A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be

We are given that a charged particle moving in a magnetic field $$\vec{B}$$ has velocity components both along $$\vec{B}$$ and perpendicular to $$\vec{B}$$.

Decompose the velocity. Let the velocity be split as $$\vec{v} = v_{\parallel}\hat{B} + v_{\perp}\hat{n}$$, where $$v_{\parallel}$$ is along $$\vec{B}$$ and $$v_{\perp}$$ is perpendicular to $$\vec{B}$$.

Effect of the perpendicular component. The magnetic force $$\vec{F} = q(\vec{v} \times \vec{B})$$ acts only on $$v_{\perp}$$. Since this force is always perpendicular to $$v_{\perp}$$, it causes the particle to move in a circular path in the plane perpendicular to $$\vec{B}$$.

Effect of the parallel component. Since $$\vec{v}_{\parallel} \times \vec{B} = 0$$, there is no magnetic force along $$\vec{B}$$. The particle moves with constant velocity $$v_{\parallel}$$ along the field direction.

Combined motion. The superposition of circular motion (perpendicular to $$\vec{B}$$) and uniform linear motion (along $$\vec{B}$$) gives a helical (spiral) path with its axis along the direction of $$\vec{B}$$.

The correct answer is Option B: helical path with the axis along magnetic field B.