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 of the motion, the proton will move in a circular path of radius

Options

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

Explanation

The kinetic energy of the proton,
`E=(1)/(2)mv^(2)=(1)/(2)m(q^(2)B^(2)r^(2))/(m^(2))`
`=(q^(2)B^(2)r^(2))/(2m)` [`:'` Velocity of the proton, `v=(qBr)/(m)`]
`:.r^(2)=(2mE)/(q^(2)B^(2))or,r=(sqrt(2Em))/(qB)`