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nursejp nursejp
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11 years ago
I am trying to implement a RFID system, in the near field range, the 125kHz system uses magnetic induction.

Every diagram and example and formula i have found gives details of induction coils being 'circular' (i.e. wrapped around a bar), and using the formula:
Field Strength = ((2 x (Number of turns) x current)/Pi) x ((radius of coil)^2 / ((Radius of coil)^2   distance^2)^3/2)

It shows that by increasing the radius or the number of turns in the coil, you can increase the magnetic field strength.

SO that brings be back to my question:  If i had an elliptical 'coil' instead of a standard circular coil, what would the effect be on the magnetic field strength?
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11 years ago
Assuming that the circumference is the same (say, as if you were to "squash" the circular coil a little), I would expect the magnetic induction at the center to increase, but I don't think that that is the number that is important for your application.  First let me explain the first part of my statement.  imagine that you squeeze the cool so much that the wires almost touch, let's say that the cross section is rectangular with one side approaching zero, the other that of half the circumference of the coil. The magnetic field at the center would be almost entirely due to the two long wires and come close to that of two infinitely long wires.  Furthermore, if the wires almost touch, that field strength appoaches infinity, unless you correct for the thickness of the wire. Now, the ellipse would do the same thing, but I won't vouch that the field strength would necessarily increase monotonically with the excentricity (as you squeezed the ellipe more), so you may want to investigate that further.  The exact solution would be rather involved, and probably is best done with a simple computer program, however, you can show that the field at the center of the ellipse is proportional to the ratio of its circumference to its area, which agrees with the physical argument I gave.

The other point I made is that what you probably want is not the magnetic induction (B) but the total magnetic flux, which is the integral of B over the cross sectional area, because that is what determines the self-inductance, L, of the coil.  I suspect that that would decrease as you made the coil elliptical, by the following simple argument.   Suppose I started with a circular coil and imposed a slight deformation, would the internal forces on the coil (due to the interaction of the magnetic field with its current) try to return the coil to its circular shape or would they try to increase the deformation?  I leave it to you to show that the forces are of a repulsive nature, so that the coil will try to return to the circular shape.  You could try to calculate the inductance of a rectangular wire. I think you will find that it is largest for a square, assuming the circumference of fixed length.

******************
I forgot to point out that I only addressed the problem of  single turn, although I believe that the conclusion applies to any number.  There is at least one more loose end, however, so I am sorry that I couldn't give you a very cohesive answer. One other suggestion from a simple practical viewpoint: If the inductance of a circular coil were not the largest  for a given amount of material, inductors would be made in other shapes than circular.
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