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

[course.git] / latex / problems / Serway_and_Jewett_8 / problem31.30.tex

diff --git a/latex/problems/Serway_and_Jewett_8/problem31.30.tex b/latex/problems/Serway_and_Jewett_8/problem31.30.tex
index b441fdd774811d324ba2e481debcfb6ea8edbe1b..d2fb1362e2b547fffae684300418a23d694735e6 100644 (file)
--- a/latex/problems/Serway_and_Jewett_8/problem31.30.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem31.30.tex
@@ -1,7 +1,7 @@
 \begin{problem*}{31.30}
 A rectangular coil with resistance $R$ has $N$ turns, each of length
 $l$ and width $w$ as shown in Figure~P31.30.  The coil moves into a
-uniform magnetic fiield $\vect{B}$ with constant velocity $\vect{v}$.
+uniform magnetic field $\vect{B}$ with constant velocity $\vect{v}$.
 What are the magnitude and direction of the total magnetic force on
 the coil \Part{a} as it enters the magnetic field, \Part{b} as it
 moves within the field, and \Part{c} as it leaves the field?
@@ -35,5 +35,40 @@ Dw.draw();
 \end{problem*}
 
 \begin{solution}
-\end{solution}
+\Part{a}
+If we define ``into the page'' as the positive direction, the flux
+through the loop will be increasing as the coil enters the field,
+which will induce a current in the counter-clockwise direction
+opposing the changing flux.  The right side of the coil and portions
+of the top and bottom sides will be in the field regions, and because
+of the current will be subject to magnetic forces directed to the
+left, down, and up respectively.  Because equal portions of the top
+and bottom side will be in the field, there will be no vertical
+component in the net force, which will be directed to the left.
+
+The magnitude of the induced \EMF\ is
+\begin{equation}
+  |\EMF| = |-\deriv{t}{\Phi_B}| = \deriv{t}{AB} = B\deriv{t}{Nxw}
+    = NBw\deriv{t}{x} = NBwv \;.
+\end{equation}
+This leads to a current of
+\begin{equation}
+  I = \frac{\EMF}{R} = \frac{Bwv}{R} \;,
+\end{equation}
+which causes a magnetic force of
+\begin{equation}
+  F_B = NIwB\sin(90\dg) = \ans{\frac{N^2B^2w^2v}{R}} \;.
+\end{equation}
 
+\Part{b}
+While the coil is completely inside the field, the flux remains
+constant, so there is no induced current and \ans{no magnetic force}.
+
+\Part{c}
+As the coil leaves the field, the flux drops back towards zero,
+indicing a clockwise current trying to keep the flux up.  Again, the
+vertical components of the resulting magnetic force cancel out, but
+the upward current in the left side will be subject to a magnetic
+force directed to the left.  The magnitude is the same as
+for \Part{a}.
+\end{solution}