Q. 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) $\mathbf{v\ = \ E\ } \times \mathbf{\ B/}B^{2}$
B) $\mathbf{v\ = \ B\ } \times \mathbf{\ E/}B^{2}$
C) $\mathbf{v\ = \ E\ } \times \mathbf{\ B/}E^{2}$
D) $\mathbf{v\ = \ B} \times \mathbf{\ E/}E^{2}$
Explanation
(a) As v of charged particle is remaining constant, it means force acting on charged particle is zero.
So, $q(\mathbf{v} \times \mathbf{B})\ = q\mathbf{E}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{v} \times \mathbf{B} = \mathbf{E}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{v}\ = \ \frac{\mathbf{E\ } \times \mathbf{\ B}}{B^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.