1. The magnetic force on a charged particle is given by $\vec F = q(\vec v \times \vec B).$ This force is always perpendicular to the velocity component that is perpendicular to $\vec B$.

2. Resolve the velocity $\vec v$ into two components:

• $v_\parallel$ along the magnetic field $\vec B$
• $v_\perp$ perpendicular to the magnetic field $\vec B$

So, $\vec v = \vec v_\parallel + \vec v_\perp.$

3. Effect of magnetic field on each component:

• For $v_\parallel$: $\vec v_\parallel \times \vec B = 0,$ so there is no force on this component. Hence the particle keeps moving uniformly along $\vec B$.

• For $v_\perp$: this component experiences magnetic force of magnitude $F = qv_\perp B,$ which acts as the centripetal force. Therefore, the particle executes circular motion in a plane perpendicular to $\vec B$.

4. Combining both motions:

• circular motion due to $v_\perp$
• uniform linear motion along $\vec B$ due to $v_\parallel$

The resultant path is a helix whose axis is along the magnetic field direction.

5. Check options:

• A: helical path with the axis along magnetic field $\mathrm{B}$ — Correct
• B: straight along the direction of magnetic field $\mathrm{B}$ — only if $v_\perp = 0$, not here
• C: circular path — only if $v_\parallel = 0$, not here
• D: helical path with the axis perpendicular to the direction of magnetic field $\mathrm{B}$ — incorrect

Therefore, the correct answer is: $\boxed{\text{A}}$