An electron moving with uniform velocity along the x-direction enters a region of uniform magnetic field along the y-direction. Which of the following quantities is (are) non-zero and remain(s) constant?
(I) Velocity of the electron
(II) Magnitude of the momentum of the electron
(III) Force on the electron
(IV) Kinetic energy of the electron
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Only (II) and (IV)
The magnetic force \( q\,\vec{v}\times\vec{B} \) on a moving charge is always perpendicular to its velocity. So it does no work. The speed stays the same, but the direction of motion keeps changing. Here \( \vec{v} \) (along x) is perpendicular to \( \vec{B} \) (along y), so the electron moves in a circle in the x-z plane.
(I) Velocity of the electron — False. Its magnitude stays the same, but its direction keeps turning along the circle, so the velocity does not remain constant.
(II) Magnitude of the momentum of the electron — True. The speed \( v \) does not change, so \( p=mv \) is non-zero and constant.
(III) Force on the electron — False. Its magnitude \( evB \) is constant, but it always points towards the centre of the circle, so its direction changes all the time.
(IV) Kinetic energy of the electron — True. The magnetic force does no work on the electron, so its kinetic energy \( mv^{2}/2 \) stays at its non-zero initial value.
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.”