A proton of mass 1.67 x 10-27 kg and charge 1.67 x 10-19 Cs projected with a speed of 2×106 ms-1 at an angle of 600 to the X-axis. If a uniform magnetic field of 0.10 T is applied along Y-axis, the path of proton is

Since the proton is entering the magnetic field at some angle other than 90o, its path is helix.Corresponding velocity of the proton along X-axis,vx=vcos60°=2×106×12=106 ms-1Due to velocity component vx, the radius of the helix is described and is given by the relationr=mvxqB=1.6×10-27×1061.6×10-19×0.10=0.1 m Now, T=2πrvx=2π×0.1106=2π×10-7 s