Arrange the following steps in the chronological order, when a charge…

The correct answer is Option (1) → (D), (A), (C), (B)

A charge q moving with velocity \( \vec{v} \) in a magnetic field feels the magnetic Lorentz force \( \vec{F} = q\,\vec{v}\times\vec{B} \). The four statements follow one from another, from this cause to its consequences.

(D) A force acts perpendicular to both the velocity of the charged particle and the magnetic field. This comes first: as soon as the particle enters the field the force acts, with magnitude \( qvB\sin 90^\circ = qvB \).

(A) The charged particle starts moving in a circular path. A force of constant size that is always perpendicular to the velocity acts as a centripetal force, so the path bends into a circle.

    \( \displaystyle qvB = \frac{mv^{2}}{r} \Rightarrow r = \frac{mv}{qB} \)

(C) The speed of the charged particle remains constant but its direction changes. The force has no component along the velocity, so it changes only the direction of motion, not the speed.

(B) Net work done by the field is zero. With constant speed the kinetic energy does not change, so by the work-energy theorem the field does no work; equivalently, the force is always perpendicular to the displacement.

    \( \displaystyle W = \Delta K = 0 \)

So the order is (D), (A), (C), (B).

NCERT (2026-27):
Motion in a Magnetic Field – “… In the case of motion of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed).”

Motion in a Magnetic Field – “… The perpendicular force, \( q\mathbf{v} \times \mathbf{B} \), acts as a centripetal force and produces a circular motion perpendicular to the magnetic field.”

Motion in a Magnetic Field – “… If the velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force.”