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 = 0.1 m and the line GS makes an angle of ${60}^{\circ }$ with the x-axis. A uniform magnetic field ‘B’ parallel to GS exists in the region outside the gun. The minimum value of B needed to make the electron hit ‘S’ is

• A. $4.74\times {10}^{-3}T$
• B. $4.41\times {10}^{-3}T$
• C. $3\times {10}^{-3}T$
• D. $2.21\times {10}^{-3}T$

Answer 1

The correct option is A $4.74\times {10}^{-3}T$

Let ‘u’ be the speed of electron.
$\Rightarrow \frac{1}{2}m{u}^2=2000eV.$
or $u=\sqrt{\frac{4000e}{m}}$
The electrons move in a helix as they come out.
To reach S, pitch of helix, GS = l
$\Rightarrow V_{11}T=l$
or u cos ${60}^{\circ }\left(\frac{2\pi m}{eB}\right)=lorB=\frac{u\pi m}{el}=\frac{\pi }{l}\sqrt{\frac{4000\left(m\right)}{e}}=4378\times {10}^{-3}T$