If a charged particle of charge to mass ratio (q/m)=α enters in a magnetic field of strength B at a speed v=(2αd)(B), then

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angle subtended by the path of charged particle in magnetic field at the center of circular path is 2π

the charge will move on a circular path and then will come out from magnetic field at some distance from the point of insertion

the time for which particle will be in the magnetic field is 2παB

angle subtended by the path of charged particle in magnetic field at the center of circular path is π/2

answer is B.

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Detailed Solution

r=mvqB=vBα=(2αd)(B)(Bα)=2di.e., the electron will move out after travelling on a semicircular path of radius r = 2d.

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