Does a solenoid get warm with inductance?

AI Thread Summary
A solenoid primarily heats up due to copper resistance, as indicated by the I^2R losses, while pure inductance does not contribute to heating. However, in practical applications, factors like eddy-current and hysteresis losses in magnetic cores can also lead to heating. The software Femm calculates parameters like voltage drop, but it does not require a specified voltage input as it focuses on inductance and current. The insulation of the wire must be selected based on the expected voltage in the circuit. Overall, understanding these factors is crucial for effective solenoid design and operation.
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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