# Magnetization and torque in an ellipsoidal disk

1. Oct 27, 2011

### buggykong

If I have an ellipsoidal disk, where the demagnetization constants $$Nd_{xx}>>Nd_{yy}>>Nd_{zz}$$. The disk lies on the y-z plane and easy axes are +z or -z.

The energy of the system (just considering demag fields) will be

$$U=Nd_{xx}sin^2\theta cos^2\phi + Nd_{yy}sin^2\theta sin^2\phi+Nd_{zz}cos^2\theta$$

The expression can be simplified to

$$U=sin^2\theta (Acos^2\phi + B)+C$$
where
$A=Nd_{xx}-Nd_{yy}, B=Nd_{yy}-Nd_{zz}, C=Nd_{zz}$

The energy landscape can be seen in the attached figure. If magnetization is at $\theta=\pi/2~and~\phi<\pi/2$, magnetization can go either to $\theta=0 ~or ~\pi$.

The effective field at $\theta=\pi/2$,
$$H_{eff}=-\nabla U= -\frac{\partial U}{\partial \theta}\hat{\theta}-\frac{1}{sin\theta}\frac{\partial U}{\partial \phi}\hat{\phi}[/itex] [tex]=-2sin\theta cos\theta (Acos^2\phi +B)(cos\theta cos\phi \hat{x} + cos\theta sin\phi\hat{y}-sin\theta\hat{z})+2Asin\theta sin\phi cos\phi (-sin\phi\hat{x}+cos\phi\hat{y})$$
$$=2Asin\phi cos\phi (-sin\phi\hat{x}+cos\phi\hat{y})$$

Torque=mx$H_{eff}$, where $m=[sin\theta cos\phi, sin\theta sin\phi, cos\theta]=[cos\phi, sin\phi, 0]$.

Torque=$2A sin\phi cos\phi [0,0,1]$. For $\phi<\pi/2, ~\theta ~will ~go ~to~ \pi ~and ~for~ \phi >\pi/2, ~\theta ~will~ go ~to ~0$.

This is opposite what the energy landscape says that for any $\phi$ at $\theta=\pi/2$,$\theta$ can go to 0 or $\pi$.

Can someone tell me what I am missing??

Thanks.

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