A proton of mass 1.67 ×  10−27 kg and charge 1.6 ×  10−19 C is projected with a speed of 2 ×  106 m/s at an angle of 60° to the X-axis. If a uniform magnetic field of 0.104 T is applied along Y-axis, the path of proton is

Path of the proton will be a helix of radius, r= mvsin θqBwhere  θ =  Angle between B→  and  v→⇒  r=1.67  ×10−27 ×  2  ×106 ×  sin  300 1.6  ×10−19 ×  0.104=0.1  mTime  period,  T=2πmqB =2π × 1.67  ×10−27 1.6  ×10−19 ×  0.104 =  2π × 10−7 s