# 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.

Navjyot Kalra
10 years ago
Hello Student,
(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