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

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 describe a parabolic path . So, a magnetic field must  be present and its direction must be perpendicular to the direction of velocity.

The magnetic force experienced by the particle $\overrightarrow{F}=q\left[\overrightarrow{V}\times \overrightarrow{B}\right].F=BqV\sin \theta$

If the particle move in horizontal direction, the force balances the force of gravity

$BqV\sin \theta =mg$      $\left[\theta ={90}^0\right]$

$BqV=mg$   (or)  $B=\frac{mg}{qv}=\frac{0.5\times {10}^{-3}\times 9.8}{2.5\times {10}^{-8}\times 6\times {10}^4}=3.27T.$