Question

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a) Write the expression for the force acting on an electron carrying a charge e moving with velocity v when it enters perpendicularly into (i) an electric field of intensity E b) Sketch two separate diagrams of the direction of field, velocity and path of ) With reference to your diagrams, state two differences between the electric (ii) a magnetic field of intensity B the electron and the force acting on it for the motion in part (a) (i) and (ii). field and magnetic field. A helium ion of charge +2e and mass 6.6 x1027 kg is accelerated by a voltage of 2400 V. The accelerated ion then moves into a perpendicular uniform magnetic field of magnitude 0.24 T. Determine (i) the radius of circular path of the helium ion. (ii) the period of revolution of the helium ion. [Given charge of electron 1.60 x 10-19 C]

Answer 1

(a) i) F^vector = e E^vector ii) F^vector = e V^vector times B^vector (b) (c) Force is parallel to E^vector and perpendicular to B^vector Path is in the plane of E^vector and is in perpendicular plane to B^vector (d) i) mu^2/R = qu/B implies R = muB/q 1/2 mu^2 = qv u = squareroot 2qv/m R = 6.6 times 10^-27 times 4 times 0.24/2 times 1.6 times 10^-19 = 6.6 times 10^-27 times 0.24/2 times 1.6 times 10^-19 times squareroot 2 times 2 times 1.6 times 10^-19 times 2400/6.6 times 10^-27 = 2.39 times 10^-3 m R = 2.39 times 10^-3 m ii) T = 2 Pi R/u = 2 Pi R/squareroot 2qv/m = 2 Pi R times squareroot m/2qv = 2Pi times 2.39 times 10^-3 times squareroot 6.6 times 10^-27/2 times 2 times 1.6 times 10^-19 times 2400 = 3.11 times 10^-2 S T = 3.11 times 10^-8 S