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A charged particle with charge $\mathrm{q}$ enters a region of constant, uniform and mutually orthogonal fields $\mathrm{\vec{E}}$ and $\mathrm{\vec{B}}$ with a velocity $\mathrm{\vec{v}}$ perpendicular to both $\mathrm{\vec{E}}$ and $\mathrm{\vec{B}}$, and comes out without any change in magnitude or direction of $\mathrm{\vec{v}}$. Then
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The correct answer is:
$\mathrm{\vec{v}=\vec{E} \times \vec{B} / B^2}$
$\mathrm{\vec{v}=\vec{E} \times \vec{B} / B^2}$
$\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{E}}$
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