Electromagnetic Forces on Parallel Current-

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1 Page 1 of 5 Tutorial Models : Electromagnetic Forces on Parallel Current-Carrying Wires Electromagnetic Forces on Parallel Current- Carrying Wires Introduction One ampere is defined as the constant current in two straight parallel conductors of infinite length and negligible circular cross section, placed one meter apart in vacuum, that produces a force of newton per meter of length. This model shows a setup of two parallel wires in the spirit of this definition, but with the difference that the wires do not have a negligible cross section. However, by successively shrinking the radius of the wires, the force would approach newton per meter of length. The force between the wires is computed using two different methods: by integrating the volume force density over the wire cross section and by integrating the stress tensor on the boundary. The results are in good agreement and are close but not equal to the number for the 1 ampere definition, as expected. Model Definition The model is built using the 2D Perpendicular Induction Currents application mode. The modeling plane is a cross section of the two wires and the surrounding air. DOMAIN EQUATIONS The equation formulation in the 2D Perpendicular Induction Currents application mode assume that the only nonzero component of the magnetic vector potential is A z. This corresponds to all currents being perpendicular to the modeling plane. The following equation is solved: where µ is the permeability of the medium and the externally applied current. is set so that the applied current in the wires equals 1 A, but with different signs. BOUNDARY CONDITIONS At the exterior boundary of the air domain, A z, is fixed to zero. The interior boundaries between the wires and the air only assume continuity, corresponding to a homogenous Neumann condition. Results and Discussion The expression for the surface stress reads where n 1 is the boundary normal pointing out from the conductor wire and T 2 the stress tensor of air. The closed line integral of this expression around the circumference of either wire evaluates to N. The minus sign indicates that the force between the wires is

2 Page 2 of 5 repulsive. The volume force density is given by The surface integral of the x component of the volume force on the cross section of a wire gives the result N. The number of mesh elements and the finite size of the geometry is what ultimately limits the accuracy in the calculations. Model Library Path: AC/DC_Module/Tutorial_Models/parallel_wires Modeling Using the Graphical User Interface MODEL NAVIGATOR 1 Select 2D in the Space dimension list. 2 Select the AC/DC Module>Statics>Magnetostatics>Perpendicular Induction Currents, Vector Potential application mode. 3 Click OK. OPTIONS AND SETTINGS 1 In the Constants dialog box, enter the following names and expressions. NAME EXPRESSION DESCRIPTION r 0.2 Wire radius (m) A pi*r^2 Wire area (m 2 ) I0 1 Wire current (A) J0 I0/A Current density (A/m 2 ) The radius r of each wire is 0.2 m, and the area is A = π r 2. The total current is I 0 = 1 ampere and the current density is J 0 = I 0 /A. 2 Give axis and grid settings according to the following table. AXIS GRID x min -6 x spacing 0.5 x max 6 Extra x y min -6 y spacing 0.5 y max 6 Extra y 0.2 GEOMETRY MODELING 1 Draw a circle C1 with radius 0.2 centered at ( 0.5, 0). 2 Draw a circle C2 with radius 0.2 centered at (0.5, 0). 3 Draw a circle C3 with radius 5 centered at (0, 0).

3 Page 3 of 5 Figure 2-1: The model geometry. PHYSICS SETTINGS Boundary Conditions The default boundary conditions are used. Subdomain Settings Apply an External current density and define Electromagnetic force variables according to the following table. SUBDOMAIN J0 -J0 Electromagnetic force variables wire1 wire2 Leave the other parameters at their default values.

4 Page 4 of 5 MESH GENERATION 1 Initialize the mesh. 2 Refine the mesh once. Figure 2-2: The mesh. COMPUTING THE SOLUTION Click the Solve toolbar button. POSTPROCESSING AND VISUALIZATION 1 The default plot shows the magnetic flux density. To visualize the magnitude of the volume force density, use a Surface plot with the expression sqrt(jez_emqa^2* (Bx_emqa^2+By_emqa^2)). Add the predefined quantity Magnetic field, norm as a Contour plot. Figure 2-3 shows the result. The volume force density is unevenly distributed within each wire due to the slight inhomogeneous magnetic field from the other wire. If you decrease the radii of the wires, the force density becomes evenly distributed. In the definition of 1 A it is assumed that the radii are negligible. The field becomes homogeneous as the radii approaches zero. The effects of the self-induced field of a wire then cancel out.

5 Page 5 of 5 Figure 2-3: Force density and magnetic field. 2 To visualize the air stress tensor on the boundary, clear the Contour check box and select the Arrow check box. Click the Arrow tab and select Boundaries from the Plot arrows on list. Click the Boundary tab in the Arrow data area and select Maxwell surface stress tensor (wire1) from the Predefined quantities list. Use a Scale factor of 0.1. Figure 2-4: Stress tensor and magnetic field. Compute the resulting force on the wires. 1 To compute the resulting force in the x direction on the left wire using the stress tensor method, open the Global Data Display dialog box and enter the expression wire1_forcex_emqa. You should get a result near 1.92e-7. 2 To alternatively compute the resulting force by integrating, open the Subdomain Integration dialog box and integrate the expression -Jez_emqa*By_emqa on subdomain 2. The result is again approximately -1.92e-7. The force in the y direction should be negligible. Confirm this by integrating Jez_emqa*Bx_emqa on subdomain 2.