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JEE MAIN_2026
05-04-26_S1
Question
The charged particle moving in a uniform magnetic field of $(3 \hat{i}+2 \hat{j}) \mathrm{T}$ has an acceleration $\left(4 \hat{i}-\frac{x}{2} \hat{j}\right) \mathrm{m} / \mathrm{s}^2$. The value of x is
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Solution
$\begin{aligned} & \vec{F}_B \propto(\vec{V} \times \vec{B}) \Rightarrow \vec{F}_B \perp \vec{B} \\ & \vec{a} \perp \vec{B} \rightarrow \vec{a} \cdot \vec{B}=0 \\ & 12-x=0 \Rightarrow x=12\end{aligned}$
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