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

Question

A deutron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 m, in a plane perpendicular to magnetic field B. The kinetic energy of a proton that describes circular orbit of radius 0.5 m in the same plane with the same magnetic field is (A) 200 keV (B) 50 keV (C) 100 keV (D) 25 keV

Tardigrade Admin · Accepted Answer

In this case magnetic force provides necessary
centripetal force

i e, q v B = \frac{m v ^{2}}{r}

Radius of path

r = \frac{m v}{Bq} = \frac{2 m E}{qB}

r = \frac{2 m E}{Bq} = \frac{2 m _{1} E _{1}}{Bq}

or

E_{1} = \frac{m E}{m _{1}}

= \frac{( 2 m _{1} )}{m _{1}} \times 50 k e V [∵ m = 2 m_{1}]

= 100 k e V

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

200 keV

50 keV

100 keV

25 keV

In this case magnetic force provides necessary centripetal force ie,qvB=rmv2​ Radius of path r=Bqmv​=qB2mE​​ r=Bq2mE​​=Bq2m1​E1​​​ or E1​=m1​mE​ =m1​(2m1​)​×50keV[∵m=2m1​] =100keV

Q. A deutron of kinetic energy $50 k e V$ is describing a circular orbit of radius $0.5 m$ , in a plane perpendicular to magnetic field $B$ . The kinetic energy of a proton that describes circular orbit of radius $0.5 m$ in the same plane with the same magnetic field is