# 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 \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.