# I want to build a solenoid.

Discussion in 'The Projects Forum' started by Ziggey, Nov 4, 2015.

1. ### Ziggey Thread Starter New Member

Nov 4, 2015
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So, I want to build a solenoid. I have been looking around and have not found an equation to use for multi-layer solenoids. I have used the equation B=μnI
μ=k(μo) k=200 (μo) = 4(pi)*10^-7

n = N/L
N=number of turns (86 a row 3 rows)
L = Length of core (8in) = .203m

I =1000

My problem starts with current. I'm using a transformer to first change my power into ac and second to bump it up to 120v. I have .12 ohms resistance in my coil. So with doing the math I came up with 1000 amps. Is this the correct way of doing it?

So if I continue I have

μ = 200*(4(pi)*10^-7) = 2.513*10^-4 t/amp m
n = 86/.203 = 423.645 turns/m
I=1000a (from above)

so if I put it all together I get

B = 2.513*10^-4 T/amp m *1000 amp * 423.645 T/m
B = 157.721 T

Is this correct? I feel like it isn't for some reason.
Thank you,
Ziggey

Jul 18, 2013
10,832
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More efficient if you use DC If this is an AC coil you have to include the inductive reactance, this is effectively in series with the resistance of the coil.
What is its function?
Max.

3. ### Ziggey Thread Starter New Member

Nov 4, 2015
15
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Well it's a small scale of my project. I need it to repel aluminum into the air. I believe that can only be done with ac....

Jul 18, 2013
10,832
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Small scale projects and you are talking 1000 amps!!?
The .1Ω you quote is most likely the resistance of the coil which is rather meaningless on its own when dealing with an AC supply.
Not sure of your actual application?
Max.

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5. ### Ziggey Thread Starter New Member

Nov 4, 2015
15
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Yes. The .1 ohms is the wire, but it is the only resistance in the circuit. I just took the voltage divided by resistance and that's hOw I got my amps. Like I said I was not positive if that's correct. I want to run my transformer of a battery. But was unsure of how/if it changed the amps in the circuit. So I just used the transformer as a power supply in my circuit. I'm fairly sure it's incorrect but I would like to correct the issue and continue do I asked for help

Jul 18, 2013
10,832
2,502
You cannot use just the resistance to calculate current in a AC inductive circuit, the coil has an inductive reactance component, also in Ohms, which is effectively in series with the wire resistance, the solenoid or coil is designed for a particular AC voltage level.
Max.

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7. ### Ziggey Thread Starter New Member

Nov 4, 2015
15
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So what your saying then to fix my problem I need to get the inductance of the coil ? How do I figure that into the equation then? Or is it way more complex than that?

8. ### strantor AAC Fanatic!

Oct 3, 2010
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I hope you're over simplifying your explanation here. You do know that transformers only work with AC input and AC output? (you can't supply a transformer directly from a battery)

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9. ### BR-549 Well-Known Member

Sep 22, 2013
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Draw us a picture.....................i. e. a circuit.

Jul 18, 2013
10,832
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The inductance will vary from minimum when not shifted to maximum after the armature has shifted over.
You can either find the inductive reactance empirically at a certain voltage and frequency, using an ammeter, or measure the inductance and calculate for the frequency.
Max.

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Jun 22, 2012
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12. ### wayneh Expert

Sep 9, 2010
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Could be. I just saw one like that in the Museum of Science and Industry in Chicago.

13. ### Ziggey Thread Starter New Member

Nov 4, 2015
15
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@ Strantor I'm sorry I have mislead you I have a power inverter.
@Max ok thank you, I have found the equation that I will use to find the inductance and then figure out the amps going through my circuit with that. I will then redo my equation for figuring out the magnetic strength of my coil.

@Dodgydave yes that is essentially what I am building. I just want to build it with a core so it will intensify the strength of the magnetic field.

@BR All I have is a simple AC inductor circuit I will get you a picture in a few bit.

14. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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Thats how an Electric meter works, using eddy currents to spin an Aluminium disc,

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15. ### Ziggey Thread Starter New Member

Nov 4, 2015
15
0
I did some snooping around the internet and came across two formulas.

XL=2πfL I believe this would be used to figure out how much "resistance" my inductor will create, but I believe that I need to use this equation (L=μAln^2) to figure out the inductance I need for the first equation.

Then one I have that figure out I should be able to use the resistance created by my inductor to accurately calculate how much current is going to my circuit. Once I have that current I should be able to plug that number into this equation B=μnI and I will be able to figure out the EMF of my inductor.

So with all that being said does this all sound mathematically correct?

16. ### Papabravo Expert

Feb 24, 2006
10,340
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Not quite. The total resistance of an inductor has a real part that is not dependent on frequency as well as an imaginary (reactive) part that is dependent on frequency. So the total impedance of the inductor will be:

$Z_L = R_L + jX_L=R_L+j 2 \pi fL$

You can measure the DC resistance, with an ohmmeter, or calculate it from the ohms/unit length of your wire times the length of n turns.

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17. ### Ziggey Thread Starter New Member

Nov 4, 2015
15
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Ok so just to make sure I need to add in the resistance of the wire and then I will have the total resistance of my inductor...

Well thank you for the help.
Ziggey

18. ### Dodgydave AAC Fanatic!

Jun 22, 2012
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Oh yes, the thicker the wire the lower the resistance will be, and the opposite will also apply,

Wire size depends on how many turns you can get wound on the former, thick wire may mean less turns, thin wire more turns...decisions decisions.

Jul 18, 2013
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In practice using AC the wire resistance will be fairly insignificant AFA current is concerned.
Max.

20. ### wayneh Expert

Sep 9, 2010
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Agreed. The inductance/frequency component to the impedance will generally dominate. Every wall wart transformer relies on this: They would explode if hooked to DC at the same voltage as the rated AC voltage. In reality they draw almost no current (if not loaded on the secondary), because of the impedance due to their high inductance at AC line frequency.

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