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Question

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

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Solution

Radius of the charge in a perpendicular magnetic field is r=2mKqBKq2m
Ratio of kinetic energy for proton and deutron is
KpKd=(qpqd)2×mdmp=(11)2×21=21
Kp=2×50=100 keV

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