Inductor/Electromagnet used to lift up a magnet? Joules to Newtons or Joules to  - Page 1
Forum section: electronics / beginners

author: renzoms
author_rank: Regular Contributor
timestamp: February 03, 2021, 02:27:48 am
content: With interest in using an electromagnet to lift up a magnet.I dove in. I will state general information. The set up is unknown but I'm interested in the physics/electrical engineering involved in calculating the amount of ______ needed to lift up a mass (magnet with magnetic strength x).I reviewed "electromagnets". I see an inductor produces a magnetic field, a summation of the contributions of electrons and we get some equations.Regarding inductors, Energy is .5LI². We can adjust the power/energy/magnetic flux in an inductor by varying the voltage/current/turns, etc.Also since V = dϕ/dt, maybe we can integrate velocity to find the magnetic flux or something but idk if magnetic flux is relevant to lifting a magnet on the ground.So I found a link to the an equation expressing the force between two magnets in Newtons homework and exercises - calculate force between two magnets - Physics Stack ExchangeI see magnetic force is measured in Gauss.I've nearly finished relearning Classical Mechanics, so I'm thinking about using F = ma, and I'm not sure how the inductor/"electromagnet" ties into this nicely.I don't know how to convert an inductors properties into Gauss, nor express it as a function of distance, and I think you guys can help!I considered using F = kqq/R or some other attraction law, but still idk how to relate an inductor and it's Joules or other properties to "magnetic strength" I can use in a kinematics analysis.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: February 03, 2021, 03:12:40 am
content: I don't know how strong your math is. The general case requires some fairly serious math. But the basis of it all is fairly simple. The force on a segment of wire (dL) carry a current I in a magnetic field of strength B is given byF=ILXB (I times L cross B)This is a vector equation, L and B are both vector . For a coil you integrate this over the length of the coil (not the the vector changes at each length increment).

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 03, 2021, 05:38:32 am
content: Is this lifting a mass purely magnetically, or turning on a magnetic field to clamp onto the mass then lifting it mechanically?And, does the magnet have any support or guide, or is it free?In both cases, there will be two energy costs, and probably a power cost. That is, some energy is required to begin performing the task, then power is required to continue lifting it. The two energy costs can be divided into magnetic field and mechanical work. The power is ultimately due to the fact that, maintaining the magnetic field, requires current flow, and presently, room temperature conductors have resistance, so drop some voltage, and voltage*current is power.A magnetic field does mechanical work, when a force is developed over a displacement: W = F.x. Force depends on a difference of magnetic field intensities, and the field itself, filling space, acts something like a pressure. So, if we have one intensity at one end of a magnet or armature, a different intensity at the other, and a given cross-section, we can find the force on that object. The pressure is the Maxwell stress, B^2 / (2 mu_0) in air. For typical flux densities (say, under 1T), this corresponds to modest air pressures, ~1 atm, so it can be compared favorably to a suction head, or compressed air cannon.Electrically, what happens is, first some energy is applied (flux times current) and the magnetic field increases; as the objects move, some flux is induced in the coil, drawing more energy. This can be modeled as an increase in inductance (reluctance motor), or opposition by an internal voltage (EMF). If this is a one-time motion, then when everything settles down, the object will be held at that displacement, so no more work is being done but we're still delivering current, and will draw flux from the coil if we try to cease that current -- the magnetic field is storing some energy. Likewise, that current flows through the resistance of the coil, so we are dissipating power while holding it, even though we're doing no useful mechanical work. (Zero-resistance superconductors do exist, so this cost can, in principle, be saved -- assuming we have something to keep them cold enough to work.)This generally details the situation, but I don't know that it's basic enough. It probably doesn't help that there's no specific setup to explain (hence, general). I've used terms covered by introductory physics; ask for further explanation, or look it up. I'm... not sure how effective further explanation may be, there's good reason to cover this sort of thing in a whole course -- it takes practice and observation to understand these rather abstract terms.Tim

author: Berni
author_rank: Super Contributor
timestamp: February 03, 2021, 06:38:52 am
content: The number depends a lot on what the scenario is.Is the coil an ideal superconductor? If not what are the exact specifications of the coil dimensions and material.Do you include the energy required at the beginning to create the magnetic field?Do you include the energy you get back after destroying the field afterwards?Basically you have to put energy into the coil to create a magnetic field, this field sucks up the magnet and its movement induces a voltage in the coil that causes more energy to be pulled from the power source (This depends how far in the magnet got sucked). Then to drop it you need to destroy the magnetic field by pulling its energy back out, returning all of the energy. Then the magnet falls and generates a voltage again that gives you back the energy you put in to lift it.You could sidestep the problem by leaning onto the conservation of energy. You know how much energy it takes to lift a weight a certain height. So if you use the usual ideal scenario that physicists love, use a superconductor coil and have the magnet in a vacuum, then the energy that will get pulled by the coil as the magnet lifts is the same as the energy taken to lift the magnet since there is no other place the energy can go.The energy used to build up the magnetic field can be calculated from the coils inductance and amount of current used to create it. Its just a classical inductor.Then if you have a real copper coil instead of a superconducting one you get restive losses along the whole way that need to get factored in and make things more difficult. For things like solenoid coils these losses are significant, hence why they tend to get hot.