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Coil EMF volt vs current

Discussion in 'Electronic Design' started by Mike Lennis, May 3, 2007.

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  1. Mike Lennis

    Mike Lennis Guest

    Given the same solenoid, and assuming a fixed available wattage, what
    proportion of 1) voltage vs. 2) current produces the highest possible
    field strength?

    How is this determined?

    Thank you,

    Mike Lennis
     
  2. Sound like a homework question!
     
  3. Guest

    Field strength is entirely determined by the current through the coil.
    The voltage across the coil reflects both the rate of change of
    current, which generates a voltage across the inductance of the coil,
    and the instanteous current through the coil which generates a voltage
    drop across the resistance of the coil.

    At very high frequencies, one has to take into account the current
    flowing throuhg the capacitance between the turns of the coil, which
    by-passes both the resistance and the inductance of the coil, and has
    no effect on the filed strength.

    This is a rather ill-posed question, which suggests that it isn't
    homework.
     
  4. Greg Neill

    Greg Neill Guest

    What determines the steady-state current in the
    solenoid? (What are the electrical properties of
    the solenoid?)
     
  5. Phil Allison

    Phil Allison Guest

    "Mike Lennis

    ** The highest current level that is within the available wattage is it.

    So, with coil resistance R and wattage W

    W is given by I squared R

    So I = sq rt ( W/R )

    And V = I.R



    ........ Phil
     
  6. MooseFET

    MooseFET Guest

    This is not really correct. The current that passes through a
    capacitance between the windings still makes a magnetic field. It is
    just a lot smaller because it travels a shorter path.


    The rules are straight forward. When a current flows, it makes a
    magnetic field around its path. The effects of multiple current flows
    add linearly.
     
  7. John Larkin

    John Larkin Guest

    The coil has a resistance, and that resistance fixes the relationship
    between the current and the voltage. So you can't vary them
    independently, so your question is sort of meaningless.

    As the voltage goes up, the current goes up, and there is a single,
    unique voltage:current that dissipates the available watts.

    John
     
  8. Phil Allison

    Phil Allison Guest

    "John Larkin"
    Mike Lennis

    ** Looks like a cute, trap question to bugger gullible students.


    ** As given by the most basic equations for DC power and ohms law.




    ........ Phil
     
  9. Mike Lennis

    Mike Lennis Guest

    Yes, very sorry. The fixed coil resistance was an error on my part.

    What I am driving at is this. Given the same signal source, consumed
    wattage can be increased by 1) lowering the coil resistance OR 2)
    applying higher voltage.

    Which of the two is preferrable for maximum field strength?

    For example, in the extreme case, one could have microvolts (low
    resistance coil) and tons of current, or heaps of voltage (high
    resistance coil) and negligible current. Yet the wattage is the same.

    What is the trade-off point for these two variables, ie. practical
    solution?

    Thanks again.

    Mike Lennis

    Thanks again.

    Mike Lennis
     
  10. John Larkin

    John Larkin Guest


    Ampere-turns make magnetic field. For a given available wire winding
    volume, the external surface area determines the cooling, so the watts
    you can dump into a given coil geometry, for a given temperature rise,
    is fixed.

    So, imagine you replace a N=100 turn coil with a 200 turn coil. The
    cross-section area of the wire must drop by 2:1 to fit into the same
    space. Resistance per foot of wire doubles, and you have twice the
    wire length, so net resistance goes up by 4:1. Since power is I^2 * R,
    we need to drop the current in half to get back to the same power
    dissipated. So N*I is unchanged!

    So, it doesn't matter. In real life, very fine wire will lose more
    winding area to the insulation, so you may start to lose pull with
    lots of very fine wire.

    Square wire helps, both to pack more copper into the available space
    and because the heat flow will be better. Anything to improve cooling
    will allow you to dump more watts into the coil, too. High temperature
    insulation will allow more watts, too.

    Then there's the old Austin-Healey overdrive solenoid trick: apply
    lots of power to pull in the solenoid, and then once it's seated, back
    off to some lower holding current.

