Force Between Two Parallel Currents — Question 2

Question 2

Repulsive magnetic support is stable vertically when lateral motion is constrained

A fixed horizontal long wire carries I1 toward +z. A second parallel wire directly above it has mass per length lambda and carries I2 toward -z. A frictionless vertical guide keeps the wires directly above one another. Their separation is h; gravity is downward. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the equilibrium height and classify vertical stability.

2. For I1=I2=20.0 A, lambda=.00200 kg/m and g=9.81 m/s squared, find the height and small-oscillation angular frequency.

Question 2 - Solution

Repulsive magnetic support is stable vertically when lateral motion is constrained

(a) Method and calculation

$$f_y=\frac{C}{h}-\lambda g,\quad C=\frac{\mu_0 I_1I_2}{2\pi},\quad h_0=\frac C{\lambda g}.$$$$\left.\frac{df_y}{dh}\right|_{h_0}=-\frac C{h_0^2}<0,\quad\lambda\ddot\eta=-\frac C{h_0^2}\eta\Rightarrow\Omega=\sqrt{g/h_0}.$$

(b) Interpretation and further calculation

$$C=8.00\times10^{-5}\,\mathrm N,\quad\boxed{h_0=.00407747\,\mathrm m}.$$$$\boxed{\Omega=49.05\,\mathrm{rad/s}}.$$

Physical check

This is stability along the imposed vertical guide. Sideways motion is excluded; no claim of free three-dimensional levitation is made.