A charge particle is moving in a uniform magn | vedclass

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Question

As $\overrightarrow{ F }= q (\overrightarrow{ v } 	imes \overrightarrow{ B })$

$\overrightarrow{ a }=\frac{ q }{ m }(\overrightarrow{ v } 	imes \

Vedclass · Accepted Answer

$6$

Vedclass · Answer

As $\overrightarrow{ F }= q (\overrightarrow{ v } 	imes \overrightarrow{ B })$

$\overrightarrow{ a }=\frac{ q }{ m }(\overrightarrow{ v } 	imes \overrightarrow{ B })$

So, $\overrightarrow{ a }\,and\,\overrightarrow{ B }$ are $\perp$ to each other

Hence, $\vec{a} \cdot \vec{B}=0$

$(\alpha \hat{i}-4 \hat{j}) \cdot(2 \hat{i}+3 \hat{j})=0$

$\alpha(2)+(-4)(3)=0$

$\alpha=\frac{12}{2} \Rightarrow \alpha=6$

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.

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