A charged particle enters a region of uniform magnetic field perpendicular to its velocity.
(A) Magnetic force acts on charged particle, perpendicular to its velocity.
(B) The particle starts moving in a circular path.
(C) The velocity of the charged particle changes continuously.
(D) The kinetic energy of the particle remains constant.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B), (C) and (D)
The Lorentz force on the particle is
\( \displaystyle \vec{F} = q\left(\vec{v} \times \vec{B}\right) \)
(A) Magnetic force acts on charged particle, perpendicular to its velocity. — Correct. A cross product is perpendicular to both \( \vec{v} \) and \( \vec{B} \).
(B) The particle starts moving in a circular path. — Correct. A force of constant magnitude \( qvB \) always perpendicular to the velocity acts as a centripetal force, giving a circle of radius \( r = mv/qB \).
(C) The velocity of the charged particle changes continuously. — Correct. The speed is constant, but the direction of motion changes at every instant, so the velocity changes.
(D) The kinetic energy of the particle remains constant. — Correct. A force perpendicular to the velocity does no work, so the speed and kinetic energy stay the same.
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). … The perpendicular force, \( q\,\mathbf{v}\times\mathbf{B} \), acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if \( \mathbf{v} \) and \( \mathbf{B} \) are perpendicular to each other (Fig. 4.5).”
