Chapter 32, Problem 043 An electron with kinetic energy Ke travels in a circular

Question

Chapter 32, Problem 043 An electron with kinetic energy Ke travels in a circular path that is perpendicular to a uniform magnetic field, which is in the positive direction of a z axis. The electron's motion is subject only to the force due to the field. The orbital magnetic dipole moment of the electron has a magnitude ?-K/B and its direction is opposite that of the field. Suppose an ionized gas consists of 4.5 x 1020 electrons/m3 and the same number density of ions. Take the average electron kinetic energy to be 6.1 × 10-20 J and the average ion kinetic energy to be 6.2 × 10-21·Calculate the magnetization of the gas when it is in a magnetic field of 1.6T. Number Units the tolerance is +/-496 Click if you would like to Show Work for this question: Open Show Wor

Explanation / Answer

Solution :

The magnitization is given by

M = ( n / B) [ Ke + Ki ]

Ke - average electron kinetic energy

Ki - average ion kinetic energy

n - number of electrons per unit volume

M = [(4.5 x 10^20 ) / 1.6 T ] [6.1 x 10^-20 + 6.2 x 10^-21]

M = 18.9 A /m

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