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Does Maxwell Tensor include Lorentz Forces ?

Electromagnetic problem

An analytical case is presented in order to show that the airgap Maxwell Tensor method is taking into account the Lorentz forces (sometimes called Laplace forces).Let us have a straight electrical wire surrounded by the void and traversed by a current. The wire is oriented toward z-axis, as illustrated in the previous figure.

    \begin{equation*} \textbf{I} = \textrm{I}\textbf{e}_z \end{equation}

The magnetic flux generated by the wire can be expressed in the polar referential as:

    \begin{equation*} B_I (r,\theta) = \mu_0 \frac{\textrm{I}}{ 2 \pi r } \textbf{e}_\theta \end{equation}

This wire is immersed in an external homogeneous electromagnetic field

    \begin{eqnarray*} B_e (r,\theta) &=& B_e \textbf{e}_y \\ &=& B_e \sin(\theta) \textbf{e}_r + B_e \cos(\theta) \textbf{e}_\theta \end{eqnarray}

such that the total electromagnetic field is

    \begin{equation*} \textbf{B}(r,\theta) = \textbf{B}_I (r,\theta) + \textbf{B}_\textrm{e}(r,\theta) = B_e \sin(\theta) \textbf{e}_r + \left( B_e \cos(\theta) + \mu_0 \frac{\textrm{I}}{2 \pi r} \right) \textbf{e}_\theta \end{equation}

which can be written as:

    \begin{equation*} \textbf{B}(r,\theta) = B_r \textbf{e}_r + B_\theta \textbf{e}_\theta \end{equation}

Lorentz force

According to the Lorentz force acting on a section L of the wire is given by:

    \begin{equation*} \textbf{F}_l = \int \textbf{I} \times \textbf{B} dl = - \textrm{I} \textrm{L} B_e \textbf{e}_\textrm{x} \end{equation}

Maxwell Tensor

According to the Maxwell Tensor method, the total magnetic force \textbf{F}_m can be computed by integrating the Maxwell stress Tensor on a closed cylinder of height L and radius R:

    \begin{equation*} \textbf{F}_m = \iint \left( \textbf{T}_m \cdot \textbf{n} \right) \textrm{dS} \end{equation}

    \begin{equation*} = \iint \left( \frac{B_r^2 - B_theta^2}{2 \mu_0} \textbf{e}_r + \frac{B_r B_theta}{\mu_0} \textbf{e}_\theta \right) \textrm{dS} \end{equation}

such that the total force in the x-direction is:

    \begin{equation*} F_x = \textbf{F}_m \cdot \textbf{e}_x \end{equation}

    \begin{equation*} = \frac{\textrm{R} \textrm{L}}{ \mu_0} \int_0^{2 \pi} \frac{B_r^2 - B_\theta^2}{2} \cos(\theta) - B_r B_\theta \sin(\theta) \textrm{d}\theta \end{equation}

    \begin{equation*} = \frac{\textrm{R} \textrm{L}}{ \mu_0} \int_0^{2 \pi} B_e^2 \sin(\theta)^2 \cos(\theta) - B_e^2 \cos(\theta)^3 - 2 \frac{\mu_0 \textrm{I} B_e}{2\pi R} \cos^2(\theta) - \frac{\mu_0^2 \textrm{I}^2}{4 \pi R^2} \cos(\theta) - B_e^2 \cos(\theta) \sin(\theta)^2 - \frac{\mu_0 \textrm{I} B_e}{ 2 \pi R} \sin(\theta)^2 \textrm{d} \theta \end{equation}

    \begin{equation*} = \frac{\textrm{L} \textrm{R}}{ \mu_0} \left( 0 - 0 - \pi \frac{\mu_0 \textrm{I} B_e}{2 \pi \textrm{R} } - 0 - 0 - \pi \frac{\mu_0 \textrm{I} B_e}{2 \pi \textrm{R}} \right) \end{equation}

    \begin{equation*} = - \textrm{L} \textrm{I} B_e \end{equation}

Following the same method in the y-direction leads to:F_y = 0

Conclusion

Then the Lorentz’s forces and the Maxwell forces are equal in this case:

    \begin{equation*} \textbf{F}_m = \textbf{F}_l \end{equation}

It confirms that the integration of Maxwell stress tensor on a closed surface is more general than the Lorentz force method.

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