The current induced in a wirewound Inductor is analyzed.
The current distribution and the resistance are solved.
This will be an axisymmetric analysis.
Unless specified in the list below, the default conditions will be applied.
Item 
Settings 
Analysis Space 
Axisymmetric 
Model unit 
mm 
Item 
Settings 
Solvers 
Magnetic Field Analysis [Gauss] 
Analysis Type 
Harmonic analysis 
Options 
Deselect Inductance Calculation. 
The frequency is swept from 1MHz to 10MHz in 1MHz interval.
Tab 
Setting Item 
Settings 
Harmonic analysis 
Frequency 
Minimum: 1×10^6[Hz] Maximum: 10×10^6[Hz] 
Step 
Linear step: Division number: 9 
The axisymmetric analysis is performed. For a model which is symmetric around Z axis,
the simulation is done for the XZ section only. It is virtually 2D analysis.
The calculation is much faster than 3D analysis. The axisymmetric analysis is well suited for this exercise.
A core is placed in center, and the coil is wound on the core.
The coil is formed by 16 turns in Z direction and 4 layers in X direction. It is 64 turns in total.
The right half section of the model is analyzed.
Body Number/Type 
Body Attribute Name 
Material Name 
Mesh Size 
063/Sheet 65/Sheet 
Body_Attribute_001 to 064 CORE 
008_Cu * CORE 
0.01

* Available from the Material DB
Set the current as follows.
Body Attribute Name 
Tab 
Settings 
Body_Attribute_001 to 064 
Current 
Waveform: AC Current: 1[A] Turns: 1[Turns] Direction: +Y direction 
No setting.
The following are solved.
1) Magnetic flux lines
2) Current density distribution
3) Frequency characteristics of the inductor’s resistance
4) Inductance
1) Magnetic flux lines
The figure below shows the magnetic flux lines at 10MHz.
2) Current density distribution
The figure below shows the current distribution. The current flows uniformly at 1MHz. At higher frequencies, the current flows unevenly especially in the outer windings.
3) Frequency characteristics of the inductor’s resistance
The chart below is the inductor’s resistance. It is higher at higher frequency.
How to calculate the resistance
Resistance R can be calculated from the loss P [W] displayed on the output window:
R = 2*P, where P = 1/2*R*I*I and I = 1[A] peak.
4) Inductance
The inductance can be calculated as follows. The magnetic energy Em, the current I and the inductance L satisfy
Em = 1/2*L*I*I As Em = 1.563e5[J] is obtained at I = 1[A] through the simulation,
L = 31.26[uH]