A particle having a mass of 0.5 g carries a charge of $2.5\times {10}^{-8} \mathrm{C}$. The particle is given an initial horizontal velocity of $6\times {10}^4 {\mathrm{ms}}^{-1}$. To keep the particle moving in a horizontal direction

detailed solution

Correct option is A

In the absence of a magnetic field, the particle will experience gravitational force mg. As a result, the particle will not continue moving in the horizontal direction but will describe a parabolic path. So, a magnetic field must be present and its direction must be perpendicular to the direction of the velocity. The magnetic force experienced by the particle is given by F→=q(v→×B→).The magnitude of the force is F=qvBsin⁡θ. If the particle is to move in the horizontal direction, this force must balance the force of gravity, i.e., mg=qvBsin⁡θThe minimum value of B corresponds to sin⁡θ=1 or θ=90∘. Thus, mg=qvBor B=mgqv=0.5×10−3×9.82.5×10−8×6×104=3.27THence, the correct options are (1) and (3).