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. `(2Em)/(qB)`
B. `(sqrt(2Em))/(qB)`
C. `(sqrtEm)/(2qB)`
D. `sqrt((2E)/(mB))`

Answer

B

Explanation

As the proton moves in a circular path of radius r in the perpendicular magnetic field B,
`:. (mv^(2))/(r )= qvB or r= (mv)/(qB)`
`:.` Kinetic energy of the proton, `E= (1)/(2)mv^(2)`
or `2Em = m^(2) v^(2) or mv= sqrt(2mE)`
`:. r= (sqrt(2mE))/(qB)`