The charged particle moving in a uniform magnetic field of (3 i + 2...

The magnetic force on a moving charged particle is given by $\vec{F} = q(\vec{v} \times \vec{B})$.

Since the acceleration $\vec{a}$ is in the direction of the magnetic force, it must be perpendicular to the magnetic field $\vec{B}$.

Therefore, the dot product of acceleration and magnetic field is zero:

$\vec{a} \cdot \vec{B} = 0$

Substituting the given vectors:

\left(4\hat{i} - \dfrac{x}{2}\hat{j}\right) \cdot (3\hat{i} + 2\hat{j}) = 0$<br><br>$(4)(3) + \left(-\dfrac{x}{2}\right)(2) = 0$<br><br>$12 - x = 0$<br><br>$x = 12$<br><br>Answer:$12