Completed rec9 solutions

[course.git] / latex / problems / Young_and_Freedman_12 / problem28.60.tex

diff --git a/latex/problems/Young_and_Freedman_12/problem28.60.tex b/latex/problems/Young_and_Freedman_12/problem28.60.tex
index 570f2a7ec36f3641a00a24de5ba5b9bf71337e3b..bbda9f7ebe7cf8177a44e62c26461eb9aee4a1d4 100644 (file)
--- 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}