A particle having a charge of 5.0 μC and a mass of 5.0 × 10^-12 kg is projected with a speed of 1.0 km/s in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin^-1 (0.90) . Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.

Sol. q = 5 μF = 5 × 10^–6 C, m = 5 × 10^–12 kg, V = 1 km/s = 10^3 m/’ ⇒ = Sin^–1 (0.9), B = 5 × 10^–3 T We have mv’2 = qv’B r = mv’/qB = mv sin θ/qB = 5 * 10^-12 *10^3 * 9/5 * 10^-6 + 5 * 10^3 + 10 = 0.18 metre Hence dimeter = 36 cm., Pitch = 2πr/v sin θ vcos θ = 2 * 3.1416 * 0.1 * √1 – 0.51/0.9 = 0.54 metre = 54 mc. The velocity has a x-component along with which no force acts so that the particle, moves with uniform velocity. The velocity has a y-component with which is accelerates with acceleration a. with the Vertical component it moves in a circular crosssection. Thus it moves in a helix.