Force of a magnetized cylinder on a ferromagnetic surface

[CharlieCW]

Homework Statement

A long straight cylinder with radius $a$ and length $L$ has an uniform magnetization $M$ along its axis.

(a) Show that when its flat extreme is placed on a flat surface with infinite permeability (i.e. a ferromagnet), it adheres with a force equal to:

$$F=8\pi a^2 L M^2\left [\frac{K(k)-E(k)}{k}-\frac{K(k_1)-E(k_1)}{k_1}\right ]$$

where $k=2a/ \sqrt{4a^2+L^2}$ and $k=a/ \sqrt{a^2+L^2}$.

(b) Find the limit for $a<<L$

Homework Equations

For a perfect ferromagnet ($\mu \rightarrow \infty$)

$$\vec{H(r)}=0 \ \ \ \ , \ \ \ \ \vec{B(r)}=\mu_0 \vec{M(r)}$$

Ficticious magnetic charge densities

$$\rho (r) *=-\nabla \cdot \vec{M} \ \ \ \ , \ \ \ \ \sigma (r_S) *=\vec{M(r_S)}\cdot \vec{n(r_S)}$$

Associated free current densities

$$\vec{j_M}=\nabla \times \vec{M(r)} \ \ \ \ , \ \ \ \ \vec{K_m(r_s)}=\vec{M(r_S)}\times \vec{n(r_S)}$$

Magnetic fields

$$\nabla\times\vec{H(r)}=\vec{j_M} \ \ \ \ , \ \ \ \ \nabla\cdot\vec{H(r)}=-\nabla\cdot\vec{M(r)}$$

Magnetic energy and force

$$U_B=\int d^3 r \int_{0}^{B}\vec{H}\cdot\vec{B}$$

$$F=-\left ( \frac{\partial U_B}{\partial \vec{R}} \right )$$

The Attempt at a Solution

After several attemps over the past week, I figured the best approach would be to use the method of images for magnetostatics and consider mirror charges on the opposite side of the ferromagnetic surface that will attract the magnetized cylinder with a magnetic force that depends on the ficticious magnetic charges (or free associated currents to $M$, both approaches are equivalent).

In Zangwill's book for electrodynamics, there's a similar exercise in page 435 that describes the coulum-like force betweem two coaxial cylindrical rods with constant magnetizations, which are essentially replaced by for equivalent ficticious magnetic charges at their extremes in the configuration (-)(+) (+)(-).

Thus my idea is to replace the cylinder with two equivalent cylindrical rings on its bases, placing two mirror charges beneath the ferromagnetic surface, and calculating the force between the rings with an associated surface current which can be easily calculated from the given magnetization (it's worth noting that there are no volumetric charges as $\nabla \times \vec{M}=0$ for a constant magnetization). Now the vector potential of a filametary current ring of radius R is an known result given in Zangwill (p. 325):

$$\vec{A(\rho,z)}=\vec{\phi}\frac{\mu_0 IR}{2}\int_{0}^{\infty} (dk J_1(k\rho)J_1(kR))e^{-k|z|}$$

where $J$ and $J_1$ are elliptic integrals. Now the force between two rings separated a distance $h$ can be calculated using mutual inductance $L_M$, defined as:

$$L_M=\frac{1}{I^2}\int d^3 r \vec{j_1(r)}\cdot\vec{A_2(r)}$$

Substituting

$$L_M=\mu_0 \pi R^2 \int_{0}^{\infty} dk J_1(kR)exp(-kh)$$

And the force is:

$$\vec{F}=I^2 \frac{dL_M}{dh}\vec{z}=\mu_0 \pi I^2 R^2 \frac{d}{dh}\int_{0}^{\infty} dk J_1(kR)exp(-kh)$$

Now if we consider the case of mirror magnetic charges in the configuration (-)(+) (+)(-), we get:

$$F_{tot}=F_{(-)(-),1}(h=2L)+F_{(-)(+),2}(h=L)+F_{(+)(+),3}(h=0)$$

However not only this gives me three terms, but there's no way to express the result as a sum of elliptic integrals and not as a product of elliptical integrals. Moreover, I'm not even sure if there could be a better approach, so if you can suggest a better one I will really appreciate it.

 

[mark726]

One possible approach could be to use the magnetic energy and force equations given in the homework statement. The magnetic energy in this case would be:

$$U_B=\int d^3 r \int_{0}^{B}\vec{H}\cdot\vec{B}=\mu_0 \int d^3 r \vec{M}\cdot\vec{B}$$

Now, we can rewrite this in terms of the magnetization and the magnetic permeability of the ferromagnet:

$$U_B=\mu_0 \int d^3 r \vec{M}\cdot(\vec{B_0}+\mu_0\vec{M})$$

Where $\vec{B_0}$ is the external magnetic field. Using the fact that the magnetic field inside the ferromagnet is equal to the external field plus the contribution from the magnetization, we can rewrite this as:

$$U_B=\mu_0 \int d^3 r \vec{M}\cdot\vec{H}=\frac{1}{2}\mu_0 \int d^3 r \vec{M}\cdot\vec{H}=\frac{1}{2}\mu_0 \int d^3 r \vec{M}\cdot(\vec{H_0}+\mu_0\vec{M})$$

Now, we can use the fact that the magnetic field inside the ferromagnet is equal to the external field plus the contribution from the magnetization, and substitute in the expression for the magnetic energy:

$$U_B=\frac{1}{2}\mu_0 \int d^3 r \vec{M}\cdot(\vec{H_0}+\mu_0\vec{M})=\frac{1}{2}\mu_0 \int d^3 r \vec{M}\cdot\vec{H_0}+\frac{1}{2}\mu_0 \int d^3 r \vec{M}\cdot(\mu_0\vec{M})$$

Now, we can use the fact that the magnetic field inside the ferromagnet is equal to the external field plus the contribution from the magnetization, and substitute in the expression for the magnetic energy:

$$U_B=\frac{1}{2}\mu_0 \int d^3 r \vec{M}\cdot\vec{H_0}+\frac{1}{2}\