28.23.tex b/latex/problems/Young_and_Freedman_12/problem28.23.tex
index e844ac454add56157779c807b69990b209aa9631..cb598017d481db65257bf69b05ab1b9d5beb1a61 100644 (file)
--- 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}