Explore BrainMass
Share

Explore BrainMass

    Ratio of magnetic moment to angular momentum

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    See the attached file.
    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).

    © BrainMass Inc. brainmass.com October 10, 2019, 4:38 am ad1c9bdddf
    https://brainmass.com/physics/angular-momentum/ratio-magnetic-moment-angular-momentum-470373

    Attachments

    Solution Preview

    Please see the attached file for the complete solution explanation.

    (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.

    $2.19