A proton of mass m and charge q is moving in a plane with kinetic energy E. if there exists a uniform magnetic field B, perpendicular to the plane motion. The proton will move in a circular path of radius

Options

A. `sqrt(2Em)/(qB)`
B. `sqrt(Em)/(2qB)`
C. `sqrt(Em)/(2qB)`
D. `sqrt(2Eq)/(mB)`

Answer

A

Explanation

Given kinetic energy =E mass =m magnetic fielf =B charge =q
We know that , `F=qvB sin theta`
(motion of a charged particle in a uniform magnetic field )
if `theta =90^(@)`
Then `F=qVB`
we also know that (centripetal force )
`F=(mv^(2))/(r )`

From eqs i and ii we get `qvB=(mv^(2))/(r ),r =(mv)/(qB)`
`rarr` The radius of circular path `r=(msqrt(2E)/(m))/(qB)=sqrt(2EM)/(qB)`
`[thereforeE=1/2 mv^(2),v=sqrt(2E)/(m)]`