AC Magnetic Field (solenoid) ID OD L AmpTurns Relationship

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To create a .4 Tesla AC magnetic coil, the user is currently facing limitations with their solenoid's dimensions and wire specifications, achieving only 0.27 Tesla at 75 Amps. The magnetic field strength is directly proportional to Amp Turns, but the user seeks clarity on how inner diameter (ID), outer diameter (OD), and length affect the field. Suggestions include using thicker wire to reduce heat generation or considering parallel windings to manage current better. However, increasing wire size may complicate cooling efficiency. Safety precautions are emphasized due to the risks associated with high currents and energy storage in inductive systems.
gosain_sanjay
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I am trying to make a .4 Tesla AC (50Hz) magnetic coil. Originally I used ID-38MM, OD-155MM & Length of Solenoid 50MM. Using 3.5MM X 1.5MM wire 380 Turns. I get 0.27 Tesla at 75Amp. but the wire allows me to work only for around 5 Min with Forced air cooling.
I do not know the relationship how the field will depend on ID,OD & Length (just that the field is directly proportional to Amp Turns).
It would be great if some could give a conditions for maximum field.

Any guidance will be great help since i was not able to find anything on NET.
 
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If no one has an answer can anyone suggest a Book for Ac coils ?
 
Hi gosain_sanjay! There are a number of inter-related factors to play with. Is 75A the maximum current you are able to supply? In which case, you can only change coil dimensions and number of turns to gain a stronger field. If your arrangement gets too hot, then you will have to use thicker wire or else use two parallel windings.

If you double the wire's cross-sectional area, you halve its resistance and the amount of heat produced falls by half, providing current remains the same. However, with the bulkier windings, there is probably less effective cooling so the temperature may not fall by as much as you would hope.

With such a large current in a large inductance, there is considerable energy stored and this can be hazardous. If a wire should break or a connection come apart, this energy will be released in one giant flash splattering molten copper around. Tight-fitting safety glasses are a must.
 
I'm working through something and want to make sure I understand the physics. In a system with three wave components at 120° phase separation, the total energy calculation depends on how we treat them: If coherent (add amplitudes first, then square): E = (A₁ + A₂ + A₃)² = 0 If independent (square each, then add): E = A₁² + A₂² + A₃² = 3/2 = constant In three-phase electrical systems, we treat the phases as independent — total power is sum of individual powers. In light interference...

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