A charge particle is moving in a uniform magnetic field $$(2\hat{i} + 3\hat{j}) \text{ T}$$. If it has an acceleration of $$(\alpha\hat{i} - 4\hat{j}) \text{ m s}^{-2}$$, then the value of $$\alpha$$ will be

A charged particle moves in a uniform magnetic field $$\vec{B} = (2\hat{i} + 3\hat{j}) \text{ T}$$. Its acceleration is $$\vec{a} = (\alpha\hat{i} - 4\hat{j}) \text{ m/s}^2$$. We need to find $$\alpha$$.

The magnetic force on a charged particle is given by:

$$\vec{F} = q(\vec{v} \times \vec{B})$$

Since $$\vec{F} = m\vec{a}$$, the acceleration $$\vec{a}$$ is proportional to $$\vec{v} \times \vec{B}$$.

The magnetic force is always perpendicular to the magnetic field. Therefore, the acceleration must also be perpendicular to $$\vec{B}$$:

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

$$(\alpha\hat{i} - 4\hat{j}) \cdot (2\hat{i} + 3\hat{j}) = 0$$

$$2\alpha + (-4)(3) = 0$$

$$2\alpha - 12 = 0$$

$$\alpha = 6$$

The correct answer is Option B: $$6$$.