The magnetic force experienced by the proton is given by
[tex]F=qvB \sin \theta[/tex]
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and [tex]\theta[/tex] the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so [tex]\sin \theta=1[/tex] and we can ignore it in the formula.
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
[tex]F=ma[/tex]
So we have
[tex]ma=qvB[/tex]
from which we can find the magnitude of the field:
[tex]B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T [/tex]
