Question

2) A charge is accelerated from rest through a potential difference V and then enters a uniform magnetic field oriented perpendicular to its path. The field deflects the particle into a circular arc of radius R. If the accelerating potential is tripled to 3V, what will be the radius of the circular arc?

Answer 1

when charge is accelerated through a potential difference V

let the speed be v

using Work energy theorm

q * V = 0.5 * m * v^2

v = sqrt(2 * q * V /m)

the radius of circular path , R = m * v /(q * B)

R = m * (sqrt(2 * q * V /m)) /(q * B)

the radius of path is directly proportional to the square root of potential difference

when V' = 3 V

the new radius , R' = sqrt(3) * R

so the correct option is C) sqrt(3) R