Completed rec9 solutions

[course.git] / latex / problems / Young_and_Freedman_12 / problem28.30.tex

diff --git a/latex/problems/Young_and_Freedman_12/problem28.30.tex b/latex/problems/Young_and_Freedman_12/problem28.30.tex
index 778235f18f386c494cca71f4aeb4fe5c9495f555..b7177fa64536e315afc692807da4bc2f1847b4ee 100644 (file)
--- 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}