Choose the correct answer from given options
A charge particle after being accelerated through a potential difference 'V' enters in a uniform magnetic field and moves in a circle of radius r. If V is doubled, the radius of the circle will become
The relation between the accelerating potential and the accelerating voltage is given as: `r = sqrt(2mqV)/(qB)`
As the potential is doubled the radius of curvature becomes `sqrt(2)` times.
Hence, the correct answer is option `sqrt(2r)`.