Paramagnetism — Question 10

Question 10

Slow magnetic relaxation separates static susceptibility from an AC response

A weak paramagnet obeys the phenomenological relaxation law $\tau\dot M+M=\chi_0H_0\cos\omega t$, with positive tau and static susceptibility chi0. Seek the steady periodic response; ignore heating and nonlinear effects.

1. Find amplitude and phase lag of M relative to H.

2. Find the in-phase and lagging quadrature coefficients, and determine the frequency where the quadrature coefficient is largest.

Question 10 - Solution

Slow magnetic relaxation separates static susceptibility from an AC response

(a) Method and calculation

$$M=A\cos(\omega t-\delta),\quad\boxed{A=\frac{\chi_0H_0}{\sqrt{1+\omega^2\tau^2}},\quad\tan\delta=\omega\tau}.$$$$M=H_0[\chi^{\prime}\cos\omega t+\chi^{\prime\prime}\sin\omega t].$$

(b) Interpretation and further calculation

$$\boxed{\chi^{\prime}=\frac{\chi_0}{1+\omega^2\tau^2},\quad\chi^{\prime\prime}=\frac{\chi_0\omega\tau}{1+\omega^2\tau^2}}.$$$$\frac{d}{du}\frac{u}{1+u^2}=\frac{1-u^2}{(1+u^2)^2}\Rightarrow\boxed{\omega_{\max}=1/\tau,\quad\chi^{\prime\prime}_{\max}=\chi_0/2}.$$

Physical check

This explicitly supplied relaxation model is dynamic; a phase lag does not imply a permanent remanent magnetization after equilibrium at zero field.