Magnetic field due to currentHard
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 ⇒ ⇒
Create a free account to view solution
View Solution FreeMore Magnetic field due to current Questions
A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, it′s direction being parallel ...Infinite number of straight wires each carrying current I are equally placed as shown in the figure. Adjacent wires have...A particle of charge -16 × 10-18 C moving with velocity 10 ms-1 along the x-axis enters region where a magnetic fie...Charge q is uniformly spread on a thin ring of radius R. the ring rotates about its axis with a uniform frequency f Hz. ...A ring of mass m and radius R is set into pure rolling on horizontal rough surface, in a uniform magnetic field of stren...