Match List-I with List-II
| List-I | List-II |
| Motion of a charge particle | Path |
| (A) Initial velocity of charge particle parallel to B | (I) Circular path |
| (B) Initial velocity of charge particle perpendicular to B | (II) Helical path |
| (C) Initial velocity of charge particle inclined to B | (III) Straight path |
| (D) Initial velocity of charge particle perpendicular to E | (IV) Parabolic path |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
A magnetic field exerts the force \( \vec{F} = q\,\vec{v}\times\vec{B} \), which acts only on the component of velocity perpendicular to \( B \). An electric field exerts \( \vec{F} = q\vec{E} \) in a fixed direction.
(A) Initial velocity parallel to B — this matches (III) Straight path. The angle between \( \vec{v} \) and \( \vec{B} \) is zero, so the force is zero and the particle keeps moving in a straight line.
(B) Initial velocity perpendicular to B — this matches (I) Circular path. The force is always perpendicular to the velocity and has constant size, so it acts as a centripetal force.
(C) Initial velocity inclined to B — this matches (II) Helical path. The component along \( B \) is unchanged (straight motion) and the component perpendicular to \( B \) goes round in a circle, so the path is a helix.
(D) Initial velocity perpendicular to E — this matches (IV) Parabolic path. The electric force is constant and along \( E \), at right angles to the initial velocity, as for a projectile under gravity, so the path is a parabola.