Add solutions for Serway and Jewett v8's chapter 31.

[course.git] / latex / problems / Serway_and_Jewett_8 / problem31.45.tex

diff --git a/latex/problems/Serway_and_Jewett_8/problem31.45.tex b/latex/problems/Serway_and_Jewett_8/problem31.45.tex
index 5acdf74a3c9d2164ba6caea8161eaa9d25967d12..183e4c2828e40fc906c7d40d0fa95f2b885a32d9 100644 (file)
--- a/latex/problems/Serway_and_Jewett_8/problem31.45.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem31.45.tex
@@ -55,5 +55,21 @@ draw((xr, R.terminal[1].y) -- R.terminal[1]);
 \end{problem*}
 
 \begin{solution}
+Picking up as the positive flux direction, the induced \EMF\ is
+\begin{equation}
+  \EMF = -\deriv{t}{\Phi_B} = -\deriv{t}{NAB} = -NA\deriv{t}{B}
+    = -NA\frac{-2B}{\Delta t} = \frac{2NAB}{\Delta t} \;.
+\end{equation}
+This \EMF\ drives a current through the resistor
+\begin{align}
+  0 &= \EMF - IR \\
+  I &= \frac{\EMF}{R} = \frac{-2NAB}{R\Delta t} \;.
+\end{align}
+Current is defined as the charge passing through a cross section of
+your circuit per unit time, so the charge entering one end of the
+resistor is
+\begin{equation}
+  \Delta q = \frac{\Delta q}{\Delta t} \Delta t = I \Delta t
+    = \frac{2NAB}{R} = \ans{0.880\U{C}} \;.
+\end{equation}
 \end{solution}
-