\begin{problem*}{22.3}
A proton travels with a speed of $v = 3.00\E{6}\U{m/s}$ at an angle of
$\theta = 37.0\dg$ with the direction of a magnetic field of $B =
0.300\U{T}$ in the $+y$ direction.  What are \Part{a} the magnitude of
the magnetic force on the proton and \Part{b} its acceleration?
\end{problem*} % problem 22.3

\begin{solution}
We'll pick the \ihat\ direction so that \vect{v} has a positive $x$-component.

\Part{a}
\begin{equation}
 F_B = q v B \sin\theta = (1.60\E{-19}\U{C})\cdot(3.00\E{6}\U{m/s})\cdot(0.300\U{T})\cdot\sin37.0\dg
     = \ans{8.67\E{-14}\U{N}}
\end{equation}
and the direction of the force is in the \khat\ direction.

\Part{b}
Using $\vect{F} = m \vect{a}$ we have
\begin{equation}
 \vect{a} = \frac{\vect{F}}{m} = \frac{8.67\E{-14}\U{N}}{1.67\E{-27}\U{kg}}
 = \ans{5.19\E{13}\U{m/s$^2$}}
\end{equation}
\end{solution}
