JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

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

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Solution

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

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

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