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Magnetization and torque in an ellipsoidal disk

  1. Oct 27, 2011 #1
    If I have an ellipsoidal disk, where the demagnetization constants [tex] Nd_{xx}>>Nd_{yy}>>Nd_{zz}[/tex]. 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

    [tex]U=Nd_{xx}sin^2\theta cos^2\phi + Nd_{yy}sin^2\theta sin^2\phi+Nd_{zz}cos^2\theta [/tex]

    The expression can be simplified to

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

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

    The effective field at [itex]\theta=\pi/2[/itex],
    [tex] 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})[/tex]
    [tex]=2Asin\phi cos\phi (-sin\phi\hat{x}+cos\phi\hat{y})[/tex]

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

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

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

    Can someone tell me what I am missing??

    Thanks.
     

    Attached Files:

  2. jcsd
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