|
Input files
|
ShortSolenoid.MIN, ShortSolenoid.PIN, MuScan.BAT, ShortSolenoidScan.SCR, ShortSolenoidScan.PIN
Download ShortSolenoid.zip
|
|
Description
|
Models of a short solenoid inductor. The 50 turn winding has length W = 2.4 cm, inner radius 1.0 cm and outer radius 1.2 cm (average radius R = 1.1 cm). An inner core of radius 1.0 cm and length 2.4 cm can be assigned values of relative magnetic permeability. Calculations are performed with an input current of 1.0 A, so the inductance is given by 2*U (total field energy).
|
|
Results
|
The field inside an infinite length air-core solenoid is Bz = Mu0*J. With J = 50/0.024= 2083 A/m, the predicted field is Bz = 2.625E-3 T. The top figure shows the calculated field distribution in the short solenoid with a maximum value of Bz = 2.142E-3 T. The following formula gives an approximate inductance value for a short air-core inductor: L = pi*Mu0*R^2*N^2/(W + 0.9R). The predicted value for the current geometry is L = 35.23 uH. The PerMag calculation of the total field energy leads to a value of 32.26 uH. The inductance can be increased by wrapping the coil on a ferrite core. The second figure shows the field distribution for MuR = 100.0. The graph shows the variation of inductance with the relative magnetic permeability. The inductance is not proportional to MuR because the system is dominated by the field energy in the air return path. For MuR = 1.0, the region inside the coil contains 63.7% of the field energy. At MuR = 100.0, the value drops to 3.6%.
|
|
Comments
|
The open-ended solenoid makes inefficient use of the ferromagnetic material compared to a solenoid wound on a toroidal core. Another disadvantage is that the inductance is sensitive to external conditions. Note that the formula giving the approximate inductance is sometimes written in terms of Mu rather than Mu0. This is misleading because the inductance is not proportional to MuR.
|
