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Desciption of Force

A charged par...

Question

A charged particle with charge $q$ enters a region of constant uniform and mutually orthogonal fields $\to E$ and $\to B$ with a velocity $\to v$ perpendicular to both $\to E$ and $\to B$ and comes out without any change in magnitude and direction of $\to v$ . Then,

A

\to v = \to E \times \to B / B^{2}

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B

\to v = \to E \times \to E / B^{2}

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C

\to v = \to E \times \to E / E^{2}

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D

\to v = \to B \times \to E / E^{2}

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Solution

The correct option is A $\to v = \to E \times \to B / B^{2}$
Since there is no change in magnitude and direction of particle velocity. So, Force due to magnetic field is equal to force due to electric field.
$\Rightarrow q (\to V \times \to B) = - q \to E$
$\Rightarrow \to V \times \to B = - \to E$
$\Rightarrow \to V = \to E \times \to B / B^{2}$

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