Diamagnetism — Question 3

Question 3

A diamagnetic bead moves toward lower field magnitude

A small isotropic bead has volume V and weak susceptibility chi less than zero. Neglect its disturbance of the applied field. Its effective mechanical potential is $U=-\chi VB^2/(2\mu_0)$. Along x, $B^2=B_0^2+ax$, with a positive.

1. Find the magnetic force and the work it does when the bead moves a distance d in +x.

2. Find the quasistatic external work for that displacement and the dependence on the sign of B.

Question 3 - Solution

A diamagnetic bead moves toward lower field magnitude

(a) Method and calculation

$$F_x=-\partial_xU=\boxed{\frac{\chi Va}{2\mu_0}<0}.$$$$\boxed{W_{\mathrm{mag}}=F_xd=\frac{\chi Vad}{2\mu_0}<0}.$$

(b) Interpretation and further calculation

$$W_{\mathrm{ext}}=\Delta U=-W_{\mathrm{mag}}=\boxed{\frac{|\chi|Vad}{2\mu_0}>0}.$$$$B\to-B\Rightarrow B^2\text{ unchanged}\Rightarrow\boxed{F_x\text{ unchanged}}.$$

Physical check

The gradient of field magnitude squared sets the force; reversing the field direction alone does not reverse a diamagnetic force.