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\def\b{ \color{red} \begin{equation}}
\def\e{\end{equation} \color{black}}
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\documentclass{article}
\usepackage{graphicx}
\usepackage{color}


\begin{document}



\subsection{Induced axial current}
\vskip3.in\hskip1in
\special{bmp:jet5.bmp y=3in }
\vskip.2in

Consider a transverse field component $B_y$

The magnetic flux between transverse positions $-x$ to $x$ and $dz$ is
\begin{equation}
d \Phi_y = 2x dz B_y(z)
\end{equation}
As a liquid metal jet passes axially down such a field at a velocity
$v={dz / dt}$, axial voltage gradients will be generated
\begin{equation}
G=x {dB_y \over dz} v
\end{equation}
If the metal electrical conductivity $\kappa$ is low enough so that
the resulting current has a negligible effect on the field, then the
axial current density $i_z$ will be
\begin{equation}
i_z= G\sigma
\end{equation}
\begin{equation}
{\rr i_z~=xv\kappa{dB_y(0,0,z) \over dz}}
\end{equation}


\subsection{Transverse elliptical distortion}
%\vskip3.3in
%\special{bmp:jet6.bmp y=3in }
\vskip.4in

If the jet is not on the solenoid axis, the axial induced currents
interacting with the transverse fields will generate distorting forces
on the jet. These transverse forces per unit volume $dx~ dy~ dz$ are
\begin{equation}
f_x~=~ i_z~B_y~
=~x~v~\kappa~B_y~ {dB_y \over dz}
\end{equation}

This force will distrot the cross section. Assuming that the liquid in incompresible, we must find the induced pressures and motions 
$\Delta \vec{r}$
within the cross section that are driven by this force, with the constraint that the diverence of these motions
is zero:
$$ Div (\Delta \vec{r})~=~0 $$

Defining 
$$ F_o~=~v~\kappa~B_y~ {dB_y \over dz}$$
so that the magnet force per unit volume:
$$ f_x({\rm magnetic})~=~ x~F_o$$

\subsubsection{Magnetic Forces}
\vskip2.in
\special{bmp:dist1.bmp y=2in }
\vskip.2in

The pressure on the surface of the jet will be independent of azymuth $\phi$:
$$p(r_o)_{\rm circular}=p_{\rm atm}~+~T/r_o ~,$$
 T being the surface tension.
If the initial cross section is circular, we can consider pressures within the cross section:
$$ p~=~p_o~-~r^2~{F_o \over 4}$$
where $p_o$ is set by the constraint the above surface pressure at $r=r_o$. This bulk pressure will induce radial pressure forces:
$$f_r({\rm pressure})~=~{dp \over dr}~=~-~{r~F_o \over 2}$$
so
$$f_x({\rm total})~=~F_o~\(x~-~{r~\cos(\phi) \over 2 }\)~=~ {x~F_o \over 2}$$

$$f_y({\rm total})~=~-~{F_o~r~\sin(\phi) \over 2 }~=~ -{y~F_o \over 2}$$
and the accelerations:
$${d^2x,y\over dz^2}~=~{f_{x,y} \over v~\rho}$$
will give dispacements:
$$\Delta x~=~ \int \int {f_{x,y} \over v~\rho}~dz^2 ~=~x~{\kappa \over 2~\rho}~\int \int~B_y~{dB_y\over dz}~dz^2 $$
$$\Delta y~=~ \int \int {f_{x,y} \over v~\rho}~dz^2 ~=~-y~{\kappa \over 2~\rho}~\int \int~B_y~{dB_y\over dz}~dz^2 $$

Such motions are quadrupole ( see fig ***) and, as required, give
$Div~(dx,dy)~=~0$. The resulting ellipticity $\epsilon = {\Delta x_o\over r_o}~=~-{\Delta y_o\over r_o}$:

{\rr
\begin{equation} \epsilon(z)~=~~{r_o~\kappa \over 2~ \rho}~\int \int B_y~ {dB_y \over dz}~dz^2  \end{equation}
} 

\subsubsection{Surface Tension Forces}
\vskip2.in
\special{bmp:dist2.bmp y=2in }
\vskip.2in


Once the cross section becomes somewhat elliptical (we consider only a small ellipticity), then the pressure at the surface is no longer independent of the asymuthal angle $\phi$, but is given by:

$$p(r=r_o)~=~ p_{\rm atm}~ +~ {T \over r_0}~(1~ -~ \epsilon)~\cos(2~\phi)$$

Consider, in addition to those given above for the circular case, pressures within the cross section:

$$p(x,y)_{\rm elliptical}~=~p_o~+~{T~\epsilon \over r_o}~\(\cos^2(2~\phi)~-~\sin^2(2~\phi)\)
~=~p_o~+~{T~\epsilon \over r_o^3}~( x^2~-~y^2)$$

which has the correct values at $r=r_o$.

Defining 
$$S_o~=~{2~T~\epsilon \over r_o^3},$$
the resulting bulk forces are:

$$f_x({\rm elliptical})~=~{dp\over dx}~=~x~S_o$$
$$f_y({\rm elliptical})~=~{dp\over dy}~=~ -y~S_o$$
which are, once again, quadrupole forces that will generate quadrupole motions with Div($\vec{dr}$)=0.

$$\Delta x,\Delta y~=~\int\int {(x,~-y)\over v~\rho}~S_o~dz^2$$

\subsubsection{Magnetic and Surface Tension Forces}

Adding these surface tension dispalcements to the forces derived for the circular case:

$$\Delta x,\Delta y~=~(x,~-y)~\int\int \( {\kappa \over 2~\rho}~B_y~{dB_y\over dz}~+~{2~T~\epsilon \over v~\rho~ r_o^3} \) 
~dz^2 $$

and the resulting ellipticity $\epsilon = {\Delta x_o\over r_o}~=~-{\Delta y_o\over r_o}$:

{\rr
\begin{equation} \epsilon(z)~=~\int\int \( {r_o~\kappa \over 2~\rho}~B_y~ {dB_y \over dz}~+~{2~T~\epsilon \over v~\rho~r_o^2}   \)
~dz^2  \end{equation}
} 

\end{document}
