From 2b7e6897bb83843024c55f87cc7724d66407d2db Mon Sep 17 00:00:00 2001
From: "W. Trevor King" <wking@tremily.us>
Date: Fri, 1 Jun 2012 17:11:45 -0400
Subject: [PATCH] Fix B_P -> B_Q in Serway and Jewett v8's 30.61.c solution.

---
 latex/problems/Serway_and_Jewett_8/problem30.61.tex | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/latex/problems/Serway_and_Jewett_8/problem30.61.tex b/latex/problems/Serway_and_Jewett_8/problem30.61.tex
index e760134..371e29a 100644
--- 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}
-- 
2.26.2