Consider a squeezing flow of an incompressible Newtonian fluid in the presence of a magnetic field of a constant density ρ and viscosity μ squeezed between two large planar parallel plates separated by a small distance 2H and the plates approaching each other with a low constant velocity V, as illustrated in Figure 1, and the flow can be assumed to quasisteady [1–3, 39]. The NavierStokes equations [3, 4] governing such flow in the presence of magnetic field, when inertial terms are retained in the flow, are given as [38]
\mathrm{\nabla}V\cdot u=0
(2.1)
and
\rho [\frac{\partial u}{\partial t}+(u\cdot \mathrm{\nabla})u]=\mathrm{\nabla}\cdot T+J\times B+\rho f,
(2.2)
where u is the velocity vector, ∇ denotes the material time derivative, T is the Cauchy stress tensor,
and
{A}_{1}=\mathrm{\nabla}u+{u}^{T},
J is the electric current density, B is the total magnetic field and
{B}_{0} represents the imposed magnetic field and b denotes the induced magnetic field. In the absence of displacement currents, the modified Ohm law and Maxwell’s equations (see [40] and the references therein) are given by [38]
J=\sigma [E+u\times B]
(2.3)
and
divB=0,\phantom{\rule{2em}{0ex}}\mathrm{\nabla}\times B={\mu}_{m}J,\phantom{\rule{2em}{0ex}}curlE=\frac{\partial B}{\partial t},
(2.4)
in which σ is the electrical conductivity, E is the electric field and {\mu}_{m} is the magnetic permeability.
The following assumptions are needed [38].

(a)
The density ρ, magnetic permeability {\mu}_{m} and electric field conductivity σ are assumed to be constant throughout the flow field region.

(b)
The electrical conductivity σ of the fluid is considered to be finite.

(c)
Total magnetic field B is perpendicular to the velocity field V and the induced magnetic field b is negligible compared with the applied magnetic field {B}_{0} so that the magnetic Reynolds number is small (see [40] and the references therein).

