In this example, we consider a solenoid conductor with finite thickness and infinite length. This allow us to ignore the z components in our equations. We admit that there is only a radial expansion.

1. Problem

Since we cannot build an infinite geometry, we will use the same as in section \ref{sec:pb-validation-mag-fe}.
The volumic force \(\mathbf{f}\) take into account the Lorenz forces but not the thermal dilation, hence \(\mathbf{f}=\mathbf{j}\times\mathbf{B}\). The current density \(\mathbf{j}\) and the magnetic field \(\mathbf{B}\) can be seen as input parameters or given by our thermoelectric and magnetostatic models.

The materials properties of the solenoid are presented in the table \ref{tab:mat-le-fe}.

Name Description Value Unit

\(E\)

Young modulus

\(128.10^{9}\)

\(Pa=kg.m^{-1} .s^{-2}\)

\(\nu\)

Poisson’s ratio

0.33

-

We impose Neumann conditions on all boundary of the solenoid, except for the top and bottom, where we will use Dirichlet boundary conditions set to 0 to modelize the clamped side of the magnet.

The analytical solution of an infinite solenoid differs from this case because of the geometry used. We will then only compare the radial displacement at \(z=0\), to be as far as possible from the top and bottom of the solenoid.

An analytical solution can be found in \cite{Wilson87} or \cite{Montgomery69}, and interested readers can find more details in them. The idea to find the analytical solution is to express our quantities in cylindrical coordinates and use the fact that the only non null component of the current density is its angular component \(j_\theta\). We can find the radial component \(u_r\) of \(\mathbf{u}\) as solution of:

\[\begin{equation*} \frac{\partial}{\partial r}\left( r \frac{\partial u_r}{\partial r} \right) - \frac{u_r}{r} = - \frac{(1+\nu)(1-2\nu)}{E(1-\nu)}r j_{\theta} b_z \end{equation*}\]

We want to express the solution as

\[\begin{equation*} u_r=C_1 r + \frac{C_2}{r} + u_p(r) \end{equation*}\]

where \(u_p\) is a particular solution. Using boundary conditions we can find the two coefficients \(C_1\) and \(C_2\):

\[\begin{alignat*}{2} C_1 &= \frac{(1+\nu)(1-2\nu)}{E(1-\nu)}j_{\theta}r_i &&\left[ \left( b_z(r_i) + \frac{\Delta b_z}{\alpha - 1}\right) \left( \frac{2 - \nu}{3} \right) \left( 1 - \frac{r_e^2}{(r_e^2 - r_i^2)}\left( 1 - \frac{r_e}{r_i} \right) \right) \right. \\ &&&+ \left. \frac{\Delta b_z}{\alpha - 1}\left( \frac{2\nu - 3}{8} \right) \left( 1 - \frac{r_e^2}{(r_e^2 - r_i^2)} \left( 1 - \left( \frac{r_e}{r_i} \right)^2 \right) \right) \right]\\ C_2 &= \frac{-r_i^3 r_e^2 j_{\theta}}{(r_e^2 - r_i^2)}\frac{(1+\nu)}{E(1-\nu)} &&\left[ \left( b_z(r_i) + \frac{\Delta b_z}{\alpha - 1}\right) \left( \frac{2 - \nu}{3} \right)\left( 1 - \frac{r_e}{r_i} \right) +\frac{\Delta b_z}{\alpha - 1}\left( \frac{2\nu - 3}{8} \right)\left( 1 - \left( \frac{r_e}{r_i} \right)^2 \right) \right] \end{alignat*}\]

where \(\alpha=\frac{r_e}{r_i}\) and \(\Delta b_z=b_z(r_i)-b_z(r_e)\).
Using the expressions for \(\mathbf{j}\) and \(\mathbf{B}\), we can find the particular solution \(u_p\). In case of a Bitter coil, as in section \ref{sec:pb-validation-te-fe}, the current distribution is in \(r_i/r\), we find:

\[\begin{equation*} u_p(r) = \frac{(1+\nu)(1-2\nu)}{E(1-\nu)}r_i j_{\theta} \left[ \frac{\Delta b_z}{\ln(\alpha)}r\ln\left(\frac{r}{r_1}\right)^2 - \left(2b_z(r_1)+\frac{\Delta b_z}{\ln(\alpha)}\right)r\ln\left(\frac{r}{r_1}\right)\right] \end{equation*}\]

whereas if the current density is uniform in the coil, we have:

\[\begin{equation*} u_p(r) = \frac{(1+\nu)(1-2\nu)}{E(1-\nu)}r_i j_{\theta} \left[ -\frac{r_i}{3} \left( b_z(r_i) + \frac{\Delta b_z}{\alpha - 1} \right)\left( \frac{r}{r_i}\right)^2 + \frac{r_i}{8}\frac{\Delta b_z}{\alpha - 1} \left( \frac{r}{r_i}\right)^3 \right] \end{equation*}\]

Finally, returning to cartesian coordinates, the displacement \(\mathbf{u}\) can be written as:

\[\begin{equation*} \mathbf{u} = \begin{pmatrix} \cos(\theta)u_r(r) \\ \sin(\theta)u_r(r) \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{x}{\sqrt{x^2 + y^2}}u_r(\sqrt{x^2 + y^2}) \\ \frac{y}{\sqrt{x^2 + y^2}}u_r(\sqrt{x^2 + y^2}) \\ 0 \end{pmatrix} \end{equation*}\]

2. Results