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Q.

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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a

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

b

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

c

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

d

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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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