Two charged particles A and B , initially at rest, are accelerated through the same distance d by uniform electric fields E_A and E_B respectively. They then enter a uniform magnetic field perpendicular to their velocities and move in circular paths of identical radius. If the ratio of their masses is m_A : m_B = 1 : 9 and the ratio of their charges is q_A : q_B = 1 : 2 , then the ratio of the electric field strength

Step-by-step solution

From the work-energy theorem, the kinetic energy K gained by a charge q accelerated by a uniform electric field E over a distance d is given by K = qEd . The radius of the circular path described by a charged particle in a uniform magnetic field B is R = 2mK qB . Substituting the expression for kinetic energy into the radius formula, we get: R = 2m(qEd) qB = 2mEd/q B Squaring both sides, we have: R^2 = 2mEd qB^2 Since the radius R , distance d , and magnetic field B are identical for both particles, the quantity mE