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Q: What is the energy used to flip the magnetization of a material? I understand that the direction of the magnetization in a material is determined by the direction of the magnetic dipole moment of that material, but can someone give an explicit calculation of the energy needed to flip the magnetization of a material. If this is energy, how would we calculate it? A: The energy required to flip the magnetization of a material by some angle $\theta$ is $$E = K_1 \cos \theta + K_2 \sin \theta$$ where $K_1$ and $K_2$ are the two components of the anisotropy energy of the material and are usually related to the strength of the interaction between the magnetic dipoles of the material and the atoms/molecules in the material. The ratio $K_1/K_2$ determines whether the material is uniaxial or biaxial. The components of the anisotropy energy are related to the two components of the magnetic dipole moment of the material (by using the $\hat{z}$ axis as the axis of rotation). It is also possible to calculate the anisotropy energy using the components of the magnetization of the material. For instance, the anisotropy energy of MnSi is approximately $$K = K_1 \cos^2 \theta + K_2 \sin^2 \theta$$ whereas the components of the magnetization of MnSi are $$\mu_x = \sin \theta \cos \theta, \mu_y = \sin \theta \sin \theta, \mu_z = \cos \theta$$ Here $K_1 = 3.88 \times 10^5 \mbox{erg/cm}^3$ and $K_2 = 8.0 \times 10^3 \mbox{erg/cm}^3$ (J. M. D. Coey, Phys. Rev. B, 41, 7630 (1990)). A: If you want to know how much energy is needed to flip the magnetization of a material you first need to define what “flipping” means. For example, if you mean that a material has a certain magnetization in one direction and it is supposed to have a different magnetization in a second