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A charge particle is moving in a uniform magnetic field $(2 \hat{i}+3 \hat{j}) \mathrm{T}$. If it has an acceleration of $(\alpha \hat{i}-4 \hat{j}) \mathrm{m} / \mathrm{s}^2$, then the value of $\alpha$ will be :
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The correct answer is:
6
Given that uniform magnetic field, $\vec{B}=(2 \hat{i}+3 \hat{j}) T$
Acceleration $\vec{a}=(\alpha \hat{i}-4 \hat{j}) \mathrm{m} / \mathrm{s}^2$ We know that
$F=q(\vec{v} \times \vec{B}) \Rightarrow m \vec{a}=q(\vec{v} \times \vec{B})$
Here, $\vec{a} \perp \vec{B}$, so, $\vec{a} \cdot \vec{B}=0$
$(\alpha \hat{i}-4 \hat{j})(2 \hat{i}+3 \hat{j})=0 \Rightarrow 2 \alpha-12=0 \Rightarrow \alpha=6$
Acceleration $\vec{a}=(\alpha \hat{i}-4 \hat{j}) \mathrm{m} / \mathrm{s}^2$ We know that
$F=q(\vec{v} \times \vec{B}) \Rightarrow m \vec{a}=q(\vec{v} \times \vec{B})$
Here, $\vec{a} \perp \vec{B}$, so, $\vec{a} \cdot \vec{B}=0$
$(\alpha \hat{i}-4 \hat{j})(2 \hat{i}+3 \hat{j})=0 \Rightarrow 2 \alpha-12=0 \Rightarrow \alpha=6$
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