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 600 with the  x-axis as shown in figure. A uniform magnetic field 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.

Kinetic energy of electron,  K=12mv2=2 keV  Speed of electron,  v=2Km=2×2×1.6×10−169.1×10−31ms−1   =2.65×107 ms−1Since the velocity v→ of the electron makes an angle of  θ=600 with the magnetic field  B→, the path will be a helix.So, the particle will hit ‘S’ if  GS=np, Here, n=1,2,3,......  and p= pitch of helix =2πmqBvcosθ ⇒ GS=n 2π mvcosθqB ⇒ B=n 2π mvcosθq(GS)But for ‘B’ to be minimum,  n=1 Bmin=2πmvcosθq(GS)Substituting the values, we have Bmin=2π9.1×10−312.65×107121.6×10−19(0.1)=4.73×10−3T .