Question

An electron gun $\text{G}$ emits electrons of energy $2\text{keV}$ travelling in the positive $x$ - direction. The electrons are required to hit the spot of $\text{S}$ where $\text{GS}=0.1\text{m}$, and the line $\text{GS}$ makes an angle of ${60}^{\circ }$ with the $x$ - axis as shown in figure. If a uniform magnetic field $\overrightarrow{B}$ parallel to $\text{GS}$ exists in the region outside the electron gun , then minimum value of $B$ needed to make the electron hit $\text{S}$ is $k\text{mT}$ Find the value of k (Give the nearest integer value)

Answer 1

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

$\therefore$ Speed of electron, $v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\times 2\times 1.6\times {10}^{-16}}{9.1\times {10}^{-31}}}{\text{ms}}^{-1}$

$=2.65\times {10}^7{\text{ms}}^{-1}$

Since the velocity $\left(\overrightarrow{v}\right)$ of the electron makes an angle of $\theta ={60}^{\circ }$ with the magnetic field $\overrightarrow{B}$. the path will be a helix.

So, the particle will hit $\text{S}$ if $\text{GS}=np$, Here, $n=1,2,\mathrm{3....}$

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

$\Rightarrow \text{GS}=\frac{n2\pi mv\cos \theta }{qB}$

$\Rightarrow B=\frac{n2\pi mv\cos \theta }{q\left(GS\right)}$

But for $B$ to be minimum we take, $n=1$

$B_{min}=\frac{2\pi mv\cos \theta }{q\left(GS\right)}$

Subtituting the values, we have

$B_{min}=\frac{(2\times \pi )(9.1\times {10}^{-31})(2.65\times {10}^7)\left(\frac{1}{2}\right)}{(1.6\times {10}^{-19})\left(0.1\right)}$

or $B_{min}=4.73\times {10}^{-3}\text{T}\approx 5\text{mT}$

Accepted answer : $5$