A charged particle enters a uniform magnetic field at an angle of 45^…

The correct answer is Option (3) → helical with axis along magnetic field

The magnetic force \( \vec{F} = q\,\vec{v}\times\vec{B} \) acts only on the part of the velocity that is perpendicular to \( \vec{B} \). So split the velocity into two components.

    \( v_{\parallel} = v\cos 45^\circ \;(\text{along } \vec{B}) \)
    \( v_{\perp} = v\sin 45^\circ \;(\text{perpendicular to } \vec{B}) \)

The parallel component feels no force, so the particle drifts uniformly along the field. The perpendicular component makes it go round in a circle in the plane perpendicular to \( \vec{B} \). The two motions together give a helix whose axis is along the magnetic field.

Option (1) linear — Incorrect. A straight path needs \( \vec{v} \) parallel or antiparallel to \( \vec{B} \) (angle \( 0^\circ \) or \( 180^\circ \)), so that the force is zero.

Option (2) circular — Incorrect. A closed circle needs \( \vec{v} \) perpendicular to \( \vec{B} \); here \( v_{\parallel} \ne 0 \) carries the particle along the field.

Option (3) helical with axis along magnetic field — Correct. Circular motion about \( \vec{B} \) plus uniform drift along \( \vec{B} \) is a helix with its axis along the field.

Option (4) helical with axis perpendicular to magnetic field — Incorrect. The drift is along \( \vec{B} \), so the axis of the helix is along the field, not perpendicular to it.

NCERT (2026-27):
Motion in a Magnetic Field – “… The particle will describe a circle if v and B are perpendicular to each other …. If velocity has a component along B, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion ….”