When a charged particle moves in a uniform magnetic field, its kineti…

The correct answer is Option (3) → remains unchanged

The magnetic force \( \vec{F} = q(\vec{v} \times \vec{B}) \) is always perpendicular to the velocity, so it does no work on the particle. By the work-energy theorem the speed, and hence the kinetic energy, cannot change; the force only changes the direction of motion.

Option (1) goes on increasing — Incorrect. No work is done by the magnetic force, so the kinetic energy cannot keep increasing.

Option (2) may increase or decrease depending on angle between its velocity and magnetic field — Incorrect. The angle only decides the shape of the path (straight line, circle or helix); the speed stays the same in every case.

Option (3) remains unchanged — Correct. Zero work means the kinetic energy is constant.

Option (4) may increase or decrease depending upon the sign of charge — Incorrect. The sign of the charge only reverses the sense in which the particle turns; the force is still perpendicular to the velocity.

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).”