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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twice the original radius

2 times the original radius

half the original radius

(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

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