A charged particle of mass m and charge q is accelerated through a potential difference of V volts. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particle. Find the radius of circular path moved by the particle in magnetic field.

Explanation

`qV=1/2mv_1^2 implies v_1 =sqrt((2qV)/m)`
To meet them again and again:
`v_2=v_1cos theta=sqrt((2qV)/m) cos theta`

Time interval : `T=(2pim)/(qB)`
Distance after which they meet=1pitch=`v_2T`
`=sqrt((2qV)/m)cos thetaxx(2pim)/(qB)=pisqrt((8Vm)/(qB^2))cos theta`