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If a proton enters perpendicularly a magnetic fieldwith velocity v, then time period of revolution is T.If proton enters with velocity 2v, then time periodwill be(2) 2 T(4) 4 T(3) 3 T |
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Answer» In a circular motion under magnetic field, the centripetal force is provided by the magnetic force. (mv2)/r = qvB r = (mv)/(Bq) T = (2Πr)/v Putting r = (mv)/(Bq) T = (2Πm)/(Bq) Time period is independent of the velocity of charged particle. |
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