A charged particle is moving in a uniform magnetic field at right ang…

The correct answer is Option (1) → Its momentum changes but kinetic energy remains the same

The magnetic force \( \vec{F} = q(\vec{v}\times\vec{B}) \) is always perpendicular to the velocity. It therefore does no work on the particle, \( W = \vec{F}\cdot\vec{d} = 0 \), and by the work-energy theorem the kinetic energy stays constant, which means the speed is constant.

The force does change the direction of the velocity (the particle moves in a circle). Momentum \( \vec{p} = m\vec{v} \) is a vector, so it changes even though its magnitude does not.

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).”