The region between y =0 and y=d constains a magnetic field `vecB = Bvecz`. A particle of mass m and charge q enters of the region with a velocity `vecv = v veci`. If `d =(mv)/(2qB)` the acceleration of the charged particle at the point of its emergence at the other side is :

A charged particle of mass m and charge q enters a magnetic field B with a velocity v at an angle theta with the direction of B. The radius of the resulting path is

A charged particle of mass m and charge q enters a magnetic field B with a velocity v at an angle theta with the direction of B . The radius fo the resulting path is

A particle of mass 'm' and carrying a charge q enters with a velocity 'v' perpendicular to a uniform magnetic field. The time period of rotation of the particle

A particle of mass m and charge Q moving with a velocity v enters a region on uniform field of induction B Then its path in the region is s

A particle of mass m and charge Q moving with a velocity v enters a region on uniform field of induction B Then its path in the region is s

A particle with charge q is moving with a velocity vecv in a magnetic field vecB . If the force acting on the particle is vecF then,

A particle of mass m and charge q enters a magnetic field B perpendicularly with a velocity v , The radius of the circular path described by it will be

A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of deviation of the particle as it comes out of the magnetic field if ihe width d of the region is (2mv)/(qB)

A non-uniform magnetic field vecB=B_(0)(1+y/d)(-hatk) is present in region of space in between y=0 & y=d .The lines are shown in the diagram.A particle of mass m and positive charge q is moving.Given an initial velocity vecv=v_(0)hati .Find the components of velocity of the particle when it leaves the field.