A particle of charge -q and mass m enters a uniform magnetic field `vecB` (perpendicular to paper inward) at P with a velocity `v_0` at an angle `alpha` and leaves the field at Q with velocity v at angle `beta` as shown in fig. A. `alpha`=`beta`B. `v=v_0`C. `PQ=(2mv_0sin alpha)/(Bq)`D. The particle remains in field for time `t=(2m(pi-alpha))/(Bq)`

A particle of charge -q and mass m enters a uniform magnetic field `ve

1.	A particle of charge -q and mass m enters a uniform magnetic field `vecB` (perpendicular to paper inward) at P with a velocity `v_0` at an angle `alpha` and leaves the field at Q with velocity v at angle `beta` as shown in fig. A. `alpha`=`beta`B. `v=v_0`C. `PQ=(2mv_0sin alpha)/(Bq)`D. The particle remains in field for time `t=(2m(pi-alpha))/(Bq)`
Answer» Correct Answer - (a,b,c,d) Option (b) is obvious `r=(mv_0)/(qB)` `PQ=2r sin alpha=2(mv_0)/(qB) sin alpha` `alpha=beta` Time taken `=T=(2pir)/v` `T=(2pim)/(qB)` For t time, `t=T/(2pi)(2pi-2alpha)=(2m)/(qB)(pi-alpha)`

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