Electrically speaking, is there a difference between solid wire and cable?

Factual Questions
1

I ask because I’ve been mucking about trying to make a solenoid, using cable from an old power cord, and I’m not getting any results. All the websites I’ve seen make this seem like a middle-school level project, and I like to think I’m at least as competent as a 13 year old, so I was wondering if the cable was my problem.

Of course, it also may not help that I don’t have a voltmeter. But the batteries I’m using (AAs and 9v) are brand new, and at least the AAs work in my cordless mouse, so I don’t think that’s the problem…

2

Not for DC. Stranded wire such as found in typical “zip cord” used for lamp and small appliance power cords is very, very slightly more efficient at high AC frequencies because of something called the skin effect. At line frequency (60 Hz in the US and 50 Hz in Europe) and certainly at DC, there is no difference electrically. In the case of line cords, stranded wire is used not because it’s any more efficient–as noted, it isn’t–but because it’s more flexible.

Now, on to your solenoid. How many turns are you using? Typical 18 AWG power cord is pretty thick, relatively speaking, and it’s hard to get a large number of turns in a reasonable volume, so you may be underwinding. For a good solenoid at ~9 VDC you’ll need to get at least a couple hundred turns on your form; otherwise your magnetic field strength won’t be high enough to pull the iron in. I’d go with magnet wire, which is designed just for this purpose (hence the name). You can get small rolls of it at Radio Shack and similar places.

3

Solenoid? AC or DC? What’s the pull-in voltage?

If you’re trying to energize a solenoid with a battery, and it’s not working, it’s probably because the battery has too much internal resistance. (The open-circuit voltage might be O.K., but the battery’s internal/series resistance might be too high.)

To correct this situation when using a battery, you’ll need to use a battery with a lower internal resistance (e.g. NiCds), parallel batteries, and/or use a higher voltage. You must be careful with the latter, however, as too high a voltage might over-heat the coil during steady state. You may also want to experiment with using a high capacity/low ESR capacitor in parallel with the battery. The battery + capacitor would supply the pull-in current, and the battery (alone) would supply the steady state power.

4

Oops. Didn’t see that you were trying to *build *a solenoid. In that case, pay attention to Q.E.D.'s post.

5

You need more oomph. I remember making a solenoid as a kid, and it used to drain a 6v lantern battery in a few minutes. A 9v battery is hopeless, and a AA (or several) isn’t much better. Try 4 D cells in series, and be prepared to have it get HOT. It will work.

6

Hum. So, I tried again using cable from a cell phone charger - it’s much smaller, so I got around a hundred loops, but it’s still not doing anything. With beowulff’s post, I’m getting worried that I’ve simply discharged my 9v battery without noticing, although given that it hasn’t worked at all, I don’t really see how that would work.

Anyways, here’s the setup, let me know if I’m doing it wrong - I’m using a ballpoint pen for a tube, and a bolt for the plunger (I did check that it’s ferrous!). I’ve got one piece of cable spiraling up the inch or so of tube I’m using - I get about 25 loops going up. Then I’m running it back down, still going clockwise, twisting it into another piece of wire, and going back up and down. If I recall my physics correctly, the wire’s generating a circular magnetic field, and coiling means the ‘down’ bits are both in the center (depending on which end is positive, of course). So, I’m always going clockwise around my tube looking down. But, since I’m spiraling it, there is a slight tilt going up and an opposite tilt going down. Is that going to cause problems? The only other thing I can think of is that my connections to the battery aren’t very good (I’ve tried looping around the 9v’s terminals, just holding the wires up to them, taping them on, and putting a loop inside the terminals, with no effect). It’s been a really long time since I’ve done anything with electricity (sophomore year of high school?), help me out here!

Alternatively, where can I get a really, really small solenoid? Like, smaller than a AA battery?

7

The winding geometry sounds about right, although it can probably stand to have a few more turns. You probably did, however, drain the 9 V battery with your previous experiments (touch the terminals to your tongue; if you don’t feel a strong tingling sensation, it’s kaput.) Even though the iron didn’t pull in, there’s still current flowing–quite a lot of it, too, since the DC resistance of a few feet of lamp cord is much less than an Ohm. As beowulff suggests, try it with a 6V lantern battery, or alternatively, stop by a Radio Shack and pick up a 4-cell D-battery holder.

