An electron gun G emits electron of energy 2ke V 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° with the x –axis as shown in the fig. A uniform magnetic field parallel to GS exists. Find parallel to GS exists in the region outside the electron gun. Find the minimum value of B needed to make the electrons hit S.

Question

Navjyot Kalra · Answer

Hello Student,

Please find the answer to your question

(a) Let us resolve the velocity into two rectangular components v₁( = vcos 60°) and v₂ ( = vsin 60°). v₁ component of velocity is responsible to move the charge particle in the direction of the magnetic field whereas v₂ component is responsible for revolving the charged particle in circular motion. The overall path is helical. The condition for the charged particle to strike S with minimum value of B is that Pitch of Helix = GS

∴ B = 2 πm / q x 0.1 x √2E / m x cos 60°

= 2π / q x 0.1 x √ 2mE x cos 60° = 2 x 3.14 / 1.6 x 10^-19 x 0.1

= √ 2 x 9.1 x 10^-31 x 2 x 10³ x 1.6 x 10^-19 x ½

= 149.8 /10^-19 x 0.316 x 10^-23 = 47.37 x 10^-4

= 4. 737 x 10^-3 T

Thanks
Navjot Kalra
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