Coil current changes while working

Question

If I wind a coil around a hollow plastic tube and apply voltage it will create a magnet that is capable of pulling an iron based plunger into the tube. I imagine the plunger will stop in the magnetic field's center. What current fluctuations in the electromagnet can I expect, and where in axial relation to the the coil? In other words will the magnet's current change as the plunger position changes? Will it increase or decrease as the plunger moves towards the magnetic center? How big a change can I expect?

Thank You

Answer 1

Interesting question.

If you're exciting the coil with DC, and the plunger is free and not at the center of the coil, then as the plunger is pulled into the tube, the inductance of the coil will increase which will limit the instantaneous current that can be supplied to the coil by the driving constant voltage source, which will limit the speed at which the plunger moves toward the center of the magnetic field.

Also, because of the mass of the plunger - unless you've critically damped its energy against the friction between the plunger and the ID of the tube - it'll oscillate for a while before it settles down at - more or less - the magnetic field's sweet spot.

If you exercise the coil with AC it gets more complicated. Do you want to go there?

Answer 2

I've never even considered the question before, but here are my initial thoughts about it.

Lots of things are changing as you insert an iron rod. One thing that's certainly changing is the magnetic path length, as iron acts as a kind of short-circuit and reduces the effective magnetic path length. As that happens, the inductance of your solenoid rises (changes.)

Your solenoid starts out as an air-core, with lower inductance and a magnetic field that spreads out into space quite a distance. I'll assume that the resistance in the wire (and any series resistance in your voltage source) has become dominant in limiting the current at the start. So the value of \$I\$ is set at some value to begin and there is a certain energy stored in the magnetic field at this time.

Once the plunger starts to be inserted axially, the ampere-turns (each turn contributes some force and the force of each combines) in the field will exert a force on it. Since you aren't asking about the force, I won't mess with that part.

The usual equation for an inductor looks like:

$$V=L\frac{\textrm{d}I}{\textrm{d}t}$$

But in this case we happen to know that the inductance also changes with time. So, it's more like:

$$V=L_t\frac{\textrm{d}I_t}{\textrm{d}t} + I_t\frac{\textrm{d}L_t}{\textrm{d}t}$$

which, with \$V\$ constant, becomes something like:

$$\frac{\textrm{d}L_t}{\textrm{d}t} I_t +L_t\frac{\textrm{d}I_t}{\textrm{d}t}=V$$

And, if you can estimate \$\frac{\textrm{d}L}{\textrm{d}t}\$ as a constant rate, becomes a simple 1st order diff eq. But yeah, the current will vary while motion is taking place. But ultimately, the final condition for the current remains limited by the same coil resistance.

Answer 3

It is important to recognize that the coil inductance has energy E=1/2Li^2 but the force is only proportional to the Change in Energy.

Unless the speed is great, the current will only slightly reduce and increase with motion oscillations about center and reduce as the copper temperature rises from excessive current.

Winding Resistance is the main trade-off affecting current with turns ratio , wire gauge , fill factor and diameter affecting force and thus velocity and thus reduction in current.

But when it has stopped with no change in energy applied V*I, there is no net force when unrestricted.

It is also important to recognize that adding turns in fact reduces force due to the rise in DCR of the current relative to the effect of increasing B field when operating in a fixed voltage, which may seem counter-intuitive. Of course the current reduces with more turns. But reducing turns is a very inefficient way of increasing force when you get down to only a few turns.