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 and direction of $v$. Then,

A

$v=E×B/B_{2}$

B

$v=E×E/B_{2}$

C

$v=E×E/E_{2}$

D

$v=B×E/E_{2}$

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Updated on : 2022-09-05

Solution

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Correct option is A)

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. $⇒q(V×B)=−qE$ $⇒V×B=−E$ $⇒V=E×B/B_{2}$

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