Completed rec9 solutions

[course.git] / latex / problems / Young_and_Freedman_12 / problem28.12.tex

diff --git a/latex/problems/Young_and_Freedman_12/problem28.12.tex b/latex/problems/Young_and_Freedman_12/problem28.12.tex
index 4b0ce191539e5a3e5621fadae7411f92057ebe8f..30df69accb0e1508024910145ce331af40f204e5 100644 (file)
--- 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}