Q.

A proton of energy 8eV is moving in a circular path in a uniform magnetic field. The energy of an  α- particle moving in the same magnetic field and along the same path will be-

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a

4eV

b

2eV

c

8eV

d

6eV

answer is C.

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Detailed Solution

r=mvqB=pqB=2mkqB                          p22m=k;    p=2mkIt r and B are samek∝qm                kpkα=q2q4mm=1                       kp=kα.
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