Step 1: Recall the magnetic force rule.
A moving charge in a magnetic field feels a force given by $F=qvB\sin\theta$. Here $q$ is the charge, $v$ is its speed, $B$ is the field strength, and $\theta$ is the angle between the velocity and the field. We use this because the question is about the force on a moving proton.
Step 2: Notice why the angle matters.
The force depends on $\sin\theta$. This means the direction of motion compared to the field is the deciding factor, not just the size of $v$ or $B$.
Step 3: Read the given direction.
The proton moves parallel to the field. Parallel means the two directions are along the same line, so the angle between them is $\theta=0^\circ$.
Step 4: Find the sine of the angle.
The sine of zero degrees is zero. \[ \sin 0^\circ=0 \]
Step 5: Put it into the formula.
Even though $q$, $v$, and $B$ are nonzero, multiplying by $\sin 0^\circ$ wipes everything out. \[ F=qvB\times 0=0 \]
Step 6: Make sense of it.
A magnetic field can only push a charge sideways to its motion. When the charge moves straight along the field, there is no sideways push at all, so the force is zero. This is option 1. \[ \boxed{0\ \text{N}} \]