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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

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detailed solution

Correct option is D

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


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