A particle of mass m and charge q is accelerated by a potential difference V volt and made to enter a magnetic field region at an angle `theta` with the field. At the same moment , another particle of same mass and charge is projected in the direction of the field from the same point. Magnetic field of induction is B.
What would be the speed of second particle so that both particles meet again and again after a regular interval of time, which should be minimum?

Options

A. `sqrt((Vm)/q) (2pi)/B cos theta`
B. `sqrt((2Vm)/(3q) (2pi)/B cos theta`
C. `sqrt((2Vm)/q) (2pi)/B cos theta`
D. `2/3 sqrt((Vm)/q) pi/m cos theta`

Answer

c

Explanation

The first parallel will have a helical path and the second particle will move rectilinearly along the field. For the two particles to meet again and again,
`V_(||)T=v'T`
where v' is the speed of the second
particle.
`:. v'=v_(||)=vcos theta`
`1/2mv^2=qV implies v=sqrt((2qV)/m)`
`:. v'=sqrt((2qV)/m) cos theta`
`T=(2pim)/(qB)`
Distance travelled =pitch
Distance `=sqrt((2qV)/m) cos thetaxx(2pim)/(qB)`
Distance `=sqrt((2Vm)/q) (2pi)/B cos theta`
.