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Analysis of Force Equation

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 or direction of $\to v$ . Then

A

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

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B

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

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C

\to v = \to E \times \to B / 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 B $\to v = \to E \times \to B / B^{2}$
Force acting on a particle under both electric and magnetic field is
$\to F = q \to E + q (\to v \times \to B) = 0$ (since neither direction nor magnitude changes)
Taking cross product with $\to B$ on both sides,
$0 = q \to E \times \to B + q (\to v \times \to B) \times \to B$
$= q \to E \times \to B + q ((\to v . \to B) \to B - \to v B^{2})$
Since $\to E . \to B = 0$ (perpendicular to each other),
$\to v = \to E \times \to B / B^{2}$

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