a deutron of kinetic energy 50KeV is describing a circular orbit of radius 0.5m ina plane perpendicular to magnetic field B vector. the kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same Bvector is

Kinetic energy of a charge particle moving in a magnetic field is given by..K=mv^2/r..&v=rqb/m..So,K=q^2r^2B^2/2m....we have K(d)=50keV...by taking the ratio of K(p) to K(d),we get.....K(p)/K(d)=[q(p)]^2 r^2B^2/2m(p)×2m(d)/q(d)^2

Kinetic energy of a charge particle moving in a magnetic field is given by K=q^2r^2B^2/2m...K(p)/K(d)=[q(p)]^2r^2B^2/2m(p)×2m(d)/[q(d)]^2r^2B^2.......q(d)=q(p)...m(d)=2m(p)...By putting value,K(p)=2K(d)...K(p)=100keV