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I'm thinking about making an instrument to show the force on a wire carrying electric current for students in lab. A normal wire which levitates from ground after it has current flowing in it.

We have the following equations

$$F = ILBsin(a)$$

$$W = ALg$$

$A$ being the mass of unit length of the wire. We will make the magnetic field perpendicular to the current flow so $a=\pi/2$ and $sin(a)=1$ and I want the wire to be suspended above the ground so in equilibrium we must have:

$$F = W$$

$$ILB = ALg$$

$$IB = Ag$$

Since $g$ and $A$ are constant in this device what I will need to do is play with magnetic field and electric current.

I'm not very familiar with the Earnshaw theorem but I know if I use an AC current this is doable. I can use normal power outlets (220 V in this case) but I don't know how to create the uniform magnetic field needed.

Are natural magnets (like Neodymium magnets) better or a magnetic field generated by another wire or coil?

Is there any easier way to do this?

What are notes to keep in mind in this experiment?


EDIT:

I figured it's better to use Helmholtz coils to generate a uniform magnetic field.

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  • $\begingroup$ Using a Alternating Current you'll induce an alternating magnetic field. Now way to levitate above a permanent magnet. Using a second current carrying wire instead of the magnet will be a good choice. $\endgroup$ Commented May 17, 2017 at 6:11

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I work in the Physics department at my university, and part of my job is overseeing the lecture demonstrations for the physics classes, and we have several demonstrations that show the force of a current carrying wire in a magnetic field. A rather simple demonstration is a coil of wire connected to an AC source, that is put over a sheet of copper. The magnetic field produced by the wire induces an opposing field in the copper plate, causing the coil to levitate. This is one way to demonstrate the effects, but not necessarily the only way.

Some things you will need to keep in mind are that if you choose to run AC, you must ensure that whatever is creating the opposing B field is perfectly out of phase so that the fields are always opposing. I would recommend using some sort of transformer between the wall power and the demonstration, not only will this allow you to better control the voltage, it will serve as a pint source of failure in case anything goes wrong (its internal fuse will blow before you trip a breaker)

On a more practical note, you must be very careful with your wiring and choice of materials. To get a good effect, you will probably be pushing a lot of current through the wires. Be very careful that nothing will suddenly catch on fire when you start the demo. And as always, make sure everything properly grounded and insulated, just so you don't ruin your day by having 220V wall power coursing through your body.

I hope this helps, let me know if you would like any more descriptions of some apparatus that you might want to emulate, I can always provide pictures when I'm at work today.

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  • $\begingroup$ My comment was to short jumped. Nice answer. $\endgroup$ Commented May 18, 2017 at 20:29
  • $\begingroup$ Thank you for your reply. I thought about this demo for a while and I think the right way to show this the way I want is to use Helmholtz coils to generate a magnetic field and pass a current carrying wire between them. I used the equation for Helmholtz coil and the one I mentioned above (IB=Ag) and I got this: (Sorry about typing formulas like this) I(wire)*I(coil)*N(coil)/(R(coil)*A) = 1.09*10^7 So if I have a coil of N=1000 and R=7 cm and a wire of A=50 g/m I can use ~6 A current for the wire and coil both. I know the theory, But what would cause trouble when I do it in real world? $\endgroup$
    – Alireza
    Commented May 20, 2017 at 22:12
  • $\begingroup$ What if I use DC current in both the wire and the coils? $\endgroup$
    – Alireza
    Commented May 20, 2017 at 22:14
  • $\begingroup$ Both AC and DC should work if you are using a separate coil to generate the opposing field. Probably the biggest real world challenge is finding a power supply to run everything, and making sure your wires can take the current for each one. $\endgroup$ Commented May 21, 2017 at 22:35

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