Question

An electron in a cathode-ray beam passes between $2.5-\mathrm{cm}-$long parallel-plate electrodes that are $5.0\mathrm{mm}$apart. A $1.0\mathrm{mT}$, 2.5-cm-wide magnetic field is perpendicular to the electric field between the plates. If the potential difference between the plates is $150\mathrm{V}$, the electron passes through the electrodes without being deflected. If the potential difference across the plates is set to zero, through what angle is the electron deflected as it passes through the magnetic field?

Short answer

We found the required $\theta =0.147\mathrm{rad}=8.43°$

Step-by-step solution

Given Information

We can start solution with equilibrlum between magnetic and electric force
$\begin{array}{rcl}{F}_E& =& F_B\\ qE& =& qvB\text{(expressions for forces)}\\ v& =& \frac{E}{B}\left(\text{express}v\right)\\ E& =& \frac{\Delta V}{d}\text{(homogenous electric field between parallel plates)}\\ v& =& \frac{\Delta V}{Bd}\left(\text{substitute}E\right)\end{array}$

When potential difference between plates is set to zero, magnetic field deflects electron into circular path which follows from equilibrium between magnetic and centripetal force:
$\begin{array}{rcl}{F}_{cp}& =& F_B\\ \frac{m{v}^2}{r}& =& evB\text{(expression for forces)}\\ r& =& \frac{mv}{eB}(expressr)\\ r& =& \frac{m\Delta V}{B^{2}ed}(substitutev)\end{array}$

Calculation

From the geometry of the problem we can see that,

$\begin{array}{rcl}\sin \theta & =& \frac{l}{r}\\ \theta & =& \arcsin \frac{l}{r}\\ \theta & =& \arcsin \frac{l{B}^{2}ed}{m\Delta V}(substituter)\\ \theta & =& \arcsin \frac{0.025·0.{001}^2·e·0.005}{9.11·{10}^{-31}·150}\\ \theta & =& 0.147\mathrm{rad}=8.43°\end{array}$