Magnets — Question 9

Question 9

A torsion fiber can stabilize an antiparallel magnet only below a threshold

A torsion fiber has energy $\kappa\theta^2/2$, where theta=0 places a permanent moment antiparallel to a uniform field B. The magnetic energy is $mB\cos\theta$ and rotational inertia is J.

1. Find the small-angle equation and the threshold for stability at theta=0.

2. For mB just above kappa, expand through cubic order in the equilibrium torque equation to estimate the two nearby tilted equilibria.

Question 9 - Solution

A torsion fiber can stabilize an antiparallel magnet only below a threshold

(a) Method and calculation

$$U=\tfrac12\kappa\theta^2+mB\cos\theta,\quad J\ddot\theta=-\kappa\theta+mB\sin\theta.$$$$\boxed{\omega^2=(\kappa-mB)/J;\quad\theta=0\text{ stable for }\kappa>mB}.$$

(b) Interpretation and further calculation

$$0\simeq(mB-\kappa)\theta-\frac{mB}{6}\theta^3.$$$$\boxed{\theta_*\simeq\pm\sqrt{6(1-\kappa/(mB))}}\quad(0<mB-\kappa\ll mB).$$

Physical check

The two small tilted equilibria have positive local curvature to leading order; the expansion is invalid far above threshold.