An electron beam passes through a region of magnetic field of \( 2\times10^{-3}\,\mathrm{T} \) and an electric field of \( 3.2\times10^{4}\,\mathrm{V\,m^{-1}} \), both acting perpendicular to each other. If the path of the electron remains undeviated calculate the speed of electron: [Mass of electron = \( 9.1\times10^{-31}\,\mathrm{kg} \)]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.6\times10^{7}\,\mathrm{m/s} \)
The beam goes straight only if the electric force and the magnetic force on an electron are equal and opposite, as in a velocity selector.
\( eE = evB \Rightarrow v = \displaystyle\frac{E}{B} \)
\( \displaystyle v = \frac{3.2\times10^{4}\,\mathrm{V\,m^{-1}}}{2\times10^{-3}\,\mathrm{T}} = 1.6\times10^{7}\,\mathrm{m\,s^{-1}} \)