A charged particle enters into a uniform magnetic field with velocity `v_(0)` perpendicular to it , the length of magnetic field is `x=sqrt(3)/(2)R`, where `R` is the radius of the circular path of the particle in the field .The magnitude of change in velocity of the particle when it comes out of the field is

A charged particle enters a uniform magnetic field with velocity v_(0) = 4 m//s perpendicular to it, the length of magnetic field is x = ((sqrt(3))/(2)) R , where R is the radius of the circular path of the particle in the field. Find the magnitude of charge in velocity (in m/s) of the particle when it comes out of the field.

Discuss the motion of a charged particle in a uniform magnetic field with initial velocity perpendicular to the magnetic field.

If a charged particle enters perpendicular in the uniform magnetic field then

A charged particle enters a uniform magnetic field perpendicular to it. The magnetic field

A charged particle enters a uniform magnetic field perpendicular to it. The magnetic field

A charged particle entering a magnetic field from outside in a direction perpendicular to the field

The path of a charged particle entering parallel to uniform magnetic field will be