A positively charged particle moving along +X axis, enters a region of uniform magnetic field directed vertically upward in +Z direction. The particle will ____
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → move in a circular path with a uniform speed
The magnetic force on a moving charge is \( \vec{F} = q\,\vec{v}\times\vec{B} \), which is always perpendicular to both \( \vec{v} \) and \( \vec{B} \).
\( \displaystyle \vec{F} = q v B\,(\hat{\imath}\times\hat{k}) = -q v B\,\hat{\jmath} \)
The force lies in the horizontal X-Y plane and is perpendicular to the velocity, so it acts as a centripetal force and the particle moves in a circle in that plane. A force perpendicular to the velocity does no work, so the kinetic energy and the speed stay constant.
Option (1) move in a circular path with an increased speed — Incorrect. The magnetic force does no work, so the speed cannot increase.
Option (2) move in a circular path with a uniform speed — Correct. The force is perpendicular to the velocity: circular motion at constant speed.
Option (3) move in a circular path with a decreased speed — Incorrect. The magnetic force does no work, so the speed cannot decrease.
Option (4) get deflected in a vertically upward direction — Incorrect. The force is along \( -\hat{\jmath} \), perpendicular to \( \vec{B} \); it has no component along the Z direction.
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 – “… 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.”