The magnitude of the force that a magnetic field exerts on a charged particle is given by the equation:
F = qvB sin(theta)
where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and theta is the angle between the velocity vector and the magnetic field vector.
In this case, the proton has a positive charge of +1.6 x 10^-19 C, and it is moving eastward with a velocity of 5.0 km/s. The magnetic field is pointing northward with a strength of 0.20 T.
The angle between the velocity vector and the magnetic field vector is 90 degrees, since the velocity is eastward and the magnetic field is northward.
Plugging these values into the equation, we get:
F = (1.6 x 10^-19 C)(5.0 x 10^3 m/s)(0.20 T) sin(90)
F = 1.6 x 10^-19 N
So the magnitude of the force that the magnetic field exerts on the proton is 1.6 x 10^-19 N.
The direction of the force can be determined using the right-hand rule. If you point your right thumb in the direction of the proton's velocity (eastward), and your fingers in the direction of the magnetic field (northward), then the direction of the force vector is perpendicular to both, pointing downward. Therefore, the direction of the force on the proton is southward.