Magnetic field due to currentMedium
Question
A particle of charge -q and mass m enters a uniform magnetic field 
(perpendicular to paper inwards) at P with a velocity v0 at an angle α and leaves the field at Q with velocity v at angle β as shown in fig., then
(perpendicular to paper inwards) at P with a velocity v0 at an angle α and leaves the field at Q with velocity v at angle β as shown in fig., then Options
A.α = β
B.v = v0
C.PQ = 
D.Particle remains in the field for time t = 
Solution
In magnetic field the path is circle and the motion is uniform.
∴ v = v0
PQ = 2R sin α =
sin α ⇒ 
⇒
∴ v = v0
PQ = 2R sin α =
sin α ⇒ ⇒
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