Question

A particle of mass m with charge q moving with a uniform speed v normal to a uniform magnetic field B , describe a circular path of radius r. Derive expression for the (i) time period of revolution and (ii) kinetic energy of the particle

Answer 1

dear student,
Please find below the solution to the asked query:



the magnetic force on a moving charge q with velocity v m/s perpendicular to its motion is qvb

now qvb is the force that provides the required centripetal force,
hence qvb = mv​2/r

or r = mv/qb

now, the time period is 

T =2π​r/v

or T= [2π​(mv/qb)]/v = 2πm/qb

Kinetic energy of the particle will be mv2/2, as motion of object under magnetic field does not change the magnitude of the velocity.




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