Does Maxwell Tensor include Lorentz Forces ?

Electromagnetic problem

JPEG - 7 kb 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 \textbf{I} = \textrm{I} \textbf{e}_{z} as illustrated in the Figure.

The magnetic flux generated by the wire can be expressed in the polar referential as: B_{I} (r,\theta) = \mu_0 \frac{\textrm{I}}{2 \pi  r} \textbf{e}_{\theta}

This wire is immersed in an external homogeneous electromagnetic field B_e : B_{e}(r,\theta) = B_e \textbf{e}_{y} = B_e sin(\theta) \textbf{e}_{r} + B_e cos(\theta) \textbf{e}_{\theta}

such that the total electromagnetic field is: \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}

which can be written as: \textbf{B}(r,\theta) = B_r \textbf{e}_{r} + B_\theta \textbf{e}_{\theta}

Lorentz force

According to the Lorentz force acting on a section L of the wire is given by:  \textbf{F}_l =  \int \textbf{I} \times \textbf{B} dl = - \textrm{I} \textrm{L} B_e  \textbf{e}_{\textrm{x}}

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: \textbf{F}_m = \iint ( \textbf{T}_m. \textbf{n} ) \textrm{d}\textbf{S} =  \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{d}\textbf{S}

such that the total force in the x-direction is:  F_x = \textbf{F}_m.\textbf{e}_{x} = \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

 F_x = \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

 F_x = \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)

  F_x = - \textrm{L} \textrm{I} B_e

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:

\textbf{F}_m = \textbf{F}_l

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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