# Ratio of magnetic moment to angular momentum

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Suppose that an electron is a small spherical shell of mass m with a charge e spread over its surface. Show that the ratio of the magnetic moment to the angular momentum of such an electron would be e/2m, whether the electron is (a) moving in a circular orbit, or (b) spinning about a diameter. (experimental values for the above ratio are e/2m and e/m, respectively).

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(Please see the attached file for the figure)

Fig. shows an electron with charge e and mass m going round in an orbit of radius R with angular frequency ω.

A charge e passes by any point on the orbit per revolution.

Number of revolutions per sec. = Frequency f = ω/2Π (As ω = 2Π/T = 2Πf)

Total charge passing by any point on the orbit per second = e x ω/2Π

Charge flowing per sec is current. Hence, an electron moving in an orbit of radius R is equivalent to a current of magnitude eω/2Π flowing through a ring of radius R.

Area of the orbit = ΠR2

By definition, magnetic moment of a ring enclosing an area A and carrying current I is given by: M = IA

Hence, magnetic moment of the revolving electron = M = (eω/2Π)(ΠR2) = eωR2/2 ...(1)

Angular momentum of a mass going around a circular orbit of radius R with a speed v ...

#### Solution Summary

This solution explains how to solve the given problem concerning angular momentum.