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Magnetic Levitation for Energy Harvesting
Thursday, June 22, 2023
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SESSION 1
SESSION 2
CEST (GMT +2)
03:00 PM
08:00 PM
EDT (GMT -4)
09:00 AM
02:00 PM
HOME / Applications / Magnetic Field along the Axis of a Cylindrical Coil

Magnetic Field along the Axis of a Cylindrical Coil

Used Tools:

Physics

A cylindrical coil is used to create a strong magnetic field inside a domain. By wrapping the same wire many times around a cylinder, the magnetic field due to the current can become quite strong. The number of turnsspace N refers to the number of loops the cylindrical coil has. More loops will bring about a stronger magnetic field.

We will determine the magnetic field B spaceat a point Pat the axis of a cylindrical coil of length L, radius R , turns N, carrying current I. According to the Biot-Savart law, the magnetic field of a current loop is:


d B subscript Z equals fraction numerator mu subscript 0 R squared space d i over denominator 2 r cubed end fraction

Where mu subscript 0 equals 4 pi cross times 10 to the power of negative 7 end exponent space H divided by m space is the vacuum permeability.

Schematic of a cylindrical coil

Figure 1 - Schematic of a cylindrical coil


The net magnetic field on the axis of the cylindrical coil is the sum of the magnetic fields of all the loops.Divide the length of the cylinder into small elements of length d z to get the total field.The number of coil turns in a length d z space is:  

fraction numerator d N over denominator d z end fraction equals N over L

Therefore:      

d N equals N over L d z

The total current in a length d z space is:

d i equals I d N equals I N over L d z

The magnetic field contribution d B at point P spacedue to each element d z carrying current d i spaceis:

d B subscript Z equals fraction numerator mu subscript 0 space R squared over denominator 2 space r to the power of 3 space end exponent end fraction I space N over L d z space space space space space left parenthesis 1 right parenthesis

For each element of length d z spacealong the length of the cylinder, the distance z spaceand the angle alpha spacechange, while the value of space R spaceremains constant.From figure 1 we have:

r space equals fraction numerator R over denominator sin left parenthesis alpha right parenthesis end fraction

and

cos space alpha equals negative z over R space space space space space left parenthesis 2 right parenthesis


Expression (2) can be differentiated as:

negative fraction numerator d alpha over denominator sin squared alpha end fraction equals negative fraction numerator d z over denominator R end fraction

 
Which results with: 

d z space equals fraction numerator R space d alpha over denominator sin squared alpha end fraction

Formula for d z spaceand formula forspace r spacecan be substituted in the equation (1):

d B subscript Z equals fraction numerator mu subscript 0 space end subscript I space N space sin alpha over denominator 2 space L end fraction d alpha

Total magnetic field B subscript z at any point on the axis can be obtained by integrating from alpha subscript 1topi minus alpha subscript 2 :

B subscript z equals space contour integral subscript alpha subscript 1 end subscript to the power of pi minus space alpha subscript 2 end exponent space space fraction numerator mu subscript 0 space I space N space sin alpha over denominator 2 space L end fraction d alpha equals fraction numerator mu subscript 0 space I space N over denominator 2 space L end fraction contour integral subscript blank subscript alpha subscript 1 end subscript end subscript to the power of pi minus alpha subscript 2 end exponent space sin space alpha space d alpha equals fraction numerator mu subscript 0 space I space N over denominator 2 space L end fraction left parenthesis cos space alpha subscript 1 plus cos space alpha subscript 2 right parenthesis

B subscript z equals fraction numerator mu subscript 0 I space N over denominator 2 space L end fraction left parenthesis cos space alpha subscript 1 plus cos space alpha subscript 2 right parenthesis space space

Hence this expression gives the magnetic field at point on the axis of a cylindrical coil of finite length.

Model

A cylindrical coil of length 200mm, radius 5.5 mm , 100 turns, carrying current 10 A is modeled with Solidworks and simulated with Magnetostatic study 1in EMS.  Copper is assigned as a material to the cylinder body, while air covers  the inner air of the cylinder and the rest of the assembly. To get accurate magnetic field results, it is necessary to create sufficiently large air domain. To see how to assign material in EMS, see the “Computing capacitance of a multi-material capacitor” example.To learn how to define the air domain in EMS, consult the “Electric field inside the cavity of a charged sphere” example.

Solidworks model of the studied example
Figure 2 - Solidworks model of the studied example

Coil

To account for the magnetic field on the axis of the cylindrical coil, the cylinder should be used to define Wound Coil  2with 100 turns3 and RMS current magnitude per turn 4 of 10A.To prescribe the EMS Coil feature to the cylinder, it is necessary to have access to its cross-section surface. Therefore, the cylinder part should be split in two bodies. To learn how to do so, see “Magnetic field on axis of a current loop“example.Figure 3 shows the Entry and Exits ports of the Coil. In this case, Exit port is same as entry port.To show how to define the Wound Coil, see “Force in a magnetic circuit” example.

 Entry and Exits ports of the Coil

Figure 3 - Entry and Exits ports of the Coil

Mesh

Meshing is a very crucial step in the EMS simulation. Quality of the mesh in the inner air region and in the cylinder is of critical importance for accurate magnetic field calculation. To achieve a good accuracy without increasing the total number of mesh elements, a mesh control of 0.75 mm element size 5, should be applied to the inner air and to the cylinder. To learn how to do so, see “Force in a magnetic circuit” example. 

Results

To display the variation of the magnetic field along the axis of the cylinder, before running the simulation:

  1. In the Assembly, Select the ZX plane and sketch a line 6 along the z axis (axis of the coil), with a length equal to the length of the coil.
  2. Then Insert/Reference geometry/Point and add a Reference point for each end of the line.
  3. In the EMS feature tree, right click study 7and select Update geometry 8.
  4. Mesh and run the study.

Once the simulation is complete:

  1. In the EMS feature tree, Under Results 8, right click on the Magnetic Flux Density folder10 and select 2D Plot then choose Linear.         
  2. The 2D Magnetic Flux Density Property Manager Page appears.
  3.  In the Select points tab, click Import.
  4. Click OK 11.

The theoretical and EMS result of the magnetic flux density along the axis of the cylindrical coil are plotted in Figure 4.
It’s obvious that EMS result comply with the Biot-Savart law.

Comparison of EMS and theoretical results for magnetic flux density along the axis of a cylindrical coil
Figure 4 - Comparison of EMS and theoretical results for magnetic flux density along the axis of a cylindrical coil
 


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