Question

An electron gun emits electrons of energy $2\text{keV}$ travelling in positive $\text{x-}$direction. The electrons are required to hit point $\text{S}$ where $\text{GS}=0.1\text{m}$, and the line $\text{GS}$ makes an angle of ${60}^{\circ }$ with $\text{x-}$ axis as shown. A uniform magnetic field exists parallel to $\text{GS}$ in the region outside the electron gun. The minimum magnitude of magnetic field needed to make the electrons hit $\text{S}$ is

• A. $47\times {10}^{-3}\text{T}$
• B. $4.7\times {10}^{-3}\text{T}$
• C. $0.84\text{T}$
• D. $0.94\text{T}$

Answer 1

The correct option is B $4.7\times {10}^{-3}\text{T}$

For electron, kinetic energy $\left(K\right)$ can be written as

$K=\frac{1}{2}m{v}^2$

$\Rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\times (2\times {10}^3\times 1.6\times {10}^{-19})}{9.1\times {10}^{-31}}}$

$\therefore v=2.65\times {10}^7\text{m/s}$

As the angle between $\overrightarrow{B}$ and $\overrightarrow{v}$ is ${60}^{\circ }$, hence the charge particle will follow a helical path.

So the particle will hit $\text{S}$ if,

$GS=nP$

Here, $n=$ a positive integer

and, pitch of helix,$P=\frac{2\pi mv\cos \theta }{Bq}$

$\therefore GS=nP=n\times \frac{2\pi mv\cos \theta }{Bq}$

For $B$ to be minimum, $n=1$

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

$B_{min}=\frac{2\pi \times 9.1\times {10}^{-31}\times 2.65\times {10}^7\times \cos {60}^{\circ }}{1.6\times {10}^{-19}\times 0.1}$

$\therefore B_{min}=4.7\times {10}^{-3}\text{T}$

Hence, option $\left(b\right)$ is the correct answer.