Maxwell stress tensor

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The Maxwell Stress Tensor (also known as Maxwell's Stress Tensor) is used to calculate the stresses on objects in magnetic or electrical fields. It is used in many finite element programs to determine the forces on objects being analyzed.

If the magnetic and electrical fields at the surface of an object are known, the forces at that surface can be calculated, and the overall force on the object can be determined. In some cases, such as motors, the electrical fields are neglected, and stress and force calculations are made using only the magnetic fields.

In physics, the Maxwell stress tensor is the stress tensor of an electromagnetic field. In cgs units, it is given by:

$\sigma_{ij}=\frac{1}{4\pi}\bigl(E_{i}E_{j}+H_{i}H_{j}- \tfrac{1}{2}(E^2+H^2)\delta_{ij}\bigr)$ ,

where E is the electric field, H is the magnetic field and $δ i j$ is Kronecker's delta.

In SI units, it is given by:

$\sigma _{ij} = \varepsilon_0 E_i E_j + \frac{1} {{\mu _0 }}B_i B_j - \frac{1}{2}\bigl( {\varepsilon_0 E^2 + \tfrac{1} {{\mu _0 }}B^2 } \bigr)\delta _{ij}$ ,

where ε₀ and μ₀ are the electric constant and the magnetic constant respectively.

The element ij of the Maxwell stress tensor has units of momentum per unit of area times time and gives the flux of momentum parallel to the ith axis crossing a surface normal to the jth axis per unit of time.

These units can also be seen as units of force per unit of area (pressure), and the ij element of the tensor can also be interpreted as the force parallel to the ith axis suffered by a surface normal to the jth axis per unit of area. Indeed the diagonal elements give the pressure acting on a differential area element normal to the corresponding axis. Unlike forces due to the pressure of an ideal gas, an area element in the electromagnetic field also feels a force in a direction that is not normal to the element. This shear (rather than pressure) is given by the off-diagonal elements of the stress tensor.

If the field is only magnetic (which is largely true in motors, for instance), some of the terms drop out, and the equation in SI units becomes:

$\sigma _{ij} = \frac{1} {{\mu _0 }}B_i B_j - \frac{1}{2}\bigl( {\tfrac{1} {{\mu _0 }}B^2 } \bigr)\delta _{ij} N/m^2$

For cylindrical objects, such as the rotor of a motor, this is further simplified to:

$\sigma _{rt} = \frac{1} {{\mu _0 }}B_r B_t - \frac{1}{2\mu _0} { B^2 } \delta _{rt} N/m^2$

Where r is the shear in the radial (outward from the cylinder) direction, and t is the shear in the tangential (around the cylinder) direction. It is the tangential force which spins the motor. B_r is the flux density in the radial direction, and B_t is the flux density in the tangential direction.