A charged particle with charge q enters a region of constant uniform and mutually orthogonal fields vec-E- and vec-B- with a velocity vec-v- perpendicular to both vec-E- and vec-B- and comes out without any change in magnitude and direction of vec-v- Then-vec-v-vec-E-times vec-B-B-2-vec-v-vec-B-times vec-E-E-2-vec-v-vec-E-times vec-E-B-2-vec-v-vec-E-times vec-E-E-2-

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Toppr Admin · Accepted Answer

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.
$\Rightarrow q (\to V \times \to B) = - q \to E$
$\Rightarrow \to V \times \to B = - \to E$
$\Rightarrow \to V = \to E \times \to B / B^{2}$

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)=−q→E⇒→V×→B=−→E⇒→V=→E×→B/B2

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.
$\Rightarrow q (\to V \times \to B) = - q \to E$
$\Rightarrow \to V \times \to B = - \to E$
$\Rightarrow \to V = \to E \times \to B / B^{2}$