A particle leaves a velocity selector with a velocity of 106

A particle leaves a velocity selector with a velocity of 10^6m/s in the +x direction. The

particle then enters a magnetic field directed of 0.2T directed in the +z direction. The

particle travels in a semi-circle of radius 2cm before hitting a detector. If the path of the

particle was counterclockwise, what is the charge to mass ratio (q/m) of the particle?

Solution

The centripetal force on the particle is given by F= mv²/r , where \'r\' is the radius of the circular path. Sice the magnetic field is in a direction perpendicular to the charged particle, this centripetal force is provided by magnetic Lorenz force F_B= Bqv, where B is the intensity of mag field. and v is the velocity of charged particle.

hence q/m= v/Br = 2.5x10⁹ C/kg