A charged particle q is accelerated by a potential difference of V enters a region of uniform magnetic field of induction B at right angles to the direction of the field. The charged particle completes semi circle of the radius r inside the magnetic field. The mass of the charged particle is (all quantities are in SI units)

Step-by-step solution

The kinetic energy gained by the charged particle when accelerated through a potential difference V is given by K = qV . The momentum of the particle is p = 2mK = 2mqV . When the particle enters a uniform magnetic field B perpendicularly, it moves in a circular path of radius r given by r = p qB . Substituting the value of p , we get r = 2mqV qB . Squaring both sides, r^2 = 2mqV q^2 B^2 = 2mV qB^2 . Rearranging for mass m , we get m = q r^2 B^2 2V .