1) Magnitude
The magnetic force acting on a moving charge is given by:
[tex] F=qvB \sin \theta [/tex]
where
q is the charge of the particle
v is its speed
B is the magnitude of the magnetic field
[tex] \theta [/tex] is the angle between the directions of v and B.
In this problem, we have a proton, so the charge is [tex] q=1.6 \cdot 10^{-19} C [/tex]; its speed is [tex] v=5.0 \cdot 10^3 m/s [/tex], while the magnetic field is [tex] B=0.20 T [/tex]. The angle is [tex] \theta=90^{\circ} [/tex], because the proton is moving perpendicular to the magnetic field. Therefore, if we substitute these data into the previous equation, we find:
[tex] F=(1.6 \cdot 10^{-19}C)(5.0 \cdot 10^3 m/s)(0.20 T)(\sin 90^{\circ})=1.6 \cdot 10^{-16} N [/tex]
So, the magnitude of the force is [tex] 1.6 \cdot 10^{-16} N [/tex].
2) Direction
The direction of the force can be determined using the right-hand rule:
- index finger: direction of the velocity (eastward)
- middle finger: direction of the magnetic field (northward)
- thumb: direction of the force (upward).
So, the direction of the force is upward.
