The Lorentz Force - Magnetic Pressure and Tension

In the momentum equation (2.19)

$$\rho {\partial {\bf v} \over \partial t} + \rho ({\bf v} \c... ...){\bf v} = - \nabla p + {\bf j} \times {\bf B} + \rho {\bf g}.$$

the Lorentz force, ${\bf j}\times {\bf B}$ provides a link between the fluid equations and the electromagnetic equations. Given a prescribed flow, ${\bf v}$, the induction equation tells us how the magnetic field will evolve in time. As ${\bf B}$ changes, the Lorentz force will provide a back reaction on the plasma producing a force that will modify the velocity through the equation of motion.

Here we analyse the properties of the Lorentz force and give it some physical meaning. Firstly, from (2.22)

$${\bf j} \times {\bf B} = {1\over \mu}(\nabla \times {\bf B}) \times {\bf B}.$$

This may be rearranged by means of a vector identity, namely,

$$\nabla \left ( \frac{1}{2} {\bf B} \cdot {\bf B} \right ) =... ...s ( \nabla \times {\bf B} ) + ({\bf B} \cdot \nabla ) {\bf B}.$$

Hence, the Lorentz force may be expressed in the form

$${\bf j} \times {\bf B} = {1\over \mu} ({\bf B}\cdot \nabla ){\bf B} - \nabla \left ( {B^{2}\over 2 \mu} \right ).$$ (2.63)

The first term represents the effect of a tension of magnitude $B^{2}/\mu$ directed parallel to ${\bf B}$. This force appears whenever the magnetic field lines are curved. The second term represents the effect of a magnetic pressure of magnitude $B^{2}/2 \mu$ per unit area that is isotropic. This force occurs when the field strength, $\vert{\bf B}\vert$, varies from position to position. Note, from (2.19), how this term has the same form as the pressure gradient term, $- \nabla p$.

Now we show how the first term, ${1\over \mu} ({\bf B}\cdot \nabla ){\bf B}$, represents a tension force. To do this we need to investigate how the magnetic field may be represented in terms of a magnitude and a direction.

Figure 2.12: A point on a field line has a position vector ${\bf r}(s)$ and a tangent vector ${\bf T}$ in the direction of the field.

Consider Figure 2.12. If $s$ is the distance along the field line, then the position vector of a point on the field line is a function of $s$. Thus, we express

$${\bf r}(s) = x(s){\bf i} + y(s) {\bf j} + z(s) {\bf k}.$$ (2.64)

The tangent vector to the field line is given by the derivative of ${\bf r}$ with respect to $s$. Thus,

$${\bf T} = {dx\over ds} {\bf i} + {dy\over ds} {\bf j} + {dz\over ds} {\bf k}.$$ (2.65)

Note from Figure 2.13 that

$$ds^{2} = dx^{2} + dy^{2} + dz^{2},$$

and so

$$\left ({dx\over ds} \right )^{2} + \left ({dy\over ds} \right )^{2} + \left ({dz\over ds} \right )^{2} = 1.$$

This implies that ${\bf T}$ is a unit vector, say ${\bf T} = \hat{{\bf s}}$, in the direction of ${\bf B}$. Therefore, we can write

$${\bf B} = B(s) \hat{{\bf s}}.$$

Figure 2.13: The elemental distance $ds$ in terms of $dx$, $dy$ and $dz$.

Example 2.8.1   For any vector ${\bf a}(t)$ whose direction changes with respect to $t$ but whose magnitude is constant so that $\vert{\bf a}\vert = a = \hbox{constant}$ we have the following property

$${d\over dt}(a^{2}) = {d\over dt}({\bf a}\cdot {\bf a}) = {\... ...t {d{\bf a}\over dt} + {d{\bf a}\over dt} \cdot {\bf a} = 0.$$

Thus,

$${\bf a}\cdot {d{\bf a}\over dt} = 0.$$

Therefore, we have the important result that ${\bf T}$ is perpendicular to $d{\bf T} / ds$ and equivalently $d\hat{{\bf s}}/ds$ is perpendicular to $\hat{{\bf s}}$ as shown in Figure 2.14. Let $\hat{{\bf n}}$ be a unit vector that is normal to the field line. Then $\hat{{\bf n}}$ is parallel to $d\hat{{\bf s}}/ds$ and so we write

$${d\hat{{\bf s}}\over ds} = K \hat{{\bf n}} = {1\over R_{c}} \hat{{\bf n}}.$$ (2.66)

$K$ is called the curvature and $R_{c}$ is the radius of curvature.

