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I need some help with the (b) part of this problem concerning the so called http://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect" [Broken].

(a) On the search of the origin of ferromagnetism Einstein and de Haas came up with an experiment in 1915 to prove that circling electrons around nuclei are the cause. A rotating electron with a particular angular momentum L is a circular current and as such has a magnetic moment [tex]\mu[/tex]. Calculate the magnetic moment as a function of the angular momentum L, the charge q and the mass of the electron.

(b) When a piece of iron is magnetized it should according to (a) have an angular momentum [tex]L=\sum L_i \propto \sum \mu_i [/tex] If the magnetization is reversed so the angular momentum must reverse, too. Because of the conservation of angular momentum such a piece of iron should start to rotate.

A "soft iron" cylinder (Permeability [tex]\mu[/tex]=7000, Radius R=0.5cm, Density [tex]\rho = 7.9 grams/cm^3[/tex]) is magnetized so that the magnetic field on the flat side of the cylinder is B = 1 Tesla. What angular velocity should be expected if the magnetic field is reversed?

From (a) I know that the magnetic moment [tex]\mu = \frac{-e}{2m} \ L[/tex]

* Magnetic moment generally: [tex]\mu = I A [/tex] (I = electric current, A = area)

* Angular momentum: [tex]\vec L = r \times p = I \omega[/tex] (I = moment of inertia)

* Torque: [tex]\tau = \frac{dL}{dt} = r \times F = I \dot \omega[/tex]

* Torque (in terms of magnetism): [tex]\tau = \mu \times B[/tex]

* Moment of inertia of a cylinder: [tex]I = \frac{1}{2}M R^2[/tex]

* Moment of inertia circle: [tex]I = M R^2[/tex]

I tried finding the atoms density in the material: [tex]N = \frac{N_a \rho}{m_a} = 8.487 \times 10^{22} \frac{Atoms}{cm^3}[/tex]

Then I substituted all that I have in [tex]\tau = \mu B[/tex] and I got:

[tex]I \dot \omega = N \frac{-e}{2m} L_i B[/tex]

[tex]\omega = \int_{0}^{t} N \frac{-e}{2m I} L_i B dt [/tex]

...and by here I figured this couldn`t be right. I tried a lot of slightly different versions always getting [tex]\omega[/tex] = some integral and never being able to cancel enough terms in order to get a numerical answer with my given conditions.

Any ideas would be greatly appreciated.

P.S. I cannot use spins etc because we still haven`t been into quantum mechanics. The answer should be really simple and not too realistic because this is still 2nd semester (Experimental physics 2).

## Homework Statement

(a) On the search of the origin of ferromagnetism Einstein and de Haas came up with an experiment in 1915 to prove that circling electrons around nuclei are the cause. A rotating electron with a particular angular momentum L is a circular current and as such has a magnetic moment [tex]\mu[/tex]. Calculate the magnetic moment as a function of the angular momentum L, the charge q and the mass of the electron.

(b) When a piece of iron is magnetized it should according to (a) have an angular momentum [tex]L=\sum L_i \propto \sum \mu_i [/tex] If the magnetization is reversed so the angular momentum must reverse, too. Because of the conservation of angular momentum such a piece of iron should start to rotate.

A "soft iron" cylinder (Permeability [tex]\mu[/tex]=7000, Radius R=0.5cm, Density [tex]\rho = 7.9 grams/cm^3[/tex]) is magnetized so that the magnetic field on the flat side of the cylinder is B = 1 Tesla. What angular velocity should be expected if the magnetic field is reversed?

## Homework Equations

From (a) I know that the magnetic moment [tex]\mu = \frac{-e}{2m} \ L[/tex]

* Magnetic moment generally: [tex]\mu = I A [/tex] (I = electric current, A = area)

* Angular momentum: [tex]\vec L = r \times p = I \omega[/tex] (I = moment of inertia)

* Torque: [tex]\tau = \frac{dL}{dt} = r \times F = I \dot \omega[/tex]

* Torque (in terms of magnetism): [tex]\tau = \mu \times B[/tex]

* Moment of inertia of a cylinder: [tex]I = \frac{1}{2}M R^2[/tex]

* Moment of inertia circle: [tex]I = M R^2[/tex]

## The Attempt at a Solution

I tried finding the atoms density in the material: [tex]N = \frac{N_a \rho}{m_a} = 8.487 \times 10^{22} \frac{Atoms}{cm^3}[/tex]

Then I substituted all that I have in [tex]\tau = \mu B[/tex] and I got:

[tex]I \dot \omega = N \frac{-e}{2m} L_i B[/tex]

[tex]\omega = \int_{0}^{t} N \frac{-e}{2m I} L_i B dt [/tex]

...and by here I figured this couldn`t be right. I tried a lot of slightly different versions always getting [tex]\omega[/tex] = some integral and never being able to cancel enough terms in order to get a numerical answer with my given conditions.

Any ideas would be greatly appreciated.

P.S. I cannot use spins etc because we still haven`t been into quantum mechanics. The answer should be really simple and not too realistic because this is still 2nd semester (Experimental physics 2).

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