JEE Main 2026 Apr 5 Shift 1, Physics Q24: Motion of a Charged Particle in a Magnetic Field

The magnetic force on a moving charge is $\vec{F}=q(\vec{v}\times \vec{B})$, which is always perpendicular to $\vec{B}$. So the resulting acceleration must satisfy $\vec{a}\cdot \vec{B}=0$.

$\left(4\hat{i}-\frac{x}{2}\hat{j}\right)\cdot (3\hat{i}+2\hat{j})=0\Rightarrow 12-x=0\Rightarrow x=12$.