Figure: The unit vectors $\hat{{\bf s}}$ and $\hat{{\bf n}}$.

The operator ${\bf B} \cdot \nabla$ is the derivative along the direction of the field line. This is expressed as

$${\bf B} \cdot \nabla = B {d\over ds},$$

where $s$ is the distance along the field line. For example, if $\phi = \phi (x(s), y(s), z(s))$

$$\begin{eqnarray*} B{d\phi\over ds} & = & B {dx\over ds}{\partial \phi \over \pa... ...ial \phi \over \partial s }\\ & = & {\bf B} \cdot \nabla \phi. \end{eqnarray*}$$

Here we have used the equations of the field lines (2.62) to express, for example, $B dx/ds$ as $B_{x}$.

Therefore the tension term can be written as

$$\begin{eqnarray*} ({\bf B} \cdot \nabla ){\bf B} & = & B{d\over ds}(B \hat{{\bf... ...s}}{\partial \over \partial s}\left ( {B^{2}\over 2} \right ). \end{eqnarray*}$$

Therefore the Lorentz force may be written as

$${\bf j} \times {\bf B} = {B^{2}\over \mu R_{c}} \hat{{\bf n... ...\mu} \right ) - \nabla \left ( {B^{2} \over 2 \mu} \right ).$$ (2.67)

(2.67) may be interpreted in the following manner. The magnetic field exerts an isotropic pressure $B^{2}/2 \mu$ in all directions and carries a tension $B^{2}/\mu$ along the lines of force. Each small flux tube is like an elastic band under tension. Neighbouring flux tubes expand against each other with a pressure $B^{2}/2 \mu$ and equilibrium is achieved when there is a balance between magnetic tension and magnetic pressure.

Example 2.8.2   Consider the magnetic field given by ${\bf B} = B_{0} ( 0, x/a, 0 )$. The equation of the field lines show that the field lines are given by $x = \hbox{constant}$. Thus, the field lines are straight as shown in Figure 2.15 but the magnitude, indicated by the closeness of the lines of force, varies with $x$.

Figure 2.15: The magnitude of $B$ varies with position.

Since the field lines are straight we do not expect any tension force. Thus

$$({\bf B}\cdot \nabla ){\bf B} = \left ( B_{x}{\partial \ove... ...} = B_{0}{x\over a}{\partial \over \partial y}({\bf B}) = 0.$$

However, since the field strength varies,

$${B^{2}\over 2 \mu} = {B_{0}^{2}\over 2 \mu} {x^{2}\over a^{2}},$$

we do expect there to be a magnetic pressure force, namely

$$- \nabla \left ( {B^{2} \over 2 \mu } \right ) = - {B_{0}^{2}\over \mu}{x \over a^{2}}{\bf i}.$$

We could have got this answer directly from the Lorentz force written in the form ${\bf j}\times {\bf B}$.

Example 2.8.3   Consider ${\bf B} = B_{0}( y/a, x/a, 0)$. The field lines are shown in Figure 2.16.

Figure 2.16: The field lines and the expected direction of the pressure and tension forces

The magnetic tension is given by

$$({\bf B} \cdot \nabla ){\bf B} = {B_{0}^{2} \over a^{2}} x{\bf i} + {B_{0}^{2} \over a^{2}} y {\bf j}.$$

Thus, on $y=0$ the tension force is along the $x$-axis.

The magnetic pressure is given by

$$-\nabla \left ({B^{2}\over 2} \right ) = - {B_{0}^{2}\over a^{2}} x {\bf i} - {B_{0}^{2}\over a^{2}} y {\bf j}.$$

Notice that the pressure force balances the tension force so that the Lorentz force is identically zero. This is clear from ${\bf j}\times {\bf B}$ since ${\bf j} = 0$.