Question

An electron gun $G$ emits electrons of energy $2keV$ traveling in the positive x-direction. The electrons are required to hit the spot of S where $GS=0.1m$, and the line $GS$ makes an angle of ${60}^o$ with the x-axis as shown in the 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 $S$.( give ans in ____$\times {10}^{-3}$)

Answer 1

The path is helix
$V\sin 60°$gives circular path to the electron and $V\cos 60°$ moves the electron in the direction of $GS$
To reach point S, the minimum value of B is such that it completes one revolution and at the same time reaches point S
So time period for one revolution
$T=\frac{2\pi m}{qB}$ independent of velocity
Time required to cover distance
$GS=T$
$GS=V\cos 60\times T$
$GS=V\cos 60\times \frac{2\pi m}{qB}$
We know that
$KE=\frac{1}{2}m{v}^2$
$\Rightarrow V=\sqrt{\frac{2\left(Ke\right)}{m}}$
$\Rightarrow V=\sqrt{\frac{2\times 2\times {10}^3\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}(1eV=1.6\times {10}^{-19}J)$
$=2\sqrt{\frac{16}{91}\times {10}^{15}}$
$=2\times {10}^7\times 4\sqrt{\frac{10}{91}}$
$=\frac{8\times {10}^7\times \sqrt{10}}{9.539}=8.38\times {10}^6\times \sqrt{10}$
$\Rightarrow GS=8.38\times {10}^6\times \frac{1}{2}\times \frac{2\pi \times 9.1\times {10}^{-31}}{1.6\times {10}^{-19}\times B}\times \sqrt{10}$
$GS=\frac{149.844}{B}\times {10}^{-6}\times \sqrt{10}$
$\to B=\frac{1.4934\times {10}^{-4}}{{10}^{-1}}\times \sqrt{10}$
$B=4.738\times {10}^{-3}T$