Bd√qdmV
qV√Bd2m
Bd√q2mV
Bd√q2mV
The proton will move in a circular path as the force due to magnetic field will be always perpendicular to the velocity of proton. The radius of the circle is given by : R = mv/qB.. (i) Where m, v and q are mass, velocity of proton and charge of proton, The proton is accelerated with potential difference of V, it's KE is given as K.E. = qV 1/2mv² = qV ⇒ v = √(2qV/m).. (ii) substituting value of v in equation (i) from equation (ii) R = m√(2qV/m)/qB = √(2mV/qB).. (iii) from geometry Rsinα = d ⇒ sinα = d/R Substituting value of R from equation (iii) sinα = d/√(2mV/qB) = Bd√(q/2mV).