Answer
1.97 micrometers.
Work Step by Step
Assume that the acceleration is so small that the velocity vector essentially keeps the same direction, so the acceleration and the magnetic field remain perpendicular to the original motion.
This assumption leads to a constant perpendicular acceleration, so we solve the problem using the methods of projectile motion.
$$qvB=ma$$
$$a=\frac{qvB}{m}$$
The time over which the acceleration acts is found from the original speed v and the horizontal distance, L.
$$t=\frac{L}{v}$$
The initial speed in the perpendicular direction is zero. Find the perpendicular displacement.
$$d=v_{o}t+\frac{1}{2}at^2=0+\frac{1}{2}\frac{qvB}{m}(\frac{L}{v})^2$$
$$d=\frac{qBL^2}{2mv}=\frac{(18.5\times10^{-9}C)(5.00\times10^{-5}T)(1500m)^2}{2(0.00340kg)(155m/s)}$$
$$=1.97\times10^{-6}m$$
