Does a solenoid get warm with inductance?

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SUMMARY

A solenoid primarily heats due to copper resistance, as indicated by the formula ##I^2R##, while inductance itself does not contribute to heating. In alternating current (AC) scenarios, additional heating may occur due to eddy-current and hysteresis losses in the magnetic core, particularly if ferromagnetic materials are present. The software FEMM does not require voltage input for calculations, focusing instead on wire diameter, type, intensity, and number of turns, which are sufficient for determining the magnetic field generated.

PREREQUISITES
  • Understanding of electrical resistance and the formula ##I^2R##
  • Basic knowledge of inductance and its role in AC circuits
  • Familiarity with magnetic core materials and their properties
  • Experience using FEMM software for electromagnetic simulations
NEXT STEPS
  • Research eddy-current and hysteresis losses in magnetic materials
  • Learn about the impact of wire insulation on voltage ratings
  • Explore advanced features of FEMM for electromagnetic analysis
  • Study the relationship between inductance, current, and voltage in AC circuits
USEFUL FOR

Electrical engineers, physicists, and anyone involved in designing or analyzing electromagnetic systems, particularly those using solenoids and FEMM software.

BenTHS
Hi,

I'm wondering if a solenoid is only hot by copper resistance or in addition with the inductance produced into the coil too ?

Another small question : in the sofware Femm there is a computing result parameter called "voltage drop"; does it means you need ( at least ) x volt to get the magnetic field or is it necessary to use much higher voltage ?
In Femm you indicate the diameter and kind of wire, the intensity or the number of turn but not the voltage of the power supply; I'm surprised ?Thanks by advance
 
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Only resistance heats up with ##I^2R## losses. Pure L and puree C don't. But in real life and nothing is pure.
 
If the coil core is not of a ferromagnetic material-and no ferromagnetic material is in vicinity then no losses are expected. Since we are speaking about L and reactive power it is about a.c. current. In a.c. a magnetic core presents eddy-current and hysteresis losses. These losses could heat the core and from here the copper could be heated also. See-for instance:
Modeling Magnetic Core Loss for Sinusoidal Waveforms
http://www.dtic.mil/get-tr-doc/pdf?AD=ADA488218
I agree with you, in order to calculate the Fem one needs only L and I [Fem=LdI/dt].The voltage is present in the entire supplied circuit. However, the wire insulation has to be chosen according the presumable voltage.
 

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