Double wound solenoid inductance - QuickField simulation example
DC voltage is applied to the low inductance solenoid with double winding. Transient process of electric current vs. time is analyzed.
How to find double-wound solenoid inductance?
Engineering answer Typical applications Geometry
Given
Task
Solution
Result
Magnetic field energy W = 0.187 μJ, inductance L = 2* 0.187 / 1.786² = 0.117 uH.
Solenoid time constant L/R = 0.117 / 0.056 = 2.09 us.
Steady-state temperature of the coil is 50°C.
Engineering question
Set up an axisymmetric QuickField DC Magnetics problem for a double-wound solenoid and evaluate inductance from computed field results.
dual-winding solenoids, sensor excitation coils, electromagnetic actuator coils
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Simulation problem
Problem Type
Axisymmetric multiphysics problem of Transient magnetics coupled to Heat Transfer.
Conductor diameter 0.4 mm.
Copper electric conductivity σ = 56 MS/m.
Voltage U = 0.1 V.
Ambient air temperature T0 = 20°C, convection coefficient average value α = 5 W/K-m².
DC voltage is applied to the coil. Calculate the inductance, resistance and steady-state temperature.
Transient magnetic formulation in QuickField allows assign electric conductivity to blocks and connect blocks using the external electric circuit. The coil resistance can be calculated using circuit-simulation results: R = V / I.
The coil inductance can be calculated using the field simulation results: L = 2*Energy / I ².
The coil temperature is calculated in a heat transfer problem that is linked to the transient magnetics problem.
Double wound solenoid current I = 1.786 A, Joule heat P = 0.1786 W. Resistance R = 0.1 / 1.786 = 0.056 Ohm.
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