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Eddy currents in a conductive disc Application


Physics 

A cylindrical region contains homogenous, time varying magnetic field B with rightwards arrow on top equals B subscript 0 cos left parenthesis omega t right parenthesis stack a subscript z with rightwards arrow on top ; B0 is the field magnitude and stack a subscript z with rightwards arrow on top is vertical unit vector. Since there is no electric charge in the system, the changing magnetic field is the sole cause of the electric field E with rightwards arrow on top . According to Farady's equation :

 nabla cross times E with rightwards arrow on top equals negative fraction numerator partial differential B with rightwards arrow on top over denominator partial differential t end fraction    (eq.1)

or in the case where B with rightwards arrow on top only has z component:
 

 1 over r open parentheses table row cell fraction numerator partial differential open parentheses r E subscript phi close parentheses over denominator partial differential r end fraction minus end cell cell fraction numerator partial differential E subscript r over denominator partial differential phi end fraction end cell end table close parentheses stack a subscript z with rightwards arrow on top equals omega B subscript 0 sin left parenthesis omega t right parenthesis stack a subscript z with rightwards arrow on top      (eq.2)

where  r and phi  represent radial and angular coordinates of the cylindrical coordinate system (Fig. 1).

Due to the symmetry of the problem, its analysis can be simplified by noting that E subscript r does not depend on phi, i.e. fraction numerator partial differential E subscript r over denominator partial differential phi end fraction equals 0 . Equation 2 therefore, takes the following form:

1 over r fraction numerator partial differential left parenthesis r E subscript phi right parenthesis over denominator partial differential r end fraction equals omega B subscript 0 sin left parenthesis omega t right parenthesis      (eq.3)

a) Copper disc in vertical magnetic field;b) Changing magnetic field is produced inside a cylinder wrapped in coils that carry AC current;
Figure 1 - a) Copper disc in vertical magnetic field;b) Changing magnetic field is produced inside a cylinder wrapped in coils that carry AC current;
 

To solve for E subscript phi , equation 3 is multiplied by rho and then integrated from 0 to rho as:
                         r E subscript phi equals integral subscript 0 superscript r omega B subscript 0 sin left parenthesis omega t right parenthesis r partial differential r                

which leads to  
E subscript phi equals 1 half omega B subscript 0 r sin left parenthesis omega t right parenthesis space
or  left parenthesis E with rightwards arrow on top equals 1 half omega B subscript 0 r sin left parenthesis omega t right parenthesis stack a subscript phi with rightwards arrow on top right parenthesis

If a copper disc is placed inside the cylinder, as a consequence of induced electric field, eddy currents are distributed inside the disc. Current density j with rightwards arrow on top is, according to the Ohm’s law :
j with rightwards arrow on top equals sigma E with rightwards arrow on top equals 1 half omega sigma B subscript 0 r sin left parenthesis omega t right parenthesis stack a subscript phi with rightwards arrow on top

where sigma is specific conductivity of copper (sigma equals 5.7 space cross times 10 to the power of 7 space S over m ). For a magnetic field with a magnitude of B subscript 0 equals 1.58 space e minus 2 T and angular frequency omega equals 2 straight pi 60 space rad divided by straight s, magnitude of  current density is r left square bracket m right square bracket asterisk times 1.69759 space cross times 10 to the power of 8 space A over m cubed . Induced eddy currents lag the change in flux density by 90º.

Model

Eddy current distribution in a copper disc can be easily simulated in EMS as a AC Magnetic AC study. To create uniform magnetic field inside cylinder, allow certain thickness to its wall, so that you can define the wall as a wound coil.Cylinder with inner radius of 15mm and height of 50mm, should be used to define Wound Coil with 100 turns and RMS current magnitude per turn of bevelled fraction numerator 10 over denominator square root of 2 end fraction.For Current phase select 0º - this will produce cosine current profile in the coil. This current will in turn induce relatively uniform flux density of magnitude B subscript 0 equals 1.46 space cross times 10 to the power of negative 3 end exponent space T over the volume of the copper disc (radius: 3mm), placed in the center of the cylinder. To help magnetic field align vertically inside the cylinder, add Normal Flux boundary condition to cylinder caps and Tangential Flux  boundary condition to the inner face of the cylinder wall. 

Boundary conditions

To help magnetic field align vertically inside the cylinder, Normal Flux boundary condition to cylinder caps and Tangential Flux boundary condition to the inner face of the cylinder wall, should be added.
 
To do define the Normal Flux,         
  1. In the EMS manger tree right-click on the Load/Restraint  1edfolder and select Normal Flux 2ed.
  2. Click inside the Faces for Normal Flux box  3edthen select Cylinder caps .
  3. Click OK5ed .
To do define the Tangential Flux,   
  1. In the EMS manger tree right-click on the Load/Restraint  1edfolder and select Tangential Flux2ed .
  2. Click inside the Faces for Tangential Flux3ed box  then select the inner face of the Cylinder wall .  
  3. Click OK5ed .

Coils

To show how to define the Wound Coil, see “Force in a magnetic circuit” example.

Eddy Effects

To set up Eddy Effects, right click the copper disc in the EMS tree and select Turn on Eddy Effects 7ed .

Meshing

To get high resolution of current density results in the copper disc, a Mesh control of 0.1 mm should be applied on the copper disc.
To do so,

  1. In the EMS manger tree right-click on the Mesh8ed  folder and select Apply Mesh Control 9ed.
  2. Click inside the Bodies10ed box then select the inner face of the copper disc .  
  3. Under Control Parameters click inside the Element Size 11edbox and type 0.1 mm.
  4. Click OK5ed .

To mesh the model:

  1. In the EMS manger tree, right-click on the Mesh 12edicon and select Create Mesh 13ed .
  2. Click OK 5ed .

Results

To plot current density distribution in the copper disc, right click the Current Densitycurrent density in the EMS tree and select 2D Plot 2D plot. For the current density Representation select Real (instantaneous). Maximum current density occurs for 90º  (minimum at 270º ), since the eddy currents have sinusoidal time dependence (in case of a cosine coil current and magnetic field).It is enough to select only 2 points along the disc radius – one in the center and the other one by the periphery. Type in number of points you want in between and EMS will plot current density along the disc radius.

 
Eddy current density inside the copper disc
Figure 2 - 3D vector plot of Eddy current density inside the copper disc
 
Comparison of EMS and theoretical results for eddy current density
Figure 3 - Comparison of EMS and theoretical results for eddy current density
 
The agreement between the theoretical solution (r left square bracket m right square bracket asterisk times 1.69759 space cross times 10 to the power of 8 space A over m cubed) and the EMS 2D current density plot is displayed on Figure 3.