Acceleration of Charged Particle Moving in Magnetic Field

The magnetic force acting on a particle of charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by Newton's second law and the Lorentz force equation: $\vec{F} = m\vec{a} = q(\vec{v} \times \vec{B})$

Since the magnetic force is the result of a cross product involving $\vec{B}$, the force vector $\vec{F}$ (and consequently the acceleration vector $\vec{a}$) is always perpendicular to the magnetic field vector $\vec{B}$.

Therefore, the scalar product (dot product) of the acceleration $\vec{a}$ and the magnetic field $\vec{B}$ must be zero: $\vec{a} \cdot \vec{B} = 0$

Given: $\vec{B} = 3\hat{i} + 2\hat{j} \text{ T}$ $\vec{a} = 4\hat{i} - \frac{x}{2}\hat{j} \text{ m/s}^2$

Substituting the vectors into the dot product equation: $\left(4\hat{i} - \frac{x}{2}\hat{j}\right) \cdot \left(3\hat{i} + 2\hat{j}\right) = 0$

Evaluating the dot product: $(4)(3) + \left(-\frac{x}{2}\right)(2) = 0$

$12 - x = 0$

$x = 12$