The charged particle moving in a uniform magnetic field of \((3\hat{i} + 2\hat{j})\) T has an acceleration \(\left(4\hat{i} - \dfrac{x}{2}\hat{j}\right)\) m/s\(^2\). The value of \(x\) is __________.

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:

\(\left(4\hat{i} - \dfrac{x}{2}\hat{j}\right) \cdot (3\hat{i} + 2\hat{j}) = 0\)

\((4)(3) + \left(-\dfrac{x}{2}\right)(2) = 0\)

\(12 - x = 0\)

\(x = 12\)

Answer: \(12\)