# Solenoid problem

1. Jun 4, 2009

### Degel

Hey guys I have been wondering do the cross-sectional area of the copper wire affect the solenoid?So far I only know that the cross-sectional of the core material , turn of wire (N) , leakage , etc. affected it.Have been wondering for so long since the formula given in many website doesn't state the area of the copper wire used.

2. Jun 4, 2009

What kind of solenoid long or short, thick or thin and what frequencies?

3. Jun 4, 2009

### Bob S

You should normally use the thinnest wire that will carry the current you want to use. I say this because you want the turns as close as possible, and the B field inside the solenoid is equal to
B (Tesla) = u u0 NI, where NI = amp-turns per meter, u0 = permeability of free space, and u = relative permeability.
Here is a wire table for copper wire, including gauge #, diameter, ohms, etc.:
http://ken-gilbert.com/techstuff/AWG_WIRE_TABLE.html
The last two columns give amps and max amps. It gives 2.1 amps and 3.2 amps for 18 Ga wire, which I agree with.

4. Jun 5, 2009

### Degel

Actualy I am just wondering about the formula for the solenoid.

Thanks Bob for your detail info.From the table I discovered that the bigger the diameter the smaller the resistance.From the formula V=IR we can know that the smaller the resistance the bigger the current will pass through , bigger current will produce bigger B field.

5. Jun 5, 2009

### Naty1

"Hey guys I have been wondering do the cross-sectional area of the copper wire affect the solenoid?"

Not an awful lot...ampacity varies as r squared and this cuts the resistance if increased but also the number of turns, but if diameter is not restricted you can add more turns to compensate.....so the answer depends on your chosen restrictions.

Bob has a good starting point: "You should normally use the thinnest wire that will carry the current you want to use."

The total resistance of the wire, hence it's length (turns) and size (cross sectional area) should also be matched to the power source....too little resistance and too much current will flow, damaging the wire from overheating; too much resistance, and the source may not be able to provide the desired current strength.

In other words, you cannot, in general, just pick the number of turns you'd like without considering the power source limitations.

Last edited: Jun 5, 2009
6. Oct 17, 2010

### Armando J

Umm, If you cooled the solenoid couldn't you hypothetically reduce the damage to the wire itself (caused by overheating) and therefore use a higher amperage to further increase the strength of the field without damaging wire? I believe cooling it also reduces the resistance in the wire, which would mean a lesser volatage drop (correct me if I'm wrong), which would mean less heat being produced, meaning less damage? Therefore the optimum wire gauge would depend on how effectively you could cool the solenoid. Get it to superconduct (as if), and you're golden. Any suggestions as to what to use for the core material or where to find a big battery? (Open to anything here)

7. Oct 17, 2010

### Bob S

First, determine what power supply you want to use (volts, amps). This will help determine the coil resistance and Ga. of the wire. You don't want to use 10 Ga. wire for 1 amp for example, because the coil will be too big and you will be very limited in the number of turns. For 1 amp, 18 Ga. would be better, but for lots of turns, you will need lots of volts (calculate the total wire resistance and IR drop for N turns). Second, determine what type of coil cooling you want (ambient air, forced air, water). For ambient air and forced air cooling, the coil is hotter inside than on the surface. The heat conducts out by thermal contact conduction between layers. My rule of thumb has been to make sure the outside of the coil is never too hot to touch. Also, if you can smell the coil getting too hot inside, it IS getting too hot inside.

Another factor that sometimes needs to be considered is that if for any reason, the coil current is suddenly interrupted, there is a very high V = L dI/dt inductive voltage across the coil that may lead to insulation breakdown. For limiting inductive voltages such as this, fewer turns is better (L scales as N2). For example, the L dI/dt voltage on the primary of a 12-volt automotive ignition coil is several hundred volts.

Bob S

8. Oct 17, 2010

### Armando J

What other factors would come into play if I were to water cool the coil? I'm thinking of useing an antifreeze-water mixture, since that won't freeze at something below zero (farenheit), like water, and immersing the coil. I'm not an expert (hardly even a novice) on thermodynamics, but immersion is a fairly decent way to cool something, right? I wouldn't want to seperate the coils to allow more surface area for contact with the mixture, because that would cause a lesser turn density as well as some non-uniformities in the field, right? If I were going to use this coil of mine to make an elcetromagnet, what would the effect of the diameter of my core be? Say, a four inch diameter core as opposed to a half inch core (I'm thinking of useing a soft magnetic alloy for the core, if I can get some. about 50% iron and 50% nickel). Does the diameter strengthen the field? Does it increase the flux density? Also, at the pole of an electromagnet such as this, is the field completely uniform? Thanks for any help.

9. Oct 17, 2010

### Bob S

10. Oct 17, 2010

### Naty1

biggest one is the application and cost.....you have to circulate water and cool the water for continuous application...like an engine radiator and there are surely other design choices which are preferable...

you should be able to answer you own remaining questions since you were smart enough to ask them...like core diameter....think really really big, like wrapping wire around the earth and moon....how strong would the field be...what happens to the cost of wire as diameter grows....its obvious....

My guess is that a lot of solenoid coil design ends up being trial and error and experience...how to place contacts, material to use for contacts, and all the pieces BobS mentions...I've bought a few over the years...one to activate a 120 volt diesel generator when a 12 volt refrigerator called for cooling, and another to carry some 800 intermittent amps at 12 volts to combine batteries for 435 HP 8V71TI Detroit Diesels....I sure did not bother trying to design those myself.....

which reminds me...another criteria is continuous or intermittent duty...marine solenoids, at least, always include that specification....an intermittent rated solenoid would likely fail in a continuous duty application....I'm pretty sure they figure that out from actual testing not plain theory...there are so many variables....ambient temperature being another one...

11. Oct 18, 2010

### Armando J

Okay, so I've concluded that a core with a larger diameter would make a greater field (By which I mean that an object would be more strongly attracted to it) with the help of my physics teacher. But how does this play into the B= (mu not)x(I)x(turn density) equation? Or does it not? How would one go about determining the strength of a magnet without knowing the strength of any other magnets (without a gaussmeter). same question, same restrictions, but I'm looking for mu not.

12. Oct 19, 2010

### Bob S

This url gives an equation for the on-axis magnetic field (flux density) B for finite-length air-core solenoids:

http://www.netdenizen.com/emagnettest/solenoids/?solenoid

The on-axis field of multi-layer solenoids can be estimated by summing the field contribution of individual layers. There is a section in this url that discusses the effect of conductor size and resistance. There are other websites that have similar equations, some with online calculators, like

http://www.vizimag.com/calculator.htm

Bob S

13. Oct 21, 2010

### Armando J

Thanks for the help!