Question

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 $GS=0.1m$, and the line GS makes an angle of ${60}^o$ with the x axis as shown in fig. A uniform magnetic field $\overrightarrow{B}$ parallel to GS exists in the region outside the electron gun. If the minimum value of B needed to make the electrons hit S is $B_{min}=4.73\times {10}^{-X}T$. find X?

Answer 1

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 $\theta ={60}^o$ with the magnetic field $\overrightarrow{B}$, the path will be a helix.
So, the particle will hit S if $Gs=np$; Here, $n=1,2,3,....$
$P=$ pitch of helix $=\frac{2\pi m}{qB}vcos\theta$
$\Rightarrow GS=\frac{n2\pi mvcos\theta }{qB}$
$\Rightarrow B=\frac{n2\pi mvcos\theta }{q\left(GS\right)}$

But for B to be minimum, $n=1$
$B_{min}=\frac{2\pi mvcos\theta }{q\left(GS\right)}$

Substituting the values, we have:
$B_{min}=\frac{\left(2\pi \right)(9.1\times {10}^{-31})(2.67\times {10}^7)\left(\frac{1}{2}\right)}{(1.6\times {10}^{-19})\left(0.1\right)}$

$B_{min}=4.73\times {10}^{-3}T$