Answer:
A) may experience a magnetic foce, but its speed will not change
Step-by-step explanation:
When a charged particle is moving in a static uniform magnetic field, it may experience a magnetic force, whose magnitude is
![F=qvB sin \theta](https://img.qammunity.org/2020/formulas/physics/high-school/y9ixw56ymzhmz9ri0s4z3hesy0guwy28du.png)
where
q is the magnitude of the charge
v is the velocity of the particle
B is the magnetic field strength
is the angle between the direction of v and B
First of all, we notice that if v is parallel to B, then
and the force is zero: this means that no force acts on the particle if the particle travels parallel to the field, so this rules out option D.
Also, the direction of this magnetic force is perpendicular to the velocity of the particle (its direction can be found by using Fleming's left hand rule). This means that the direction of motion of the particle will change, so this rules out option B.
Moreover, the magnetic force, being perpendicular to the direction of motion, does not do any work on the particle: this means that its kinetic energy does not change, so its speed remains the same, therefore option A is the correct one.