An electron gun $G$ emits electrons of energy $2 keV$ traveling in the positive $x$-direction. The electrons are required to hit the spot $S$ where $GS=0.1 m$, and the line $GS$ makes an angle of $60°$ with the $x$-axis as shown in the figure. A uniform magnetic field $B$ parallel to $GS$ exists in the region outside the electron gun.

Find the minimum value of $B$ needed to make the electron hit $S$.

Kinetic energy of electron, $K=\frac{1}{2}m{v}^2=2 keV$

$\therefore$Speed of electron, $v=\sqrt{\frac{2K}{m}}$

                                   $$\begin{gathered} v=\sqrt{\frac{2\times 2\times 1.6\times {10}^{-16}}{9.1\times {10}^{-31}}}m/s \\ =2.65\times {10}^7 m/s \end{gathered}$$

Since, the velocity $\left(v\right)$ of the electron makes an angle of $\theta =60°$ with the magnetic field $B$. The path will be a helix. So, the particle will hit $S$ if,

$GS=np$

Here, $n=1,2,3,.....$

$p=$pitch of helix$=\frac{2\mathrm{\pi m}}{qB}v\cos \theta$

But for $B$ to be minimum, $n=1$

Hence,

 $$\begin{gathered} GS=p=\frac{2\mathrm{\pi m}}{qB}v\cos \theta \\ B=B_{min}=\frac{2\mathrm{\pi mvcos\theta }}{q\left(GS\right)} \end{gathered}$$

Substituting the values we get, 

$$\begin{gathered} B_{min}=\frac{\left(2\mathrm{\pi }\right)\left(9.1\times {10}^{-31}\right)\left(2.65\times {10}^7\right)\left({\displaystyle \frac{1}{2}}\right)}{\left(1.6\times {10}^{-19}\right)\left(0.1\right)} \\ {B}_{min}=4.73\times {10}^{-3 }T \end{gathered}$$