    John
     
  11. MooseFET

    MooseFET Guest

    Make the coil resistance zero by making it super conducting. Infinite
    voiltages are much harder to do.
     
  12. Rich Grise

    Rich Grise Guest

    Unless, of course, you've got some superconductor wire. ;-)

    Cheers!
    Rich
     
  13. BFoelsch

    BFoelsch Guest

    Y'all miss the point.

    A magnetic field is potential energy. Once the field is created it
    exists, that is why a permanent magnet works. It takes no energy to
    maintain the field. Hence, "consuming wattage" in this case means
    nothing.

    Assuming we are talking about DC fields here, all that matters from
    the solenoid point of view is ampere-turns. For a fixed voltage, given
    a wire of uniform resistance, the MMF is independent of the number of
    turns, but the consumed power is inversely proportional to the number
    of turns. Hence, (for a fixed field voltage)a solenoid with an
    infinite number of turns will develop the same field as a solenoid
    with one turn, but with infinitesimally low power consumption. It
    will, however, also have infinite inductance, so the field will take
    an infinite amount of time to build up.
     
  14. John Larkin

    John Larkin Guest

    If the winding area is constant, increasing the number of turns by a
    factor of N increases the resistance by N^2. Since power is E^2/R,
    then for constant E, power falls as N^2, not N. Field strength goes as
    1/N.
    That assumes you don't change the wire size, so it will also be
    infinitely large.

    John
     
  15. BFoelsch

    BFoelsch Guest

    True enough. I was not concerned with constant area. I presupposed a fixed
    wire size, too.
    Yup.

    I was simply trying to show that steady-state power consumption is no
    predictor of field strength. It is all wasted, once the field is created.
     
  16. jasen

    jasen Guest

    no, power consumption is about the same (unless the solenoid is larger)
    It'll also have infinite resistance, beacause the wire's thinness and length.

    More turns won't make solenoids more efficient, more copper can but that
    has limits.

    Bye.
    Jasen
     
  17. MooseFET

    MooseFET Guest

    Ignoring build up:

    If you hold the pounds of copper constant, changing the number of
    turns doesn't change the power. The resistance of the windings runs
    as the square of the number of turns and the required current
    decreases proportionally so I^2*R remains the same. It is better to
    think in terms of how much conducting material you can afford.

    Thinking about build up:

    The outter most turn can be quite a bit further from the coil center
    than the inner most. Adding turns to the outside of a coil increases
    resistance more rapidly than N because they are bigger turns. For a
    very thick coil, the resistance grows as N^2. The field from each of
    the new turns is decreasing by a 1/N curve.

    Holding current constant, we can find the rate that power and the
    field increases with N.

    <=> is proportional to.

    P = I^2 * R <=> N^2

    Field <=> integral(1/N) <=> ln(N)

    Therefor for a very thick coil:

    Field/P <=> ln(N)/ N^2

    For some large amount of wire, adding more turns stops helping. This
    is another good reason to go with super conducting wire.
     
  18. John Larkin

    John Larkin Guest

    Good points.
    Big electromagnets used to be water cooled and now mostly
    superconductive. The NMR magnets I work around have 20T fields, are
    stable to a ppm field per day, and have zero power consumption.

    John
     
  19. Phil Hobbs

    Phil Hobbs Guest

    Scaling arguments--fun.
    This assumes that the core is a small perturbation on the inductance of
    one turn at the outer edge, which is true for open cores but not for
    closed ones. For closed-loop cores, the field is determined by the
    total current threading the loop. Position doesn't matter, at least not
    until order (iirc) 1/mu**2. If "solenoid" here has the physicsy meaning
    of "a coil wound like a Miniductor", this is true, but if it means "a
    push-pull magnetic actuator whose magnetic circuit is nearly closed at
    all times" then it isn't.

    Cheers,

    Phil Hobbs
     
  20. MooseFET

    MooseFET Guest

    To add more turns, once the core is full you need a bigger core. You
    get into the same bind with a core once you start going to larger and
    larger cores. All the turns get longer when you go to a bigger core.
     
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