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Question

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

A
v=E×B/B2
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B
v=B×E/B2
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C
v=E×B/E2
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D
v=B×E/E2
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Solution

The correct option is B v=E×B/B2
Force acting on a particle under both electric and magnetic field is
F=qE+q(v×B)=0(since neither direction nor magnitude changes)
Taking cross product with B on both sides,
0=qE×B+q(v×B)×B
=qE×B+q((v.B)BvB2)
Since E.B=0(perpendicular to each other),
v=E×B/B2

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