Magnetic Materials — Question 8

Question 8

An anisotropic induced moment has a one-half factor in its energy

A small weakly magnetic crystal of volume V has principal susceptibilities chi1 and chi2 in a plane. A uniform external B makes angle theta with axis 1. Neglect demagnetization. Use $U=-V\mathbf B\cdot\boldsymbol\chi\mathbf B/(2\mu_0)$.

1. Find energy as a function of theta and the generalized magnetic torque about the normal to the plane.

2. For chi1 greater than chi2, identify the stable orientation and the work needed to rotate from 0 to 90 degrees.

Question 8 - Solution

An anisotropic induced moment has a one-half factor in its energy

(a) Method and calculation

$$U(\theta)=-\frac{VB^2}{2\mu_0}(\chi_1\cos^2\theta+\chi_2\sin^2\theta).$$$$\boxed{\tau_\theta=-\frac{VB^2}{2\mu_0}(\chi_1-\chi_2)\sin2\theta}.$$

(b) Interpretation and further calculation

$$\chi_1>\chi_2\Rightarrow\boxed{\theta=0\pmod\pi\text{ is stable}}.$$$$\boxed{W_{0\to\pi/2}=\frac{VB^2}{2\mu_0}(\chi_1-\chi_2)>0}.$$

Physical check

The induced moment varies with field, so inserting it into the fixed-dipole expression without the one-half factor would double the energy.