charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec E$ and $\vec B$ with a velocity $\vec v$ perpendicular to both $\vec E$ and $\vec B$ , and comes out without any change in magnitude or direction of $\vec v$ . Then
AIEEE 2007, Medium
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Here, $\vec{E}$ and $\vec{B}$ are perpendicular to each other and the velocity $\vec{v}$ does not change; therefore
$q E=q v B \Rightarrow v=\frac{E}{B}$
Also,
$\left|\frac{\vec{E} \times \vec{B}}{B^{2}}\right|=\frac{E B \sin \theta}{B^{2}}=\frac{E B \sin 90^{\circ}}{B^{2}}=\frac{E}{B}=|\vec{v}|=v$
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