8

You’ll get a lot more turns in the same space with magnet wire, as the insulation is only painted on. Conventional hookup wire with plastic or fabric insulation have very thick insulation compared. Just be careful not to nick the coating.

9

Many more turns of smaller gauge wire will probably do the trick.

10

Is it possible you did this:

Strip both conductors at each end of a two wire cable. Twist the two conductors at One end together. (X&Y) Wind the pair of conductors into a solenoid. Then apply the battery between the two separate conductor ends that were at the same end of the cable? (A&B)

A-------------------------------------------------------------X
B-------------------------------------------------------------Y

In that case, You have created two coils, and have created equal but opposite current flow in each. There is thus zero net magnetic field. This is why coiling an extension cord does not create a problem due to magnetic fields. (It concentrates the heat, which is why it is discouraged)

If you instead connect A&Y, and apply the battery to B&X, then you have current flowing the same direction in both coils, and it should work if you have enough turns.
To get high current in few turns of wire, you need low internal resistance in your battery, as mentioned above. One way to do this is to place a bunch of D cells in parallel. This will still give you 1.5 volts, but with N times more current capacity. Use house brand alkaline cells, and get them on sale for cheap. I used to make up such batteries to provide starting power for the glow plugs used on model airplane engines. (the batteries and #6 cells intended for such were expensive, and never went on sale)

11

Paralleled NiCd’s may be best for this application. While they don’t have the energy capacity of alkaline batteries, they have a *lot *less internal resistance (all else being equal). Using paralleled NiCds along with many many turns of small gauge wire should do the trick.

12

Ok, so the 9v still has charge - definitely a tingle there.

I’ll have to get over to RadioShack and get some magnet wire - in the meantime, can anyone help me figure out what determines the solenoid’s strength? Looking at the wiki article, I get the formula B=μ*NI/h. Wiki describes B as the path integral of loop b, which I totally don’t understand, but am assuming to be an indicator of strength? μ appears to be a constant, and N and I are number of loops and current, respectively. I don’t know what h is (wiki says height of loop) but maybe the size of the loop? That would mean bigger loops would be weaker, which makes sense, I think.

Anyways, so it looks like the variables I can control are N, I, and maybe h (if my assumptions are correct). I’m think this is for a solenoid of arbitrary length, so N is going to be the number of layers of wire wrapping? Which means more wire is good. For the power source, current is the issue. But, wouldn’t more wire increase the resistance, and lower current? I don’t know the formula for that, but is there a point of diminishing returns? Like, at some point you’ve got so many loops that your power source is mostly just heating up the wire? Bear in mind, this is half-remembered sophomore physics, here.

The upshot of all of this, I think, is that if I want to keep my battery small (and I do) I should a) use a battery with low internal resistance, like a NiCd, and b) use more loops, which means thinner, longer wire so I can pack it tighter. But should I just go crazy and cram as much wire in there as possible, or is there an ideal amount of wire? Also, I should try and make the inner diameter as small as possible. Is that correct?

Oh, and if I have a choice between a one-inch long solenoid with 5 layers of loops, and a half-inch long solenoid with 10 layers (or maybe 9, with the larger diameter) of loops, which will be stronger?

Finally, I’m only interested in making a solenoid for intermittent use (What fun is something that just stays still? Moving is much neater). So, would it be worth bringing capacitors into this? Or would that just make my head explode?

13

B is the magnetic field strength, μ is the magnetic permeability of the core, N and I are, as you correctly surmise, number of turns and current and h is the length of the winding. IIRC, entering μ in H/m, I in amperes and h in meters gets you B in Tesla if that helps any.

Caps are likely not necessary here, although to protect the circuit, you probably want to connect a diode (a general-purpose silicon such as a 1N4001 or 1N4002 will do nicely) in parallel with the solenoid winding such that it is reverse biased when DC voltage is applied. This will dissipate the high inverse voltage spike which will appear across the coil when it is de-energized.

14

You realize I have no idea what you just said, right?

15

Basically, when you turn off your diode there will be a large spike caused by the magnetic field collapsing. A diode is used to dissipate the energy; diodes (the ones referred to here at least) allow current to flow in one direction only.

This link explains it fairly well.

16

Just a thought, but it’s possible the battery and the coil are nowhere near appropriate for each other. That is, their impedances may be so different that very little power is getting from the battery to the coil.

