A charge particle is moving in a uniform magnetic field $(2 \hat{i}+3 \hat{j}) T$. If it has an acceleration of $(\alpha \hat{i}-4 \hat{j}) m / s ^{2}$, then the value of $\alpha$ will be.
JEE MAIN 2022, Diffcult
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As $\overrightarrow{ F }= q (\overrightarrow{ v } \times \overrightarrow{ B })$
$\overrightarrow{ a }=\frac{ q }{ m }(\overrightarrow{ v } \times \overrightarrow{ B })$
So, $\overrightarrow{ a }\,and\,\overrightarrow{ B }$ are $\perp$ to each other
Hence, $\vec{a} \cdot \vec{B}=0$
$(\alpha \hat{i}-4 \hat{j}) \cdot(2 \hat{i}+3 \hat{j})=0$
$\alpha(2)+(-4)(3)=0$
$\alpha=\frac{12}{2} \Rightarrow \alpha=6$
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