Motion of a charged particle having velocity v in a uniform magnetic…

The correct answer is Option (3) → perpendicular to B

The magnetic force \( \vec{F} = q\,\vec{v}\times\vec{B} \) has size \( qvB\sin\theta \) and is always perpendicular to the velocity. It acts only on the velocity component perpendicular to \( \vec{B} \).

Option (1) parallel to B — Incorrect. For \( \theta = 0^\circ \) the force is zero, so the particle moves in a straight line.

Option (2) anti-parallel to B — Incorrect. For \( \theta = 180^\circ \) the force is again zero, giving a straight line.

Option (3) perpendicular to B — Correct. For \( \theta = 90^\circ \) the force \( qvB \) has constant size and is always at right angles to \( \vec{v} \), so it acts as a centripetal force and the path is a circle of radius \( r = mv/(qB) \).

Option (4) at any angle to B — Incorrect. At a general angle the component along \( \vec{B} \) is unchanged while the perpendicular component goes round in a circle, so the path is a helix, not a circle.

NCERT (2026-27):
Motion in a Magnetic Field – “… The perpendicular force, q v × B, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if v and B are perpendicular to each other (Fig. 4.5). 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 (Fig. 4.6).”