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=q→E+q(→v×→B)=0(since neither direction nor magnitude changes)