a- J. J. Thompson calculated the ratio of electron charge e to its mass m via his relationship within the familiar framework and in terms of the familiar quantities. Thompson measured the velocity of the electron by applying a magnetic field vector B to its trajectory; the intensity of this field vector is B = 10^(-2) Tesla. Through this procedure, he could compensate the upward electric force F = eE of the electron by exerting a magnetic force on it in the opposite direction. The electric field intensity E is given by V/d, where V is the electric potential difference between the plates of the capacitor and d is the distance between them. We can take d = 0.03 m and V = 1000 Volt. Calculate v0 under the given circumstances.
b- Hall already in 1879 used the same procedure to measure the drift velocity vd of free electrons carrying an electric current in a cable. Suppose the current goes downward on this sheet, and the magnetic field vector B applied to the cable by Hall is perpendicular to this sheet and directed inward. Its magnitude B is 1.5 Tesla. Explain what happens then. The Hall electric potential difference VH that is accordingly developed between the two sides of the cable is 10x10^(-6) Volt. The thickness of the cable is 10 micrometers. Calculate vd. (Note that 1 micrometer is 10^(-6) m.)
c- The electric current intensity going through the cable is 3.0 A. Find the free electron number per unit volume.