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

[course.git] / latex / problems / Serway_and_Jewett_8 / problem31.23.tex

diff --git a/latex/problems/Serway_and_Jewett_8/problem31.23.tex b/latex/problems/Serway_and_Jewett_8/problem31.23.tex
index b2618d2fc72c90cb7fb54b5ee837a2b875a1334f..80c2eec278fe2ac06f8bec8da7f3ca9514810e33 100644 (file)
--- a/latex/problems/Serway_and_Jewett_8/problem31.23.tex
+++ b/latex/problems/Serway_and_Jewett_8/problem31.23.tex
@@ -39,5 +39,49 @@ Dl.draw();
 \end{problem*}
 
 \begin{solution}
-\end{solution}
+\Part{a}
+Moving the bar to the right increases the magnetic flux directed into
+the page by increasing the area of magnetic field enclosed by the
+loop.  Increasing the flux induces a counter-clockwise current to
+resist the change.
+\begin{align}
+  \Phi_B &= AB = lxB \\
+  |\EMF| &= |-\deriv{t}{\Phi_B}| = lB \deriv{t}{x} = lBv \\
+  0 &= \sum_\text{loop} V_i = \EMF - IR \\
+  I &= \frac{\EMF}{R} = \frac{lBv}{R} \;.
+\end{align}
+The counter-clockwise current is moving up through the sliding bar, so
+the magnetic force on the bar is directed to the left, with a
+magnitude of
+\begin{equation}
+  F_B = IlB\sin(90\dg) = IlB = \frac{l^2B^2v}{R} \;.
+\end{equation}
+
+For the bar to move to the right at a constant speed, the net force in
+the $\ihat$ direction should be zero.  We'll have to apply an external
+$F_\text{app}$ to the right to counter $F_B$.
+\begin{align}
+  0 &= \sum_i F_{i,x} = F_\text{app} - F_B \\
+  F_\text{app} &= F_B = \frac{l^2B^2v}{R}
+    = \frac{(1.20\U{m}\cdot2.50\U{T})^2\cdot2.00\U{m/s}}{6.00\U{\Ohm}}
+    = \ans{3.00\U{N}} \;.
+\end{align}
 
+\Part{b}
+The power delivered to a resister is
+\begin{equation}
+  P_R = IV = I^2R = \p({\frac{lBv}{R}})^2 R = \frac{(lBv)^2}{R}
+    = \frac{(1.20\U{m}\cdot2.50\U{T}\cdot2.00\U{m/s})^2}{6.00\U{\Ohm}}
+    = \ans{6.00\U{W}} \;.
+\end{equation}
+
+If you want to look at the problem in terms of energy conservation,
+the external force is putting energy into the system at a rate of
+\begin{equation}
+  P_\text{in} = \deriv{t}{F_\text{app} x} = F_\text{app}\deriv{t}{x}
+    = F_\text{app}v = \frac{l^2B^2v}{R} \cdot v = \frac{(lBv)^2}{R} \;.
+\end{equation}
+All this energy has to go somewhere, and the only place for it to go
+is into resistor heat, so it's no surprise that we get the same
+formula for $P_\text{in}$ that we got for $P_R$.
+\end{solution}