Q. An electron is shot in steady electric and magnetic fields such that its velocity $v,$ electric field $E$ and magnetic field $B$ are mutually perpendicular. The magnitude of $B\ $ are mutually perpendicular. The magnitude of $E$ is 1 $Vcm^{- 1}$ and that of $B\ $ is 2 T. Now if it so happens that the Lorentz (magnetic) force cancels the electrostatic force on the electron, then the velocity of the electron is
A) $50\ ms^{- 1}$
B) $2\ c{ms}^{- 1}$
C) $0.5\ c{ms}^{- 1}$
D) $200\ c{ms}^{- 1}$
Explanation
(a) $\ eE = evB\ \Rightarrow v = \frac{E}{B}\ $
Here, $E = 1\ \ Vcm^{- 1} = 100\ \ Vm^{- 1},\ B = 2\ $ T
$\therefore\ \ \ \ v = \frac{100}{2} = 50\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.