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A proton and a deuteron having the same kinetic energies enter a region of uniform magnetic field perpendicularly, Deuteron's mass is twice that of proton. Calculate the ratio of the radii of their circular paths. |
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Answer» Solution :1. The trajectory of a charged particle ENTERING perpendicular to magnetic field is circular. Cetripetal force required for circular motion of particle is provided by magnetic force QVB. `THEREFORE(mv^(2))/r=qvB` `therefore(mv)/r=qB=p/r` 2. For a motion of proton, `(m_(p)v_(p))/r_(p)=Bq=p_(p)/r_(p)""...(1)` For a motion of DEUTERON `(m_(d)v_(d))/r_(d)=B_(q)=p_(d)/r_(d)""...(2)` From equation (1) and (2), `thereforep_(p)/r_(p)=p_(d)/r_(d)` `thereforer_(d)/r_(p)=p_(d)/p_(p)""...(3)` but, `E=p^(2)/(2m)""thereforep=SQRT(2mE)` `thereforep_(p)=sqrt(2m_(p)E_(p))andp_(d)=sqrt(2mE_(d))` From equation (3), `thereforer_(d)/r_(p)=sqrt((2m_(d)E_(d))/(2m_(p)E_(p)))` But kinetic energy of both are equal so `E_(d)=E_(p)andm_(d)=2m_(p)` `thereforer_(d)/r_(p)=sqrt((2m_(p))/m_(p))=sqrt2` |
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