Question

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

Answer 1

hope it is helpful for you

Answer 2

Answer = 100 keV

Explanation:

When Charged particle move on circular path the force F on the charged particle due to magnetic field provides the required centripetal force ( = mv^2 / r ) necessary for motion along the circular path

So, mv^2 / r = qvB

Where,

m = mass of particle

v = velocity on particle

q = charge on particle

r = radius of circular path

mv^2 = Bqvr

Kinetic energy

Ek = 1/2 mv^2 = 1/2 Bqvr

= Bq × r /2 × ( Bqr / m ) = B^2 q^2 r^2 / 2m

• FOR DEUTRON

E1 = (B^2 q^2 r^2) / 2 × ( 2m )

... [ mass = 2m ]

• FOR PROTON

E2 = B^2q^2r^2 / 2m ...[ mass = m ]

E1 / E2 = 1/2

=> 50 keV / E2 = 1/2 or

=> E2 = 100 keV !