A charged particle moves perpendicular to a uniform external magnetic…

The correct answer is Option (4) → A circular path

The magnetic force \( q\,\vec{v}\times\vec{B} \) is always perpendicular to the velocity, so it changes only the direction of motion, never the speed. When \( \vec{v} \) is perpendicular to \( \vec{B} \), this force of constant magnitude \( qvB \) acts as a centripetal force in the plane normal to \( \vec{B} \).

Option (1) Concentric circles — Incorrect. The speed and the field stay the same, so the radius \( r=mv/qB \) stays fixed. The particle keeps going round one circle and does not move from circle to circle.

Option (2) A helical path — Incorrect. A helix needs a velocity component along \( \vec{B} \), which carries the particle along the field while it circles. Here the velocity has no such component.

Option (3) An arbitrary trajectory — Incorrect. In a uniform field the force has a fixed magnitude and always points towards one centre, so the path is fully determined.

Option (4) A circular path — Correct. The force \( qvB \) is constant in magnitude and always perpendicular to the velocity. So the particle moves in a circle of radius \( r=mv/qB \) in a plane perpendicular to the field.

NCERT (2026-27):
Motion in a Magnetic Field – “… First consider the case of v perpendicular to B. The perpendicular force, q v × B, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field.”