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Q.

A proton of energy E moving in a normal magnetic field in a circular path of radius R. The energy of  an α−particle moving in the same magnetic field with same radius is

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a

E

b

E/2

c

2 E

d

E/4

answer is A.

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Detailed Solution

ω=QBm, E=12m(ωR)2=Q2B2R22m∴ EαEP=(QαQP)2.(mPmα)=(2)r.(14)=1⇒ Eα=EP=E
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