Circular motion of electrons- Angular Momentum and magnetic dipole moment:
If an electron (charge ‘e’, mass ‘m’) is in uniform circular motion with angular velocity ‘ω’ the orbital angular momentum of the electron about a perpendicular axis passing through the centre of the circle is mωr^2 (=mvr). Often you will encounter questions regarding the direction of the angular momentum vector. You should note that the orbital angular momentum of the electron is directed perpendicular to the plane of the orbit.
The magnetic dipole moment of an orbiting electron is eωr^2/2 (= evr/2). [Magnetic dipole moment = IA = (e/T)×πr^2 = (eω/2π)×πr^2 = eωr^2/2=evr/2. Here I is the equivalent current, T is the orbital period and r is the orbital radius].
Now consider the following M.C.Q. which appeared in Kerala Engineering Entrance Test paper of 2006:
Two electrons (each of charge = e and mass = m) are attached one at each end of a light rigid rod of length 2r. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the angular momentum about the axis of rotation to the magnetic dipole moment of the system is
(a) 2me/3 (b) e^2/2m (c) 2×specific charge of electron (d) 5m/2e (e) 2m/e
The ‘light rigid rod’ in the problem could have been dispensed with.
Angular momentum of two electrons = 2mωr^2. As shown above, the magnetic dipole moment of two orbiting electrons = 2 eωr^2/2. The ratio of angular momentum to the magnetic dipole moment = 2m/e. [Option(e)].
Note that this result is independent of the number of electrons.
You should also note that the ratio of the magnetic moment to the angular momentum of a charged particle in angular motion is called gyromagnetic ratio.
If an electron (charge ‘e’, mass ‘m’) is in uniform circular motion with angular velocity ‘ω’ the orbital angular momentum of the electron about a perpendicular axis passing through the centre of the circle is mωr^2 (=mvr). Often you will encounter questions regarding the direction of the angular momentum vector. You should note that the orbital angular momentum of the electron is directed perpendicular to the plane of the orbit.
The magnetic dipole moment of an orbiting electron is eωr^2/2 (= evr/2). [Magnetic dipole moment = IA = (e/T)×πr^2 = (eω/2π)×πr^2 = eωr^2/2=evr/2. Here I is the equivalent current, T is the orbital period and r is the orbital radius].
Now consider the following M.C.Q. which appeared in Kerala Engineering Entrance Test paper of 2006:
Two electrons (each of charge = e and mass = m) are attached one at each end of a light rigid rod of length 2r. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the angular momentum about the axis of rotation to the magnetic dipole moment of the system is
(a) 2me/3 (b) e^2/2m (c) 2×specific charge of electron (d) 5m/2e (e) 2m/e
The ‘light rigid rod’ in the problem could have been dispensed with.
Angular momentum of two electrons = 2mωr^2. As shown above, the magnetic dipole moment of two orbiting electrons = 2 eωr^2/2. The ratio of angular momentum to the magnetic dipole moment = 2m/e. [Option(e)].
Note that this result is independent of the number of electrons.
You should also note that the ratio of the magnetic moment to the angular momentum of a charged particle in angular motion is called gyromagnetic ratio.
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