A simple model to help you visualize: The magnetic flux forms loops, you cannot have a magnetic field line terminate on nothing in free space. That means that all those flux lines inside the solenoid heading (for example) to the right. The flux line must continue outside the solenoid and loop around to join the other end of the flux loop. In doing so it heads to the left. No external flux lines -> no flux.
So what could be the issue?
1) In an infinite solenoid, there is no external magnetic field. One example of this is a Tokamak or toroidal solenoid. Like a big doughnut. The flux lines form loops within the doughnut and have no need to go outside.
- of course a mathematically ideal infinite solenoid could have been what they were discussing, in which case the field lines loop around infinity. Hard to test though.
2) The flux density outside of the linear finite solenoid is much less than inside. So while there may be X flux across the throat of the solenoid that same amount of flux (X) is spread out over a much much larger cross sectional area. The density is much less. [here the area is the cross sectional area across the solenoid.]
3) Perhaps they are talking about a shielded solenoid? This would involve having a high permittivity material external to the coils to trap the flux lines.
I am not sure if I clearly understand your new sensor, but as far as I did, these are the main physical principles:
- The sensor is a transformer with \$n_1\$ turns in the primary and \$n_2\$ turns on the secondary. This is not an LVDT, because the \$n=n_2/n_1\$ is constant,
- The displacement \$x\$ is measured as the advance of the iron rod. \$x=0\$ mainly out of the coils, \$x=1\$: mainly inside the coils,
- No magnetization/hysteresis effects are relevant (according your OP),
- \$v_1(t)\$ is sinusoidal. We could assume phasors here.
Evidently, because the relative magnetic permeability of the air and the iron are \$\mu^r_{air}=1\$ and \$\mu^r_{iron}=4000\$ aprox. respectively, the magnetic flux \$\phi\$ (in Webers) will be fully passing through the core with \$x=1\$ and almost totally passing through the air with \$x=0\$.
The shape of the flux depend on a 2D Magnetostatic FEM modelation (though the geometry is actually a 3D cylinder, you can simplify it to a rectangular 2D rod). The primary excitation is the Magnetic Field \$H\$ (in A.m) and the rod-air system measurement is the Magnetic Flux Density \$B\$ (in Tesla), or directly the voltage through the coils (depending on your package skills), doing several runs for different \$x\$ values. Any FEM package or program for 2D/3D magnetostatic such as Ansys, Comsol, CST, or any other cheaper/easier alternatives are useful and perhaps better for this case.
If you are more hardware focused, and you don´t want to model anything with FEM, i will not blame you. Because \$\mu^r_{air}<<\mu^r_{iron}\$ the approximation solution can be expressed as:
$$v_2(x)=m(x)+e(x)$$
where \$m(x)\$ is the main linear flux component -the integrated flux density through the rod- and \$e(x)\$ is the remaining error -the integrated flux density out of the rod.
Hence:
- \$m(x)=n \cdot v_1(t) \cdot x\$, that is, no flux amplified with the rod out of the coils and all the flux is amplified with the rod inside the coils. The linear model comes from Maxwell's, as a standard transformer, just in the rod segment. No magnetic nor electrical losses.
- \$e(0)=e_1, e(1)=e_2, e_1>e_2=0\$. \$e_1\$ is the integral of the flux density when the rod is fully out, and \$e_2\$ is zero, because in this case all the flux is explained by the "linear" part. \$e(x)\$ is positive, have its maximum on \$x=0\$, and tends smoothly to zero. This approximation is VERY violent, rough and mostly "empirical" (?). This components ruins the linearity, represents all the integrated flux density over the paired coils, and deviates your sensor from an LVDT, which is much more linear, because on an LVDT, the integrated flux density over the air is much less.
\$v_2(x)\$ indeed should have a maximum for \$x=1\$, a small nonzero value for \$x=0\$ and a minimum with \$x=\pm\inf\$ (with the rod far out), which shows that you make the data only for a piece of the rod inserted. The whole curve should be half bell shaped.
I am really sorry for not including pictures, but my FEM package is not in this computer. But if you really require them, we can start another question with that...
Best Answer
Learning to read a magnetic circuit is a skill that is not much taught nowadays.
You're right (in your comment to Dmitry's answer) that the same flux is present on the central pole and on the outer (split) pole.
However, notice that the total area on the central pole is one square (I'll guess one square inch) - flush with the bobbin.
Now measure the total area of the other pole - the whole back surface (about 3 square inches), the outer ends (about 2 square inches each), both sides (about 3 square inches each) and the two pole piece surfaces themselves (summing to 1 square inch). Total is somewhere around ... 14 square inches, so a very rough approximation to the flux density iwould be 1/14 of what you expect.
If you can read the currents and voltages in a circuit with low resistance wire, thinner high resistance wire and actual resistances, you can start to understand this circuit by imagining a thick iron bar as a low resistance, or having high conductivity - air or vacuum as a high resistance (i.e. low conductivity).
The actual term for resistance in magnetic circuits is "reluctance", and that for conductivity is "permeability". Air has a "relative permeability" of 1, iron in the thousands. So an iron bar conducts magnetic flux thousands of times better than an air path of the same length (until high flux densities - then it will saturate).
So the flux density is not equally distributed around the huge outer pole, it's proportional to the area of a section of that pole, and inversely proportional to the air gap length. So it'll be slightly stronger at the inner edge of those outer pole pieces where the air gap is only 1/2 inch, and a bit weaker on the bottom surface where the air gap (from the inner pole) is about 2-3 inches.
Calculating the exact flux densities can be done with calculus for simple shapes, but simulations and finite element analysis are more often used now.
Now, I hope you kept the "I" laminations? Use them as an iron bar spanning the top of the "E". As you bring them closer, you'll find the air gaps between E and I reduce - and as you reduce the gap, the flux will concentrate in those gaps - and as you reduce the air gaps, you reduce the "resistance" i.e. the reluctance, and so the "current" i.e. flux will increase dramatically, and so will the attractive force between the electromagnet and the bar. WARNING keep your fingers out of the way when you do this!
The magnetic flux can't increase infinitely high, eventually the iron will "saturate" at about 1.2 Tesla.
Now you can see how Dmitry's horseshoe magnet works, and how to improve it - bend the poles closer together to reduce the air gap. Also, look at toy electric motors - how the pole pieces are shaped to match the iron rotor (with the coil wound on it) to concentrate the flux in the small gap between the magnet's poles and the rotor.
EDIT: found quite a good introduction here...
Pay attention to the figures, read the words later... Note the following:
Having covered some of the "why", if you're really asking "what do I do about it?" add some context about what you want to achieve to the question. It should now be clear that magnetic circuits are designed for a specific purpose, and we don't know anything about what your purpose is.