1.

Electron and proton of equal momentum enter a uniform magnetic field normal to the lines of force. If the radii of curvature of circular paths be r_(e ) and r_(p) respectively, then

Answer»

`(r_(e ))/(r_(p))= (1)/(1)`
`(r_(e ))/(r_(p)) = (m_(p))/(m_(e))`
`(r_(e))/(r_(p)) = sqrt(m_(p)//m_(e))`
`(r_(e))/(r_(p)) = sqrt(m_(e )//m_(p))`

Solution :Radius of CIRCULAR path travelled by ELECTRON `r_(e ) = (p_(e))/(Be)`, radius of circular path travelled by proton `r_(p) = (p_(m))/(Be)`
As the momentum of `p_(m) and p_(e )` are equal and MAGNITUDE of charge of electron and proton are equal ALONG with same magnetic field `r_(p) = r_(e) or r_(p)//r_(e)= 1//1`


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