You could measure the resistance of the coil with an ohmmeter, or you could calculate what it should be by looking up the resistance per meter (or foot) of your wire gage and multiplying that by the number of meters (or feet) of wire in the coil.

The coil will mostly consume power, when it’s on, by being a resistor and wasting the power as heat. A resistor the size of a AA battery could waste perhaps 3 watts and still not be too hot to manage. So, a nice power supply for a coil that size would deliver 3 watts into whatever resistance the coil has. At Radio Shack they may have “power resistors”, which are resistors intended to dissipate more than about a watt. Look at how big they are and how many watts they’re rated for. Anything that size would dissipate that many watts and, while getting too hot to hold, would not get hot enough to scorch a printed circuit board or damage plastic parts near it. Let this be your rough guide to coil power dissipation.

My unquantifiably tested impression is that coils the size of AA or C cells that get hot enough you don’t want to keep holding them would make a ferrous bolt jump very excitedly around, but the further you go from that size or temperature, the less exciting the result.

If Radio Shack sells wall warts or desktop power supplies, you would get the most activity in a coil by winding it so its wire resistance would give almost the power supply’s maximum current at its voltage.

17

When you turn off the coil, the magnetic field collapses. When this happens, it induces a voltage which is opposite in polarity to the original applied voltage. How large this voltage is depends on how fast the field collapses and the inductance of the coil. To sort this voltage to ground, you connect a diode between one terminal of your coil and the other. Diodes only let current flow in one direction (as long as the peak inverse voltage rating isn’t exceeded) so you connect the diode so the terminal with the stripe (called the cathode end) painted on it is on the positive terminal when the battery is connected and the other end (obviously) on the negative terminal. Current will only flow through the diode when the applied voltage is positive on the anode (the non-stripe end) and negative on the cathode. When connected as I described, no current will flow through the diode when the battery is connected and all of it will flow through the coil, which is, of course, what you want. When you disconnect the battery, the large induced voltage is opposite in polarity which turns on the diode to conduct, so most of the current flows through diode until the voltage drops to about .7 volts–the turn-on voltage for typical silicon diodes. Clearer?

18

To add to what Q.E.D. said:

The magical equation for the solenoid is v = L di/dt, where v is the voltage across the coil, L is the inductance of the coil, i is the current through the coil, and t is time.

Let’s say you flip the switch to energize the solenoid. After a few milliseconds the armature pulls in and the current becomes steady state (DC). Life’s good. But things get ugly when you de-energize the coil.

Let’s assume the steady state current through the coil is 2 A when it is energized. When you open the switch, you are trying to force the current to go from 2 A to 0 A in zero time. Stated another way, you are trying to force the change in current over time (di/dt) to be infinity. If we plug di/dt = infinity into the magical equation for the solenoid, we see that the voltage across the coil would go to infinity.

Obviously this is not what really happens when you interrupt the current through a coil. (You cannot have infinite voltage.) The circuit is pretty “smart,” it does not want to create “infinite” conditions, and therefore it does not “allow” you to change the current in zero time. How does it do this?

When the switch is closed there is no voltage across the contacts (even when the solenoid is energized). Now start opening the switch… when the distance between the contacts is just a few angstroms, the voltage between the contacts will try to quickly skyrocket from 0 to infinity. But it will never reach infinity; an arc will form between the two contacts due to the high voltage across the contacts. For DC circuits, this arc will continue as the contacts get farther apart, and then eventually extinguish itself after the energy in the coil goes below a certain value. When you think about it, the arc must be there in order to prevent the current from changing in zero time, and to prevent the voltage from becoming infinity.

The arcing on the switch contacts will cause pitting and severely decrease their life. Hence the reason for putting the diode across the coil - it eliminates (or at least minimizes) the arcing across the switch contacts, thus making your switch last longer. And it’s not just the switch contacts… if the switch was designed to minimize arcing (e.g. vacuum relay) the arcing might instead occur across the coil, between individual coil windings, and/or between the wires.

19

Napier’s comments reminded me of something else:

It takes a lot more power to “pull in” the armature than to “hold” the position of the armature after it’s energized. This usually means there is an inrush current, and the driving circuit must allow for this overhead. I have even seen some clever designs that provide maximum power to pull in the armature, and then subtly “throttled back” the voltage after the armature is pulled in.

20

Much, thank you! I think I’ve got the basics, now I just need to go to Radio Shack and get some wire. Thanks for the detailed answers, folks.