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Question

A deutron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 m, in a plane perpendicular to magnetic field B. The kinetic energy of a proton that discribes a circular orbit of radius 0.5 m in the same plane with the same magnetic field B is

A
200 keV
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B
50 keV
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C
100 keV
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D
25 keV
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Solution

The correct option is C 100 keV
rdeutron=2mdEdBq;rproton=2mpEpBq
For same radius, B and q:
mpEp=mdEd
Ep=mdmpEd=21×50=100keV

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