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B Non-ferrous solenoid actuator -- Force calculation

  1. Jun 2, 2017 #1
    I'm trying to come up with a lightweight solenoid linear actuator, by constructing two electromagnets that telescope. The outer coil would be wrapped around a soda straw, and the inner coil would fit inside the soda straw and be wrapped around something smaller, like a bamboo skewer (which also serve as the actuator shaft).

    Well, I'm wondering about the pushing or pulling force this thing can impart. The force of an electromagnet on an object would be
    $$F=\frac{(N I)^2 \mu_0 A}{2 d^2}$$
    where ##N## is the number of turns, ##I## is the current, ##A## is the cross-section area of the coil, and ##d## is the distance between the open end of the coil and the object.

    If the object is another coil of a different diameter and windings, but with the same current, the force would be
    $$F = \frac{\mu_0 I^2}{2 d^2}(N_1^2 A_1 + N_2^2 A_2)$$
    This looks like it would blow up when ##d=0##. And what would be the force when one coil is partially inside the other? I would think it would be less, going to zero when one coil is completely centered inside the other.

    If my objective is light weight, is this even efficient, or would it be more efficient to use a strong magnet on the end of my rod?

    Also, if I'm limited to, say, 0.3 amps, it seems like I'd need a lot of windings just to get 30 or so Newtons of force, and that's right at zero distance, with hardly any force seen if the actuator moves even a little bit.
  2. jcsd
  3. Jun 4, 2017 #2


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    I think you will need a Lot of turns for that size current. The following video may help you with this issue.

    Attached Files:

  4. Jun 7, 2017 #3
    The current I chose is sort of arbitrary. I don't want to overload a lithium battery. But now I see that a 30C 2S 300mAh 7.4V LiPo Battery doesn't limit me to 0.3A, it limits me to 30×0.3A, or 9A as the maximum safe discharge current.

    In my quest to make something lightweight, I'm wondering if a torsional actuator might be better. I noticed that the magnets inside the disposable heads of my Sonicare toothbrush are incredibly powerful bits of magnetic material, about 6×3×1.5 mm, with the north-south axis perpendicular to the 6×3 face. They weigh about 1 gram each. If I stick two of them together I have a lot of trouble pulling them apart. Maybe I could put one of those on each side of a bamboo skewer, with a coil on each side, and attach the skewer to a pushrod horn.
  5. Jun 8, 2017 #4
    I just realized that the quantity ##NI## in the formula above is a constant for a given wire diameter, because wire has some resistance. 40 gauge copper wire has a resistance of about 3.5 ohms/m. Given a constant voltage source, the higher the value of ##N##, the higher the resistance, which decreases the current ##I## proportionally. When I rewrite the force equation in terms of voltage and wire length, a bunch of stuff cancels out and I'm left with
    $$F=\frac{V^2 \mu_0}{8\pi\Omega^2 d^2}$$
    where ##\Omega## in this case is the resistance per unit length.

    This means that for a given input voltage ##V## and distance to object ##d##, the force is determined completely by ##\Omega##. The number of turns and the diameter of the coil are irrelevant!

    The only reason to have a large number of turns would be to reduce the current to preserve the battery.

    For a 7.4V battery, a 40 gauge wire coil at 3.5 ohms/m, at 1mm from an object, would exert a force of 0.24 N regardless of the number of turns. I need more force, so it looks like 40 gauge wire won't work at all.

    If I want a coil that exerts 3N of force (about 2/3 lb) I need to go to a heavier gauge magnet wire. 31 gauge, at 0.43 ohms/m, would give me 3.3N. Then if I wanted to limit the current to 9 amps, I'd need about 2 meters of wire, weighing 3.6 g.

    I may be able to increase the force and lighten the weight if I used the Sonicare toothbrush magnets I mentioned earlier.
    Last edited: Jun 8, 2017
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