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# Does an inductor start with a magnetic field, or does it only retain one as long as it is powered? Answered

I am trying to figure out how to keep a reed switch on by proximity to a magnet. I read that an inductor stores the current as a magnetic force, so I was planning on just having the reed switch close to the inductor, and just use a small magnet to close the circuit initially. i was hoping the field from the powered inductor would keep the switch closed. Am I on the right track?

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A reed switch
http://en.wikipedia.org/wiki/Reed_switch
closes when it is in the presence of a magnetic field, and the source of that magnetic field can be a permanent magnet, an electromagnet, or some combination of permanent magnets and electromagnets.

A reed switch and an electromagnet together is called a reed relay.
http://en.wikipedia.org/wiki/Reed_relay

To be clear, what I mean by "electromagnet", is a magnet that you, the experimenter, can turn on and off electrically.  The strength of its magnetic field is directly proportional to the electric current running through it.  When the current running through this electromagnet is zero, the field produced by this electromagnet is also zero.  Or that's how an "ideal electromagnet" would behave.

As you point out, real electromagnets have this property of "inductance", which basically says that the current through the electromagnet has something like inertia.  Its current cannot change instanteously. Mathematically this is VL = L*(dI/dt).   When turning the electromagnet on, the external circuit must do work, W=(1/2)*L*I2, on the inductor to build up the magnetic field, and conversely when turning the electromagnet off, the inductor must do work upon the circuit, to get rid of the energy in the magnetic field.

If you're using a transistor to do the switching, the usual trick is to place a "freewheeling diode", in parallel with the electromagnet. This circuit gives the energy stored in the inductance a safe place to go. Without the freewheeling diode, the stored energy goes into a high voltage pulse that can destroy the transistor. More here:
http://en.wikipedia.org/wiki/Freewheeling_diode

Besides the fact that you cannot turn an electromagnet on or off instantaneously, the only other minor "problem" you might encounter is that of remanence.
http://en.wikipedia.org/wiki/Remanence
That is to say for some magnetic materials, a non-zero magnetic field remains even after you've turned off the electric current.  However for a "soft" magnetic magnetic material,e.g. ordinary carbon steel, this remanent field will be small compared to the field when the current is on. By the way, remanence is a property of magnetic materials. Thus an air-core electromagnet will not exhibit remanence.

Also worth mentioning: permanent magnets are made from materials which posess a large remanent field, and that's sort of the point. The field remains, "permanently", even though there's no externally supplied electric current.  The other part of "permanent" is high coercivity, which means your permanent magnet is difficult to demagnetize. I.e. you cannot dammage it just by bringing other magnets or electromagnets near it.

Back to your problem with the reed switch. Typically these switches are open for zero magnetic field. So bringing a magnet close, will make the switch close.  However you can also stick the reed switch in the middle of a null created by two permanent magnets.  That way the switch will close when you take one of the magnets away. This is explained, sort of, in this instructable:
https://www.instructables.com/id/Make-a-Magnetic-Reed-Switch-Work-Backwards-sort-o/

If you like, one of sources of magnetic field surrounding the reed switch can be an electromagnet, which is what I said in the beginning of this answer. You can make a reed relay that's normally open, or normally closed, just by sort of tweaking the magnetic field around it with a permanent magnet.

ok, assuming I have the copper wire to make an electromagnet, is it entirely necessary to have a ferrous core in it? I am planning to put this circuit in a piece of clothing, and the last thing I want is to go through security with a nail on me.That and to save space I was thinking about wrapping the reed switch in the wiring itself. I could not tell you the gage the of the wire, but I can tell you that it is enameled and comes from one of those cheep hobby motors

I think that is the way reed relays are constructed. That is, just a whole bunch of turns of wire wrapped around the hollow cylinder containing the reed switch. If so, the only thing ferrous in there is the reed switch itself. Take a look at this picture:
http://en.wikipedia.org/wiki/File:Reedrelay.jpg

Regarding the question of how much wire, and what kind of wire, and how much current, the wiki article,
http://en.wikipedia.org/wiki/Reed_switch
gave a quote of "10 to 60 [ampere*turns]", and what that means is... For example suppose the actual sensitivity of your reed switch is 50 ampere*turns, then that's equivalent to:

100 turns, and a current of 0.5 A
1000 turns, and a current of 50 mA
10 000 turns, and a current of 5 mA

That is to say, with more turns you need less current to produce the same magnetic field, and it's the magnetic field that moves the reeds.

I'm not sure why you want to roll your own, I mean wind your own, when there are plenty of good reed relays out there that come with wire already wrapped on them.

I suspect the reed relays that are out there tend to use fine wire and 1000s of turns of it, so that the necessary current is in 10s of milliamps.  Not sure what that works out to in terms of resistance and voltage, but I've got a bunch of reed relays in my junk box that have "700 Ω" printed on them.  Guess 4.9V = 700 Ω *7 mA.  Something like that.

an inductor is like an electromagnet that cancels itself - think a straight bar electromagnet, now bend the N and S pole to touch, and keep the windings going all the way around. It inhibits CHANGE in current, but allows linear current to flow.

A reed switch doesn't generate power, its merely opens and closes in the presence of an appropriately oriented magnetic field -- like a relay does, but the coil is 'outside' the reed in this case.

Someone correct me if I'm wrong, but I don't think an inductor would actuate a reed switch - the point of an inductor is just that, to stay self contained. If it sent magnetic fields flying in all directions, it would interfere with other parts of a circuit.

If you want to just have a positive feedback loop where a magnetic field keeps the reed closed once the initial magnet passes, then use an Electromagnet -- wind wire around a ferrous core, and use the reed to power the coil. Initially it will be open circuit, but the magnet passing will close the circuit, and the electromagnet will LATCH the circuit closed, feeding back on its own state.

If you just want a circuit that stays on once something triggers it, you are looking for a Latch. There are a hundred kajbillion ways to make a latch, from transistors, relays, or in this case, a reed switch -- so long as one part of the circuit can activate another part of the circuit, you've got it. Along the same lines the search terms 'flip flop' or 'bistable flip flop' come to mind

technically, any coil is an inductor, at least when compared to a resistor or capacitor. The same basic "mechanics" apply to a solenoid as to an inductor that might be used in a circuit, it's simply that the former uses an air core, whereas the latter may or may not use an air core to store the magnetic field. Of course, in an inductor, that's the whole purpose, to act as the electronic equivalent to a flywheel in mechanics. Spin it up and it wants to keep going. In the inductor, that means, get current flowing, and it wants to stay flowing even when the power source has been removed. Remove power, and it will either continue to release energy store in the magnetic field until it is spent, or develop a voltage that will cause current to continue to flow (a spark, for instance)

the inductor initially has a magnetic field of zero. Once a voltage is applied, it begins to generate a magnetic field. Once the time constant has been met (a finite "spin up" time), it achieves it's steady state magnetic field. Once power has been removed, the inductor begins to lose its magnetic field until, after a the time specified by the time constant, it drops back to zero..

Thank you, I think this has been one of the best explanations of an inductor I have found.