Q 12-04-039JEE MainJEE Main 2026 (5 Apr, Shift 1)Medium
The charged particle moving in a uniform magnetic field of $(3\hat{i} + 2\hat{j})$ T has an acceleration $\left(4\hat{i} - \dfrac{x}{2}\hat{j}\right)\ \text{m/s}^2$. The value of $x$ is
Numerical value type. Enter your answer.
Answer: 12
The magnetic force $q\vec{v} \times \vec{B}$ is always perpendicular to $\vec{B}$, so $\vec{a} \cdot \vec{B} = 0$:
$$4 \times 3 - \frac{x}{2} \times 2 = 0 \;\Rightarrow\; x = 12$$
Solution by Sreeraj P, M.Sc Physics