Fix B_P -> B_Q in Serway and Jewett v8's 30.61.c solution.

diff --git a/latex/problems/Serway_and_Jewett_8/problem30.61.tex b/latex/problems/Serway_and_Jewett_8/problem30.61.tex
index e7601341fc39ca9bb239718bf7ae7e0b5b2d2a63..371e29a299a4d7327c5939f1bd7b3d026b8b717f 100644 (file)
--- a/latex/problems/Serway_and_Jewett_8/problem30.61.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem30.61.tex
@@ -85,14 +85,14 @@ The contribution from the $x$-aligned wire will be along $-\jhat$,
 while the contribution from the $y$-aligned wire will be along
 $\ihat$.  The total magnetic field is
 \begin{align}
-  \vect{B}_P
+  \vect{B}_Q
     &=   \frac{\mu_0 I_x}{2\pi r} \cdot (-\jhat)
        + \frac{\mu_0 I_y}{2\pi r}\ihat
     = \frac{\mu_0}{2\pi r}\cdot(I_y\ihat-I_x\jhat)
     = (2.00\ihat - 3.33\jhat)\U{$\mu$T} \\
-  |\vect{B}_P| &= \sqrt{2.00^2 + (-3.33)^2}\U{$\mu$T}
+  |\vect{B}_Q| &= \sqrt{2.00^2 + (-3.33)^2}\U{$\mu$T}
     = \ans{3.89\U{$\mu$T}} \\
-  \theta_P &= \arctan\p({\frac{-3.33}{2.00}})
+  \theta_Q &= \arctan\p({\frac{-3.33}{2.00}})
     = \ans{-59.0\dg} \;.
 \end{align}
 \end{solution}