A charged particle moving in a uniform magnetic field penetrates a layer of lead and loses one half of its kinetic energy. The radius of curvature changes to
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a
twice the original radius
b
2 times the original radius
c
half the original radius
d
(1 /2) times the original radius
answer is D.
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Detailed Solution
The radius of curvature of a charged particle moving in uniform magnetic field is given byr=mvqB=2mKqBr′=2mK′qB∴ r′r=K′K=K/2K=12r′=12⋅r