An electron beam passes through a region of magnetic field of $2×10^{-3} T$ and an electric field of $3.2×10^4 V m^{-1}$, both acting perpendicular to each other. If the path of the electron remains undeviated calculate the speed of electron: [Mass of electron = $9.1×10^{-31} kg$]

$1.6×10^7 m/s$

The correct answer is Option (4) → $1.6×10^7 m/s$

Electric force on electron, $F_E=eE$

Magnetic force on electron, $F_M=evB$

and,

$F_E=F_M$ [Given]

$⇒V=\frac{E}{B}=\frac{3.2×10^4V/m}{2×10^{-3}T}$

$⇒V=1.6×10^7m/s$