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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 B 4.74×103T
Let ‘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=πl4000(m)e=4378×103T

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