2 Non-relativistic pinch model

Let us consider a coordinate system moving with the electron drift velocity  $u$ . In this coordinate system an induced electric field of magnitude $uB$ exists in the radial direction, so that the total electric field in the radial direction is given by $E+uB$ , where $E$ is the electric field in the laboratory frame of reference. With the elementary electric charge, $e$ , Boltzmann constant, $k_{B}$ , and the radial distance as measured from the cylindrical axis, $r$ , the electron density is therefore given by

(2.1) $$\begin{eqnarray}n_{e}=n_{e0}\exp \left(\frac{-e}{k_{B}T_{e}}\int _{0}^{r}(E+uB)\,\text{d}r\right).\end{eqnarray}$$

The electron density in thermal equilibrium depends upon the electric potential $V$ and is unaffected by the presence of a magnetic field. Individual particle orbits depend on the magnetic field, but macroscopic quantities are unaffected. This has been shown by Allen (Reference Allen1983) who pointed out that the Maxwell–Boltzmann distribution satisfies the Vlasov equation for a collision-free plasma in a magnetic field.

Figure 1.

The pinch effect.

We can note that the electron density is unchanged when moving from one reference frame to another in a Galilean (or Newtonian) system. Equation (2.1) therefore can be taken to give the electron density in the laboratory frame of reference. In this reference frame the ion density is given by the Boltzmann relation

(2.2) $$\begin{eqnarray}n_{i}=n_{i0}\exp \left(\frac{-eV}{k_{B}T_{i}}\right).\end{eqnarray}$$

The magnetic field is given by Ampères law, which can be written as

(2.3) $$\begin{eqnarray}B=\frac{\unicode[STIX]{x1D707}_{0}eu}{r}\int _{0}^{r}nr\,\text{d}r.\end{eqnarray}$$

Where $\unicode[STIX]{x1D707}_{0}$ is the permeability of free space. We shall solve the problem using the concept of quasi-neutrality, i.e.

(2.4) $$\begin{eqnarray}n_{i}=n_{e}=n.\end{eqnarray}$$

So that, using (2.1) and (2.2),

(2.5) $$\begin{eqnarray}V=\left(\frac{T_{i}}{T_{e}+T_{i}}\right)\int _{0}^{r}uB\,\text{d}r.\end{eqnarray}$$

The electric field in the laboratory frame of reference is given by

(2.6) $$\begin{eqnarray}E=-\left(\frac{T_{i}}{T_{i}+T_{e}}\right)uB\end{eqnarray}$$

so that the force on the positive ions, per unit volume is $-neuBT_{i}/(T_{e}+T_{i})$ , the negative sign indicating force towards the axis. The electric force acting on the electrons is $neuBT_{i}/(T_{e}+T_{i})$ , i.e. an outward force. The magnetic force acting on the electrons is $-neuB$ so that the net force on the electrons is $-neuBT_{e}/(T_{e}+T_{i})$ . The total force on a unit volume of plasma is seen to be $-neuB$ which can be identified as the $\boldsymbol{j}\times \boldsymbol{B}$ force employed in magnetohydrodynamics. The MHD model does not reveal the existence of the electric field which transmits part of the $\boldsymbol{j}\times \boldsymbol{B}$ force to the positive ions.

From (2.3) and (2.6) we can obtain the following

(2.7) $$\begin{eqnarray}E=-\frac{T_{i}}{T_{e}+T_{i}}\frac{1}{r}\int _{0}^{r}\unicode[STIX]{x1D707}_{0}eu^{2}nr\,\text{d}r.\end{eqnarray}$$

From (2.2) we have

(2.8) $$\begin{eqnarray}neE=k_{B}T_{i}\frac{dn}{dr}.\end{eqnarray}$$

So that substitution in (2.7) gives

(2.9) $$\begin{eqnarray}\frac{\text{d}}{\text{d}r}\left(\frac{r}{n}\frac{\text{d}n}{\text{d}r}\right)+\frac{\unicode[STIX]{x1D707}_{0}e^{2}u^{2}nr}{k_{B}(T_{e}+T_{i})}=0.\end{eqnarray}$$

We shall look for a solution for the particle density $n$ (electron or ion) of the following form

(2.10) $$\begin{eqnarray}n=\frac{n_{0}}{(1+bn_{0}r^{2})^{2}}.\end{eqnarray}$$

Figure 2.

Profile of density with normalised radial distance $L^{\ast }=(bn_{0})^{-0.5}$ . The density is normalised by $n_{0}$ .

This function was originally found by Bennett (Reference Bennett1934) to give the electron density in the fully relativistic case. It can readily be shown to be a solution of the present case of (2.9), where the quantity $b$ , which has the dimensions of length, is given by

(2.11) $$\begin{eqnarray}b=\frac{\unicode[STIX]{x1D707}_{0}e^{2}u^{2}}{8k_{B}(T_{e}+T_{i})}.\end{eqnarray}$$

The density profile is plotted in figure 2.

Thonemann & Cowhig (Reference Thonemann and Cowhig1950) found the solution for the particular case of $E=0$ (and hence $T_{i}=0$ ). In their experiments with a high current ( ${\sim}50$  A) mercury arc, probe measurements indicated that the radial electric field was negligible.

The result of integrating equation (2.3) using equation (2.10) to express $n$ gives

(2.12) $$\begin{eqnarray}B=\frac{\unicode[STIX]{x1D707}_{0}eu}{r}\int _{0}^{r}\frac{n_{0}r}{(1+bn_{0}r^{2})^{2}}\text{d}r=\frac{\unicode[STIX]{x1D707}_{0}eun_{0}r}{2(1+bn_{0}r^{2})}.\end{eqnarray}$$

The radial variation of the magnetic field is shown in figure 3, together with the electric field as given by (2.6).

Figure 3.

Profiles of electric and magnetic fields where the electric field is normalised by $(T_{i}/(T_{i}+T_{e}))(\unicode[STIX]{x1D707}_{0}en_{0}u^{2}L^{\ast }/2)$ and the magnetic field by $B^{\ast }=\unicode[STIX]{x1D707}_{0}en_{0}uL^{\ast }/2$ with normalised radial distance $L^{\ast }=(bn_{0})^{-0.5}$ .

Using (2.12) the magnetic field at large radii is given by

(2.13) $$\begin{eqnarray}B=\frac{\unicode[STIX]{x1D707}_{0}eu}{2br}.\end{eqnarray}$$

So that we can obtain the following expression for the current $I$ , using Ampères law,

(2.14) $$\begin{eqnarray}2\unicode[STIX]{x03C0}rH=\frac{eu\unicode[STIX]{x03C0}}{b}=I\end{eqnarray}$$

but

(2.15) $$\begin{eqnarray}I=\int _{0}^{\infty }2\unicode[STIX]{x03C0}rneu\,\text{d}r=Neu.\end{eqnarray}$$

Where $N$ is the number of electrons (or ions) per unit length. Using (2.11), (2.14) and (2.15), we can write

(2.16) $$\begin{eqnarray}2Nk_{B}(T_{e}+T_{i})=\frac{\unicode[STIX]{x1D707}_{0}}{4\unicode[STIX]{x03C0}}I^{2}\end{eqnarray}$$

which is the well-known Bennett relation.