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A charged particle of mass m and charge q is accelerated through a potential difference of V volts. It enters a region of uniform magnetic field which is directed perpendicular to the direction of motion of the particle. Find the radius of circular path moved by the particle in magnetic field.

Answer» Since the particle is accelerated through V volts, therefore
it kinetic energy will be equal to qV.
or `1/2mv^2=qV`. Therefore, `v=sqrt((2qV)/m)`.
Radius of circular path is given by `R=(mv)/(qB)=sqrt((2mV)/(qB^2))`.


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