Reorganized problems by textbook and imported my previous courses' code.

[course.git] / latex / problems / Serway_and_Jewett_4 / problem23.53.tex

diff --git a/latex/problems/Serway_and_Jewett_4/problem23.53.tex b/latex/problems/Serway_and_Jewett_4/problem23.53.tex
new file mode 100644 (file)
index 0000000..0bc23c2
--- /dev/null
+++ b/latex/problems/Serway_and_Jewett_4/problem23.53.tex
@@ -0,0 +1,46 @@
+\begin{problem*}{23.53}
+A particle with a mass of $m = 2.00\E{-16}\U{kg}$ and a charge of $q =
+30.0\U{nC}$ starts from rest, is accelerated by a strong electric
+field, and is fired from a small source inside a region of uniform
+constant magnetic field $B = 0.600\U{T}$.  The velocity of the
+particle is perpendicular to the field.  The circular orbit of the
+particle encloses a magnetic flux of $\Phi_B = 15.0\U{$\mu$Wb}$.
+\Part{a} Calculate the speed of the particle.
+\Part{b} Calculate the potential difference through which the particle
+accelerated inside the source.
+\end{problem*} % problem 23.53
+
+\begin{solution}
+\Part{a}
+For particles circling in a uniform, perpendicular magnetic field,
+\begin{align}
+ F_c &= m \frac{v^2}{r} = qvB \\
+ mv &= qrB 
+\end{align}
+
+Letting $\tau$ be the period, from $\Delta x = v \Delta t$ we have
+\begin{equation}
+ \tau = \frac{2 \pi r}{v} = \frac{2 \pi r m}{qrB} = \frac{2 \pi m}{qB}
+      = 69.8\U{ns}
+\end{equation}
+The inverse of our cyclotron frequency from Recitation 7.
+
+The flux and magnetic field give us radius by
+\begin{align}
+ \Phi_B &= AB = \pi r^2 B \\
+ r &= \sqrt{\frac{\Phi_B}{\pi B}} = \ans{2.82\U{mm}}
+\end{align}
+
+So the speed is given by
+\begin{equation}
+ v = \frac{2 \pi r}{\tau} = \frac{qB}{2 \pi m}\sqrt{\frac{\Phi_B}{\pi B}}
+   = \ans{254\U{km/s}}
+\end{equation}
+
+\Part{b}
+Conserving energy
+\begin{align}
+ K &= \frac{1}{2}mv^2 = q\Delta V \\
+ \Delta V &= \frac{mv^2}{2q} = \ans{215\U{V}}
+\end{align}
+\end{solution}