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.
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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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CENGAGE PHYSICS-MAGNETIC FIELD AND MAGNETIC FORCES-Multiple Correct Answer type
