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(B) The force on a charged particle in a magnetic field is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since $\vec{F} = m\vec{a}$,the acceleration

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$6$

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(B) The force on a charged particle in a magnetic field is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.Since $\vec{F} = m\vec{a}$,the acceleration is $\vec{a} = \frac{q}{m}(\vec{v} 	imes \vec{B})$.This implies that the acceleration vector $\vec{a}$ is always perpendicular to the magnetic field vector $\vec{B}$.For two vectors to be perpendicular,their dot product must be zero,so $\vec{a} \cdot \vec{B} = 0$.Given $\vec{a} = (\alpha \hat{i} - 4 \hat{j})$ and $\vec{B} = (2 \hat{i} + 3 \hat{j})$,we have:$(\alpha \hat{i} - 4 \hat{j}) \cdot (2 \hat{i} + 3 \hat{j}) = 0$$2\alpha - 12 = 0$$2\alpha = 12$$\alpha = 6$.

$A$ charged particle is moving in a uniform magnetic field $(2 \hat{i} + 3 \hat{j}) \text{ T}$. If it has an acceleration of $(\alpha \hat{i} - 4 \hat{j}) \text{ m/s}^2$,then the value of $\alpha$ will be:

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