\begin{problem*}{22.12}
A cyclotron designed to accelerate protons has an outer radius of $R =
0.350\U{m}$.  The protons are emitted nearly at rest from a source at
the center and are accelerated through $V = 600\U{V}$ each time they
cross the gap between the dees.  The dees are between the poles of an
electromagnet where the field is $B = 0.800\U{T}$.
\Part{a} Find the cyclotron frequency $f$.
\Part{b} Find the speed $v_e$ at which the protons exit the cyclotron and 
\Part{c} their kinetic energy $K$.
\Part{d} How many revolutions $N$ does a proton make in the cyclotron?
\Part{e} For what time $\Delta t$ interval does one proton accelerate?
\end{problem*} % problem 22.12

\begin{solution}
\Part{a}
Protons with velocities $v$ in a constant magntic field will move in
circles of radius $r$ in the plane perpendicular to the magnetic
field.  The centerward acceleration is given by
\begin{align}
 F_c = m \frac{v^2}{r} &= F_B = qvB \\
 v &= \frac{qrB}{m}
\end{align}
Their velocity can also be related to their period $T=1/f$ by
\begin{equation}
 v = \frac{dr}{dt} = \frac{2\pi r}{T} = 2\pi r f
\end{equation}
So
\begin{align}
 2\pi r f &= \frac{qrB}{m} \\
 f &= \frac{qB}{2\pi m} = 12.2\U{MHz}
\end{align}
This is the frequency of revolution for a proton \emph{anywhere}
inside the cyclotron.

\Part{b}
Using our $v(r)$ equation from \Part{a} when $r = R$, we have
\begin{equation}
 v_e = \frac{qRB}{m} = \frac{(1.60\E{-19}\U{C})\cdot(0.350\U{m})\cdot(0.800\U{T})}{1.69\E{-27}\U{kg}} = \ans{26.8\U{Mm/s}}
\end{equation}

\Part{c}
\begin{equation}
 K = \frac{1}{2} m v^2 = \frac{(qrB)^2}{2m} = \ans{6.01\E{-13}\U{J}}
\end{equation}

\Part{d}
The kinetic energy $K$ is built up from $2N$ passes through $V$ (twice
per revolution).
\begin{equation}
 N = \frac{K}{2qV} = \frac{6.01\E{-13}\U{J}}{2\cdot(1.60\E{-19})\cdot(600\U{V})}
   = \ans{3130}
\end{equation}

\Part{e}
\begin{equation}
 \Delta t = T N = N/f = \ans{257\U{$\mu$s}}
\end{equation}
\end{solution}
