Thanks for correcting it. Yes the current is what affects the intensity of the magnetic field of the electromagnet. I have been reading about buck converters and will soon try to use it instead of the high 24 V and avoid wasting power in the 10 Ohm resistor. I am thinking of using this type of buck converter:

https://www.electronicscomp.com/xl4...sQGAOrwCz3ZtTXi6YZ5t9snwGsVj1JCxoCaVUQAvD_BwE

Then for 2.5 Amps to flow through the inductor I will need a 2 Ohm, 10 Watts resistor in series with the electromagnet/inductor. The Power supplied by the buck converter would be 12.5 Watts which is much lesser than the 57 Watts wasted by the resistor.

However I also read the buck converter generates heat which could be dissipating electric power to some extent but that may not be as high as the 57 Watts dissipated by the resistor.

If you want 2.5 Amps to flow through the inductor then you need a constant current regulator.

 

If you want 2.5 Amps to flow through the inductor then you need a constant current regulator.

Yes correct. I already considered that option but buying a current regulator is too expensive for my small project so I couldn't.

 

How strong a magnetic field is required? For what purpose? How long must the magnetic field be maintained??
The more we understand the better the advice will be.

 

Yes correct. I already considered that option but buying a current regulator is too expensive for my small project so I couldn't.

Just an example:

CC buck converter

Edited: probably $1 on AliExpress

 

I think this is an example of what Bob is suggesting.
Depending on how critical control of the magnetic field strength is just feeding it with a constant voltage may be good enough.
(This is why you are being asked for the exact purpose of the magnet.) Fed with a constant voltage the current will reduce slightly as the magnet coil heats up. For example if you were using it for a nuclear magnetic resonance spectrometer just voltage control would not be good enough.

Les.

 

buying a current regulator is too expensive for my small project so I couldn't

@sab201,
You can set any current up to 5A
through your coil.
https://aliexpress.ru/item/10050031...w.productlist.search_results.2.813a4aa67fDaWC
Оnly $0.33, free delivery.

 

A fairly good but cheap current regulator can be created with a three terminal voltage regulator and a resistor. Not very efficient but certainly simple and cheap. Adding a PNP "wraparound"power transistor doubles the complexity to six components but increases the current handling capacity to several amps
AND it will still be useful to let us know about the purpose of the electromagnet.

 

A fairly good but cheap current regulator can be created with a three terminal voltage regulator and a resistor. Not very efficient but certainly simple and cheap. Adding a PNP "wraparound"power transistor doubles the complexity to six components but increases the current handling capacity to several amps
AND it will still be useful to let us know about the purpose of the electromagnet.

But for this application, it offers no advantage over a resistor.

 

Just an example:

CC buck converter

Edited: probably $1 on AliExpress

Good Day Bob,

When I connect my inductor directly across the output of the buck converter - that could act like a short circuit unless the buck converter has built in internal resistance otherwise the buck converter could overheat. My electromagnet is a small one and surely it has no considerable resistance in the windings.

So as you suggest can the resistor in series with the inductor (for drawing the required current) still be omitted while using the buck converter at 5 Volts. From my understanding all loads need to dissipate power. The only other way I thought I could supply desired current is by using an adjustable variable current source designed for laboratory purposes and that would be too expensive. Those have built in resistances to give the desired current outputs without damaging the source.

Thanks to all others for your suggestions.

 

When I connect my inductor directly across the output of the buck converter - that could act like a short circuit unless the buck converter has built in internal resistance otherwise the buck converter could overheat.

No. Stop making assumptions about things you do not know. The voltage is reduced to the point where the correct current is achieved. The coil has a resistance, that is the only resistance needed. If you short circuit the output of a current source, the voltage goes to near zero to allow the desired current. That is what is meant by constant current source, it adjusts the voltage to get the desired current.

The buck converter actually uses an inductor, not a resistor, to limit the current. That is why it dies not produce excess heat and waste power.

Have you even bothered to measure the resistance of your coil?

 

A supply with ratings suitable for the connected load will not overheat because the load is excessive. A one amp supply marked as 3 amps will certainly overheat when delivering 2.5 amps. So deal with an honest supplier.
No amount of lies about specifications will make an inadequate product able to deliver what it is not able to deliver. That is why I suggest using an honest supplier. So look at sources not selling on amazon.

 

Have you even bothered to measure the resistance of your coil?

Yes when I measured with a multimeter it fluctuated between 0 to 3 Ohms back and fourth.

Thanks for the information about the buck converter.

 

Yes when I measured with a multimeter it fluctuated between 0 to 3 Ohms back and fourth.

Then your meter is incapable of measuring the low resistance. Did you zero it by shorting the leads and pressing the zero button? You need to do that when measuring low resistances.

How about this: How many meters of what gauge wire? Then we can calculate the resistance.

 

Then your meter is incapable of measuring the low resistance. Did you zero it by shorting the leads and pressing the zero button? You need to do that when measuring low resistances.

How about this: How many meters of what gauge wire? Then we can calculate the resistance.

The core of the electromagnet is 2.5 meters long with diameter of 3 mm. I used pvc insulated 0.75 sq. mm wire to wind around this core that was approximately 10 to 15 meters of wire.

 

0.75 sq. mm wire to wind around this core that was approximately 10 to 15 meters of wire.

15 metres has a resistance of 342 milliOhms.

 

The core of the electromagnet is 2.5 meters long with diameter of 3 mm. I used pvc insulated 0.75 sq. mm wire to wind around this core that was approximately 10 to 15 meters of wire.

Now we know that the electromagnet is 2.5 M long, (about ten feet) and 3mm diameter (about 5/32 inch diameter.)
I am wonder what the application would be, and just how strong a field is required for that application. Changing a compass reading is all that comes to my mind for such an electromagnet.

Certainly this would be an interesting application to hear about.

 

Now we know that the electromagnet is 2.5 M long, (about ten feet) and 3mm diameter (about 5/32 inch diameter.)
I am wonder what the application would be, and just how strong a field is required for that application. Changing a compass reading is all that comes to my mind for such an electromagnet.

Certainly this would be an interesting application to hear about.

The application is for research purpose I am trying to learn about the behaviour of magnetic field in different shapes of the magnet, the core is a flexible wire rope.

 

Keep in mind that the magnetic field will also be affected by the field created by current flowing in the power feed wires.

 

15 metres has a resistance of 342 milliOhms.

Which would require 0.85V to get 2.5A.