Question

An electron beam is moving between two parallel plates having electric field $1.125\times {10}^{-6}N/m$. A magnetic field $3\times {10}^{-10}T$ is also applied, so that beam of electrons do not deflect. The velocity of the electron is

• A. $4225m/s$
• B. $3750m/s$
• C. $2750m/s$
• D. $3200m/s$

Answer 1

The correct option is B $3750m/s$

When charged particle acts on by both electric field and magnetic field, then Lorentz force acts on particle.
When the moving charged particle is subjected simultaneously to both electric field $\overrightarrow{E}$ and magnetic field $\overrightarrow{B}$, the moving charged particle will experience electric force $\overrightarrow{F_e}=q\overrightarrow{E}$ and magnetic force $\overrightarrow{F_m}=q(\overrightarrow{V}\times \overrightarrow{B})$, so the net force on it will be
$\overrightarrow{F}=q[\overrightarrow{E}+(\overrightarrow{V}\times \overrightarrow{B}\left)\right]$ ...(i)
which is the famous 'Lorentz force equation'. If $\overrightarrow{V},\overrightarrow{E}$ and $\overrightarrow{B}$ are mutually perpendicular in this situation, if $\overrightarrow{E}$ and $\overrightarrow{B}$ are such that
$\overrightarrow{F}=\overrightarrow{F_e}+\overrightarrow{F_m}=0$
ie, $\overrightarrow{a}=\overrightarrow{F}/m=0$
as shown in Fig. (i), the particle will pass through the field with same velocity
and in this situation, as
$F_e=F_m$
ie, $qE=qvB$
or $v=E/B$
Given, $E=1.125\times {10}^{-6}N/m$,
$B=3\times {10}^{-10}T$
$v=\frac{1.125\times {10}^{-6}}{3\times {10}^{-10}}$
$=3.75\times {10}^3$
$=3750m/s$
Note : This principle is used in 'velocity selector' to get a charged beam having a specific velocity.