Questions
A proton of energy 8eV is moving in a circular path in a uniform magnetic field. The energy of an particle moving in the same magnetic field and along the same path will be-
detailed solution
Correct option is C
r=mvqB=pqB=2mkqB p22m=k; p=2mkIt r and B are samek∝qm kpkα=q2q4mm=1 kp=kα.Talk to our academic expert!
Similar Questions
in the arrangement shown, readings of ideal voltmeters V1, V2 and V3 are 20 volt, 50 volt and 60 volt respectively. then current flowing through the current carrying branch is
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