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Question:

A proton (mass m) accelerated by a potential difference V files through a uniform transverse magnetic field B. The field occupies a region of space by width 'd'. If 'α' be the angle of deviation of proton from initial direction of motion (see figure), the value of sinα will be:

Bd√qdmV

qV√Bd2m

Bd√q2mV

Bd√q2mV

Solution:

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).