An accelration produces a narrow beam of protons, each having an initial speed of `v_(0)`. The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of `(R//2)` from the axis, as indicated in the diagram.

The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as `m_(P)` and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue.
One the potential of the sphere has reached its final, constant value, the minimum speed v of a proton along its trajectory path is given by

Options

A. `v_(0)`
B. `v_(0)//2`
C. `2v_(0)`
D. None of these

Answer

B

Explanation

From the angular momentum conservation
`mv_(0)xxR/2=mvxxRrArr v=V_(0)/2`