A charged particle of mass $m$ and charge $q$ enters a uniform magnetic field $B$ with velocity $\vec{v}$ at an ANGLE $\theta$ ($0 < \theta < 90^\circ$) to the field direction. Its trajectory is

1. Decompose the velocity into two components: $v_\parallel = v\cos\theta$ along $\vec{B}$, and $v_\perp = v\sin\theta$ perpendicular to $\vec{B}$. 2. The PARALLEL component experiences NO magnetic force (since $\vec{v}\parallel\vec{B}$ gives $\sin 0 = 0$). So $v_\parallel$ is unchanged. 3. The PERPENDICULAR component undergoes circular motion in a plane perpendicular to $\vec{B}$ with radius $r = m v_\perp/(qB)$. 4. The combination — uniform motion along the axis plus circular motion perpendicular — produces a HELIX winding around the field line. 5. Pitch of the helix: $p = v_\parallel\,T = \dfrac{2\pi m v\cos\theta}{qB}$. 6. Options A, B, D are degenerate cases ($\theta = 0$, $\theta = 90^\circ$) or wrong physics for non-zero $\theta$ between extremes. _Source: NCERT Class 12 Physics Part 1, Ch 4, §4.4.2 (Motion in Magnetic Field — helical motion), p. 9._