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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∘ 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
4.74×103T
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B
4.41×103T
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C
3×103T
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D
2.21×103T
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Solution

## The correct option is A 4.74×10−3TLet ‘u’ be the speed of electron. ⇒12mu2=2000eV. or u=√4000em The electrons move in a helix as they come out. To reach S, pitch of helix, GS = l ⇒V11T=l or u cos 60∘(2πmeB)=lor B=uπmel=πl√4000(m)e=4378×10−3T

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