A deutron of kinetic energy `50` keV is describing a circular orbit of radius `0.5` meter in a plane perpendicular to magnetic field `vecB`. The kinetic energy of the proton that describes a circular orbit of radius `0.5` meter in the same plane with the same `vecB` is

Options

A. 25keV
B. 50keV
C. 200keV
D. 100k eV

Answer

D

Explanation

For a charged particle orbiting in a circular path in a magnetic field the required centripetal force is provided by the magnetic force,
`(mv^(2))/(r ) = Bvq= v = (B qr)/(m) rArr mv^(2) = Bqvr`
`:.` Kinetic Energy, `E_(K) =(1)/(2) mv^(2) = (1)/(2) Bqvr`
`=Bq (r )/(2).(Bqr)/(m)= (B^(2) q^(2) r^(2))/(2m)`
For deuteron, `E_(1)= (B^(2) q^(2) xx r^(2))/( 2 xx 2m)` and for proton `E_(2)= (B^(2) q^(2)r^(2))/(2m)`
`:. (E_(1))/(E_(2)) = (1)/(2) rArr (50keV)/(E_(2)) = (1)/(2) rArr E_(2) = 100keV`.