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.

Options

A. `(mV)/(qB)`
B. `sqrt((2 mV)/(qB))`
C. `sqrt((2mV)/(qB^(2))`
D. `sqrt((mV)/(qB^(2)))`

Answer

c

Explanation

Work done by electric field appears as kinetic energy of charge, hence we may write the following equation: `qV = 1/2 mv^(2) rArr m^(2) v^(2) = 2 mq V rArr mv = sqrt(2mqV)` Radius of the circle is given by `R = (mv)/(qB) = sqrt(2mqV)/(qB) = sqrt((2 mV)/(qB^(2)))`
Hence, option (c ) is correct.