(d)
We assume a situation where no energy is added or extracted from the fluid by the electric field, which implies that there is no electric field present in the fluid flow region.
Under these assumptions, the magnetohydrodynamic force involved in Eq. (2.2) can be put into the form
J\times B=\sigma {B}_{0}^{2}u.
(2.5)
An axisymmetric flow in cylindrical coordinates r, θ, z with zaxis perpendicular to plates and z=\pm H at the plates. Since we have axial symmetry, u is represented by
u=({u}_{r}(r,z),0,{u}_{z}(r,z)),
when body forces are negligible, NavierStokes Eqs. (2.1)(2.2) in cylindrical coordinates, where there is no tangential velocity ({u}_{\theta}=0), are given as [38]
\rho ({u}_{r}\frac{\partial {u}_{r}}{\partial r}+{u}_{z}\frac{\partial {u}_{r}}{\partial z})=\frac{\partial p}{\partial r}+(\frac{{\partial}^{2}{u}_{r}}{\partial {r}^{2}}+\frac{1}{r}\frac{\partial {u}_{r}}{\partial r}\frac{{u}_{r}}{{r}^{2}}+\frac{{\partial}^{2}{u}_{r}}{\partial {z}^{2}})+\sigma {B}_{0}^{2}u
(2.6)
and
\rho ({u}_{z}\frac{\partial {u}_{z}}{\partial r}+{u}_{z}\frac{\partial {u}_{z}}{\partial z})=\frac{\partial p}{\partial r}+(\frac{{\partial}^{2}{u}_{z}}{\partial {r}^{2}}+\frac{1}{r}\frac{\partial {u}_{z}}{\partial r}+\frac{{\partial}^{2}{u}_{z}}{\partial {z}^{2}}),
(2.7)
where p is the pressure, and the equation of continuity is given by [38]
\frac{1}{r}\frac{\partial}{\partial r}(r{u}_{r})+\frac{\partial {u}_{z}}{\partial z}=0.
(2.8)
The boundary conditions require
\begin{array}{r}{u}_{r}=0,\phantom{\rule{2em}{0ex}}{u}_{z}=V\phantom{\rule{1em}{0ex}}\text{at}z=H,\\ \frac{\partial {u}_{r}}{\partial z}=0,\phantom{\rule{2em}{0ex}}{u}_{z}=0\phantom{\rule{1em}{0ex}}\text{at}z=0.\end{array}
(2.9)
Let us introduce the axisymmetric Stokes stream function Ψ as
{u}_{r}=\frac{1}{r}\frac{\partial \mathrm{\Psi}}{\partial z},\phantom{\rule{2em}{0ex}}{u}_{z}=\frac{1}{r}\frac{\partial \mathrm{\Psi}}{\partial r}.
(2.10)
The continuity equation is satisfied using Eq. (2.10). Substituting Eqs. (2.3)(2.5) and Eq. (2.10) into Eqs. (2.7)(2.8), we obtain
\frac{\rho}{{r}^{2}}\frac{\partial \mathrm{\Psi}}{\partial r}{E}^{2}\mathrm{\Psi}=\frac{\partial p}{\partial r}+\frac{\mu}{r}\frac{\partial {E}^{2}\mathrm{\Psi}}{\partial z}\frac{\sigma {B}_{0}^{2}}{r}\frac{\partial \mathrm{\Psi}}{\partial z}
(2.11)
and
\frac{\rho}{{r}^{2}}\frac{\partial \mathrm{\Psi}}{\partial z}{E}^{2}\mathrm{\Psi}=\frac{\partial p}{\partial z}+\frac{\mu}{r}\frac{\partial {E}^{2}\mathrm{\Psi}}{\partial r}.
(2.12)
Eliminating the pressure from Eqs. (2.11) and (2.12) by the integrability condition, we get the compatibility equation as [38]
\rho \left[\frac{\partial (\mathrm{\Psi},\frac{{E}^{2}\mathrm{\Psi}}{{r}^{2}})}{\partial (r,z)}\right]=\frac{\mu}{r}{E}^{2}\mathrm{\Psi}\frac{\sigma {B}_{0}^{2}}{r}\frac{{\partial}^{2\mathrm{\Psi}}}{\partial {z}^{2}},
(2.13)
where
{E}^{2}=\frac{{\partial}^{2}}{\partial {r}^{2}}\frac{1}{r}\frac{\partial}{\partial r}+\frac{{\partial}^{2}}{\partial {z}^{2}}.
The stream function can be expressed as [1, 3]
\mathrm{\Psi}(r,z)={r}^{2}F(z).
(2.14)
In view of Eq. (2.14), the compatibility equation (2.13) and the boundary conditions (2.9) take the form
{F}^{(iv)}(z)\frac{\sigma {B}_{0}^{2}}{r}{F}^{\u2033}(z)+2\frac{\rho}{\mu}F(z){F}^{\u2034}(z)=0,
(2.15)
subject to
\begin{array}{r}F(0)=0,\phantom{\rule{2em}{0ex}}{F}^{\u2033}(0)=0,\\ F(H)=\frac{V}{2},\phantom{\rule{2em}{0ex}}{F}^{\prime}(H)=0.\end{array}
(2.16)
Nondimensional parameters are given as [38]
{F}^{\ast}=2\frac{F}{V},\phantom{\rule{2em}{0ex}}{z}^{\ast}=\frac{z}{H},\phantom{\rule{2em}{0ex}}Re=\frac{\rho HV}{\mu},\phantom{\rule{2em}{0ex}}m={B}_{0}H\sqrt{\frac{\sigma}{\mu}}.
For simplicity omitting the ∗, the boundary value problem (2.15)(2.16) becomes [38]
{F}^{(iv)}(z){m}^{2}{F}^{\u2033}(z)+ReF(z){F}^{\u2034}(z)=0,
(2.17)
with the boundary conditions
\begin{array}{r}F(0)=0,\phantom{\rule{2em}{0ex}}{F}^{\u2033}(0)=0,\\ F(1)=1,\phantom{\rule{2em}{0ex}}{F}^{\prime}(1)=0,\end{array}
(2.18)
where Re is the Reynolds number and m is the Hartmann number.