Field Precision title

Coil force integrals in Magnum

The magnetic force on a vector current element dl carrying current I in a magnetic field with flux density B is

dF = I dl × B (1)

In Magnum applied currents are represented by a collection of filamentary elements, so you might think that it would be straightforward to implement automatic routines to find forces on drive coils. The problem is one of accuracy when the drive coils are the primary source of the field. The field varies rapidly around the filamentary elements so a simple integral using Eq. 1 may give misleading values.

On the other hand, there are situations where such an integral would be very useful. I recently had a consulting project where a strong permanent-magnet assembly exerts a force on a current loop. The loop carries less than an ampere and makes only a small contribution to the field. Accordingly, I recently added a routine for automatic coil-force integrals to MagView. The new command Coil force is in the File operations menu. Here's how it works:


Figure 1. Magnet assembly with diagnostic coil paths.

1) Suppose you have a solution for a permanent-magnet assembly or an electromagnet and want to find the force on a coil inserted in the magnet volume. Load the solution into MagView with the File operations/Load solution file command.

2) Use Magwinder to define the sensor coil. It may be a simple rectangle (as in Fig. 1) or multi-turn shape of any complexity like a coil box. Assign a coil current of 1.0 A.

3) If you want to check the force at multiple positions, define multiple Coils in the CDF file with different values for the Shift and Rotate operations, as in Fig. 1.

4) Export an element file (WND).

5) Load the element file into MagView with the File operations/Load coils command.

6) Click on the Coil force command (note that the command functions only when both a solution and coil file have been loaded). The command shows a dialog with force components for each coil in the WND file and also write values to a data file if one is open. Here is an example listing:

--- Coil forces? ---
 NCoil  Fx           Fy           Fz
 ===========================================
   1    9.4794E-03   1.2813E-09  -3.9264E-06
   2    8.9958E-03  -2.5137E-04   9.6336E-07
   3    8.9808E-03   3.3423E-04? -2.9572E-06

If the diagnostic coils carry 1.0 A, the units of force are newtons/A.

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