A magnetic circuit is a physical system whose components produce and contain magnetic flux. Flux is generally produced by permanent magnets or coils, while the flux paths consist of some ferromagnetic material of high permeability, such as iron. The example of magnetic circuit [1] in Figure 1 contains two iron parts of cross-section area $\mathrm{S}=4c{\mathrm{m2}}^{}$ . One part is stationary and the other is moving; they are separated by two air gaps of length $\mathrm{L.\; N}$-turn coil wrapped around the stationary part, carries the current I ( $N=300;I=1A$) and is responsible for the flux in the circuit.
Force on the moving part can be determined by calculating the change in the magnetic energy that would be produced by moving the part over a small distance . Force magnitude is obtained as:
This simple method is based upon the virtual displacement principle and it is often used to calculate forces in magnetic devices.
Since the permeability of iron is much greater than the permeability of air (), magnetic field in the iron (), as well as leakage flux, are neglected. Ampere’s law for the contour in Figure 1. can be written as:
where ${\mathrm{Ha}}_{}$ represents magnetic field in the air gap.
With no leakage and no filed in the iron (), the entire magnetic energy of the system is stored in the volume $V=SL$ of the air gap and can be computed as:
Assuming uniform field distribution in the air gap allows to calculate $W$ by simply multiplying energy density and volume $V$. The circuit contains 2 air gaps, therefore, after doubling the energy in eq.3 and substituting eq.2 in eq.3:
(eq. 4)
For two different air gap lengths${\mathrm{L1}}_{}=1.5mm$ and ${\mathrm{L2}}_{}=2.5mm$ , change in magnetic energy can be computed as:
(eq. 5)
Equation 5, combined with eq. 1, yields the following expression for force magnitude:
(eq. 6)
To accurately represent situation in the analytical example, where force has been computed using energies for $1.5mm$ and $2.5mm$ air gaps, Solidworks model of the circuit has been created with two air gaps of different lengths:${\mathrm{L1}}_{}=1.5mm$ and ${\mathrm{L2}}_{}=2.5mm$ (Figure 2.).
Due to the symmetry of the problem, it is enough to simulate only one half of the space (half of the magnetic circuit). The simulation uses EMS Magnetostatic study, with symmetry factor of .
Current driven Wound Coil with $300$ turns and $1A/turn$ is added to the coil region. Its Entry and Exit Ports are faces in the plane of symmetry.
The following instructions show you how to create new material library, define custom material and add material to an element of your model:
Copper is assigned to the coil region, and air to the surrounding volume.
Normal component of the flux density at the plane of symmetry ( $xy$ plane in Figure 2.) is zero (all flux lines are parallel to this plane). Therefore, Tangential Flux boundary condition should be applied on all surfaces that belongs to the plane, including air and coil cross-sections. To do so:
To add a coil to a Magnetostatic study:
Figure 4 - Entry and exit ports of the coil
EMS automatically computes the nodal force distribution without any user input. However, for a rigid body force calculation, part on which the force or torque shall be calculated needs to be defined before running the simulation.
Quality of the mesh in the air gap region is of critical importance for accurate force calculation. EMS allows user to take full control over the mesh resolution.
To mesh the model:
Right-click Study icon and selected Run , to execute simulation. Once the computation is done, the program creates five folders in the EMS manager tree. These folders are: Report, Magnetic Flux density, Magnetic Field intensity, Current density, and Force Distribution.
It is a good habit to first view the magnetic flux density in the model, including the outer air. This action gives an indication whether the outer air boundary is far enough.
By examining Figure 5. it becomes clear that the magnetic flux density is very small on the outer air boundary. Thus, the air box is large enough. Had it been otherwise, air box surrounding the magnetic circuit would have to be larger.
In the EMS Manager tree, Double-click Result Table to display force result.
Remember that, because of the symmetry about xy plane, only half of the problem has been modeled. Thus, $\mathrm{Fx}$ and components must be multiplied by a factor of , while the $\mathrm{Fz}$ component cancels out. Since $\mathrm{Fy}$ is very small compared to $\mathrm{Fx}$ , the resultant force is virtually in the X direction with a magnitude of $\mathrm{FxTotal}=2x1.507=3.014N$ . Analytical solution compares very well with the EMS result.
Analytical Virtual Work Solution | EMS Result | |
Force [N] | 3.02 | 3.014 |
[1] Electromagnetics and calculation of fields, by Nathan Ida and Joao P. A. Bastos, 2nd Edition, page 183-184. Publisher: Springer-Verlag;