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, $q(v \times B)=q E$
$ \Rightarrow v \times B=E$
Velocity of a charged particle
$\therefore v=\frac{E \times B}{B^{2}}, v =\frac{ E \times B }{B^{2}}$