--- /dev/null
+\newcommand{\dB}{d\vect{B}}
+\newcommand{\dl}{d\vect{l}}
+\newcommand{\rhat}{\hat{r}}
+
+\begin{problem*}{28.12}
+
+\begin{center}
+\begin{asy}
+import Mechanics;
+import ElectroMag;
+
+real u = 0.1cm;
+real Ysep = 5u;
+real hypot = 8u;
+real Xsep = sqrt(hypot**2 - (Ysep/2)**2);
+real Xslush = 1u;
+
+Wire wtop_seg = Wire(
+Wire wtop = Wire((), (Xslush,Ysep/2));
+
+wtop.draw();
+wtop_seg.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+Using our right hand rules for current-created \vect{B} fields, we see
+that both segments create magnetic fields that are into the page at
+$P$, so the sum will also be directed \ans{into the page}. To find
+the magnitude, we'll use the Biot-Savart law
+\begin{equation}
+ \dB = \frac{\mu_0}{4\pi}\cdot\frac{I\dl\times\rhat}{r^2}
+\end{equation}
+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}
+\end{asy}
+\end{center}
+Using our knowledge of cross products and trigonometry
+\begin{equation}
+ |\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
+magnitude and direction for the bottom segment, although you could
+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}
+ \cdot(12.0\U{A}+24.0\U{A})
+ = \ans{264\U{nT}} \;.
+\end{equation}
+\end{solution}