The charged particle moving in a uniform magn | vedclass

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Question

(B) The magnetic force acting on a charged particle is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.According to Newton's second law, $\vec{F} = m\v

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The magnetic force acting on a charged particle is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.According to Newton's second law, $\vec{F} = m\vec{a}$, so $m\vec{a} = q(\vec{v} 	imes \vec{B})$.Since the magnetic force $\vec{F}$ is always perpendicular to the magnetic field $\vec{B}$, the acceleration $\vec{a}$ must also be perpendicular to the magnetic field $\vec{B}$.Therefore, the dot product of acceleration and magnetic field must be zero: $\vec{a} \cdot \vec{B} = 0$.Given $\vec{B} = (3\hat{i} + 2\hat{j}) \ 	ext{T}$ and $\vec{a} = (4\hat{i} - \frac{x}{2}\hat{j}) \ 	ext{m/s}^2$, we calculate the dot product:$(4)(3) + (-\frac{x}{2})(2) = 0$$12 - x = 0$$x = 12$.

The charged particle moving in a uniform magnetic field of $(3\hat{i} + 2\hat{j}) \ \text{T}$ has an acceleration $(4\hat{i} - \frac{x}{2}\hat{j}) \ \text{m/s}^2$. The value of $x$ is . . . . . . .

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