A charge particle is moving in a uniform magnetic field (2 ^ i + 3...

The magnetic force on a moving charged particle is given by $\overset{\to}{F} = q\left(\overset{\to}{v} \times \overset{\to}{B}\right)$. Therefore, we know that the magnetic force is perpendicular to the magnetic field. Therefore, the resulting acceleration will also be perpendicular to the magnetic field. As the scalar product of two mutually perpendicular vectors is zero,

⇒ \overset{\to}{a} \cdot \overset{\to}{B} = 0 ⇒ \left(\alpha\overset{^}{i} - 4\overset{^}{j}\right) \cdot \left(2\overset{^}{i} + 3\overset{^}{j}\right) = 0

$⇒ 2\alpha - 12 = 0 ⇒ \alpha = 6$