A charged particle enters a uniform magnetic field such that its velocity is perpendicular to the direction of the field. The charge will move
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → in a circular path perpendicular to the direction of magnetic field
The magnetic force \( q\vec{v}\times\vec{B} \) is always perpendicular to the velocity, so it changes only the direction of motion and not the speed. When \( \vec{v} \perp \vec{B} \), this constant-magnitude sideways force acts as a centripetal force.
\( \displaystyle qvB = \frac{mv^{2}}{r} \Rightarrow r = \frac{mv}{qB} \)
The force also lies in the plane perpendicular to \( \vec{B} \), so the particle moves on a circle in a plane perpendicular to the field.
Why the other options are incorrect
Option (1) in a circular path parallel to the direction of magnetic field — Incorrect. The circle cannot lie parallel to the field, because the force has no component along \( \vec{B} \).
Option (3) in a helical path — Incorrect. A helical path needs a velocity component along \( \vec{B} \), which is zero here.
Option (4) in a straight line parallel to the direction of magnetic field — Incorrect. The particle moves in a straight line only when its velocity is parallel to \( \vec{B} \), because then the force is zero.
NCERT (2026-27):
Motion in a Magnetic Field – “… First consider the case of \( \mathbf{v} \) perpendicular to \( \mathbf{B} \). The perpendicular force, \( q\,\mathbf{v}\times\mathbf{B} \), acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if \( \mathbf{v} \) and \( \mathbf{B} \) are perpendicular to each other (Fig. 4.5).”
