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. $v=E\times B/B^2$
• B. $v=B\times E/E^2$
• C. $v=E\times B/E^2$
• D. $v=B\times E/B^2$

Answer 1

Answer: (A)

When a charged particle $q$ enters a region with velocity as $v$ of charged particle is remaining constant, it means force acting on charged particle is zero.
So, $q(v\times B)=qE$
$\Rightarrow v\times B=E$
Velocity of a charged particle
$\therefore v=\frac{E\times B}{B^2},v=\frac{E\times B}{B^2}$