A current I creates a magnetic field that exerts forces on other currents in the field, but not on the current I itself.
On the other hand, if the current is through a coil, then it has magnetic effect on itself, through self-induction.
Isn't this strange?
Not very strange at all, if you really understand Faraday's law.
The first statement is incorrect.
| A current I creates a magnetic field that exerts forces on other | currents in the field, but not on the current I itself. | | On the other hand, if the current is through a coil, then it has | magnetic effect on itself, through self-induction. | | Isn't this strange?
No.
Actually, you can't isolate "a current" so simply. In a wire with current flowing through it, there are a very large number of parallel paths through which current is flowing. Each of these paths creates own field that both adds to the total field as well as exerts force on the other currents. The end result with a sufficiently large current would be a wire that squishes itself thinner. This force would also be acting in an arc to confine the arc to a finite width. Consider a large 500-pair telephone cable used as a single conductor by connecting all 1000 wires together at each end. Now consider a single conductor wire with the same total cross-sectional area. Both can conduct the current. Both will have a magnetic field around them. Both will have the self-induction force to try to squish the cable or wire to a smaller cross-section. Try a 10MA of current through these and see the effect (that's a big 'M', not a little 'm').
If you have 1 amp going through a 100 turn coil you'll get a certain level of magnetic field from that. Now run 100 separately connected wires going around the coil just once and feed them in parallel with 100 amps. You'll get the same field strength. Put just one big thick wire around it and run the 100 amps through. Again, same magnetic field strength. This is assumping the _changes_ in the current are not taking place. If that is happening, there can be slight or great differences depending on the rate of change (frequency of AC). To understand this, think of a transformer with two 120V 60A (7.2kVA) windings. They can be wired in parallel or in series and you get the same effect if you have 60A going through. You just have to apply 240V in the series case to get the 60A, and have to have 120A available in the parallel case to get the 60A in each winding.
Think of a measure called "amp*turns".
If course physical geomtry of coils and windings can be subtle effects on the total shape of the magnetic field and just how uniformly it all adds up.
Actually a straight wire with a current has a self-inductance as well, so the difference you imply isn't there.
Furthermore, self-induction is not a force on a current.
Clarity of terminology is the first task in any physics discussion.
PD
In alt.engineering.electrical PD wrote: | On May 13, 10:46 am, qwerty wrote: |> A current I creates a magnetic field that exerts forces on other |> currents in the field, but not on the current I itself. |>
|> On the other hand, if the current is through a coil, then it has |> magnetic effect on itself, through self-induction. |>
|> Isn't this strange? | | Actually a straight wire with a current has a self-inductance as well, | so the difference you imply isn't there. | | Furthermore, self-induction is not a force on a current.
It is a force against changing the current.
| Clarity of terminology is the first task in any physics discussion. | | PD |
This is an interesting thing to think about. With respect to inductance not magnetic field. The usual calculation with inductance assumes that if you have one fat wire then it has a certain inductance. But if one subdivides the fat wire (current sheet) like a telephone cable, one now gets an inductance N squared times the single turn value, where N is the number of turns.
It's pretty clear that this gives decent results if the coil diameter is larger and the wire bundle is quite compact, but is it generally valid? I note that if you compare a telephone cable coil where all wires are in series ( more or less the standard inductor case) the current in each individual wire is exactly the same. However, when all the wires are wired in parallel, one can no longer say that each individual wire is carrying the same identical current. In fact, one should be able to somehow calculate the current in each wire and plot a current distribution across the cable. The same thing should be possible for solid conductors that are not subdivided by simply correctly dividing them mathematically.
The bottom line here is that it must be true that fundamentally the series coil and the parallel wires case are NOT simply linked by an N squared factor unless one can justify that all wire currents in the parallel case are equal.
Benj
Self-induction is not a force although it is analogous to inertia. But there are forces in coils carrying current. A great demonstration is when a circuit is created from a spring made from copper wire is completed by dipping into mercury. You get a bouncing spring.
Bill
-- Fermez le Bush--about two years to go.
No, not really. Mass isn't a force resisting a change in velocity, either.
In alt.engineering.electrical Benj wrote: | | snipped-for-privacy@ipal.net wrote: |>
|> Actually, you can't isolate "a current" so simply. In a wire with current |> flowing through it, there are a very large number of parallel paths through |> which current is flowing. Each of these paths creates own field that both |> adds to the total field as well as exerts force on the other currents. The |> end result with a sufficiently large current would be a wire that squishes |> itself thinner. This force would also be acting in an arc to confine the |> arc to a finite width. Consider a large 500-pair telephone cable used as |> a single conductor by connecting all 1000 wires together at each end. Now |> consider a single conductor wire with the same total cross-sectional area. |> Both can conduct the current. Both will have a magnetic field around them. |> Both will have the self-induction force to try to squish the cable or wire |> to a smaller cross-section. Try a 10MA of current through these and see |> the effect (that's a big 'M', not a little 'm'). | | This is an interesting thing to think about. With respect to | inductance not magnetic field. The usual calculation with inductance | assumes that if you have one fat wire then it has a certain | inductance. But if one subdivides the fat wire (current sheet) like a | telephone cable, one now gets an inductance N squared times the single | turn value, where N is the number of turns. | | It's pretty clear that this gives decent results if the coil diameter | is larger and the wire bundle is quite compact, but is it generally | valid? I note that if you compare a telephone cable coil where all | wires are in series ( more or less the standard inductor case) the | current in each individual wire is exactly the same. However, when all | the wires are wired in parallel, one can no longer say that each | individual wire is carrying the same identical current. In fact, one | should be able to somehow calculate the current in each wire and plot | a current distribution across the cable. The same thing should be | possible for solid conductors that are not subdivided by simply | correctly dividing them mathematically. | | The bottom line here is that it must be true that fundamentally the | series coil and the parallel wires case are NOT simply linked by an N | squared factor unless one can justify that all wire currents in the | parallel case are equal.
Maybe another way to think about it is "total cross section current". If a winding with 50 turns has a current of 3 amps end to end, that's a total of 150 amps "total cross section current".
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