Problem 22.49

22.49. b)
After the proton enters the region, it travels in a circular path. The force acting on it is given by
$F = q\vec{v} \times \vec{B}$
F = qvB
the proton experiences centripetal acceleration. Therefore,
$qvB = \frac{mv^{2}}{r}$
Solving for the radius of the circle, we find that
$r = \frac{mv}{qB}$
We now need the velocity v of the proton. We find this from the kinetic energy.
$K = \frac{1}{2} m v^{2}$
$v = \sqrt{\frac{2K}{m}}$
$v = 3.10 \times 10^{7} m/s$
Substituting in to our equation for the radius, we find that
r = 6.46 m
Now, we turn to the geometry of the problem. While the proton is in between the plates, it moves in a circular path. Therefore, it travels through an angle $\alpha$ while experiencing the magnetic field. Using the fact that we know the radius of the circle, and the fact that the plate is only one meter in length, we find that
$\sin \alpha = \frac{1 m}{6.46 m}$
Solving for the angle $\alpha$, we find
$\alpha = 8.90{}^{\circ}$

a)
To find the component of the momentum, first we have to find the momentum. We find that the total momentum is given by
$p = - (m_{proton})(3.10 \times 10^{7} m/s)$
so, the $\hat{y}$ component is given by
$p_{y} = p \sin \alpha$
$p_{y} = -8.00 \times 10^{-21} kg \cdot m/s$