A charge moves in a circular path perpendicular to a magnetic field. The time period of revolution is independent of

Step-by-step solution

When a charged particle moves in a circular path perpendicular to a magnetic field, the magnetic force provides the necessary centripetal force. qvB = mv^2 r The radius of the circular path is given by r = mv qB The time period of revolution is T = 2 r v Substituting the value of r , we get T = 2 m qB The time period depends on the mass of the particle, its charge, and the magnetic field. It is independent of the velocity of the particle. Answer: velocity of the particle