Magnetization and torque in an ellipsoidal disk

In summary: Asin\phi cos\phi (-sin\phi\hat{x}+cos\phi\hat{y})+2Acos^2\theta \hat{z}-2Bsin\phi cos\phi (sin\phi\hat{x}-cos\phi\hat{y})=2Asin\phi cos\phi (-sin\phi\hat{x}+cos\phi\hat{y})+2A(cos^2\theta-cos^2\phi)(-sin\phi\hat{x}+cos\phi\hat{y})-2Bsin\phi cos\phi (sin\phi\hat
  • #1
buggykong
4
0
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.
 

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  • #2


It seems that you have made a small error in your calculations. The effective field at \theta=\pi/2 should be:

H_{eff}=-\nabla U= -\frac{\partial U}{\partial \theta}\hat{\theta}-\frac{1}{sin\theta}\frac{\partial U}{\partial \phi}\hat{\phi}
=-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\theta cos\phi (-sin\phi\hat{x}+cos\phi\hat{y})+2Acos\theta (cos\theta cos\phi \hat{x} + cos\theta sin\phi\hat{y}-sin\theta\hat{z})+2Bsin\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})+2Acos\theta (cos\theta cos\phi \hat{x} + cos\theta sin\phi\hat{y}-sin\theta\hat{z})+2Bsin^2\phi (-sin\phi\hat{x}+cos\phi\hat{y})
=2Asin\phi cos\phi (-sin\phi\hat{x}+cos\phi\hat{y})+2Acos\theta (cos\theta cos\phi \hat{x} + cos\theta sin\phi\hat{y}-sin\theta\hat{z})-2Bsin\phi cos\phi (sin\phi\hat{x}-cos\phi\hat{y})
=2Asin\phi cos\phi (-sin\phi\hat{x}+cos\phi\hat{y})+2Acos\theta (cos\theta cos\phi \hat{x} + cos\theta sin\phi\hat{y}-sin\theta\hat{z})-2Bsin\phi cos\phi (sin
 

Related to Magnetization and torque in an ellipsoidal disk

1. What is magnetization in an ellipsoidal disk?

Magnetization in an ellipsoidal disk refers to the process of aligning the magnetic moments of the atoms or molecules within the disk in a specific direction, creating a net magnetic field. This can occur naturally in certain materials or can be induced through the use of an external magnetic field.

2. How is torque related to magnetization in an ellipsoidal disk?

Torque is the rotational force that is exerted on an object. In an ellipsoidal disk, the magnetization creates a magnetic dipole moment, which can experience torque when placed in an external magnetic field. This torque can cause the disk to rotate, aligning its magnetic moments with the external field.

3. What factors affect the magnetization and torque in an ellipsoidal disk?

The magnetization and torque in an ellipsoidal disk can be affected by several factors, including the strength and direction of the external magnetic field, the material properties of the disk (such as its shape and composition), and the temperature of the system.

4. Can the magnetization and torque in an ellipsoidal disk be controlled?

Yes, the magnetization and torque in an ellipsoidal disk can be controlled by adjusting the external magnetic field or by changing the material properties of the disk. For example, the strength of the magnetic field can be increased or decreased to alter the level of magnetization and torque, and the shape or composition of the disk can be modified to change its response to an external magnetic field.

5. What are some practical applications of magnetization and torque in ellipsoidal disks?

Magnetization and torque in ellipsoidal disks have several practical applications, including in data storage devices (such as hard drives and magnetic tapes), magnetic sensors, and medical imaging equipment (such as MRI machines). They can also be used in research and development for studying the properties of magnetic materials and designing new technologies.

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