An electron gun G emits electrons of energy 2 keV travelling 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 ${60}^0$ with the  x-axis as shown in figure. A uniform magnetic field $\overrightarrow{B}$ 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,  $K=\frac{1}{2}m{v}^2=2keV$
  Speed of electron,  $v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\times 2\times 1.6\times {10}^{-16}}{9.1\times {10}^{-31}}}m{s}^{-1}$   $=2.65\times {10}^{7}m{s}^{-1}$
Since the velocity $\left(\overrightarrow{v}\right)$ of the electron makes an angle of  $\theta ={60}^0$ with the magnetic field  $\overrightarrow{B}$, 
the path will be a helix.
So, the particle will hit ‘S’ if  $GS=np$, Here, $n=1,2,3,\mathrm{......}$  and $p=$ pitch of helix $=\frac{2\pi m}{qB}v\cos \theta$
 $\Rightarrow GS=\frac{n2\pi mv\cos \theta }{qB}$
 $\Rightarrow B=\frac{n2\pi mv\cos \theta }{q\left(GS\right)}$
But for ‘B’ to be minimum,  $n=1$
 $B_{\min }=\frac{2\pi mv\cos \theta }{q\left(GS\right)}$
Substituting the values, we have 
$B_{\min }=\frac{2\pi \left(9.1\times {10}^{-31}\right)\left(2.65\times {10}^7\right)\left(\frac{1}{2}\right)}{\left(1.6\times {10}^{-19}\right)\left(0.1\right)}=4.73\times {10}^{-3}T$ .