See also ...

[
Producing wound components]
[ Air coils]
[A guide to the terminology used in the science of magnetism]
[ Power loss in wound components]
[ Faraday's law]

When the coil current increases so does the magnetic field strength, H. That, in turn, leads to an increase in magnetic flux, . The increase in flux induces a voltage in the coil. It's the power needed to push the current into the coil against this voltage which we now calculate.

We choose a toroid because over its cross-sectional area, A, the flux density should be approximately uniform
(particularly if the core radius is large compared with it's cross
section). We let the flux path length around the core be equal to L_{f} and the cross-sectional area be
equal to A_{x}. We assume that the
core is initially unmagnetized and that the electrical energy (W) supplied
to the coil will all be converted to magnetic field energy in the core (we
ignore eddy currents).

W = v×i dt joules

Faraday's law gives the voltage as Substituting: Now, N×i = F
W = H×L_{f}×A_{x} dB joules

W_{d} = (H×L_{f}×A_{x}
dB)/(L_{f}×A_{x})

W_{d} = H dB
joules m^{-3} |
Equation EFH |

W_{d} = B/µ dB

W_{d} = B^{2}/(2µ)
joules m^{-3} |
Equation EFB |

W_{L} = L×I^{2}/2 joules

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Now, the point is that there will be no net force on the iron, no matter how strong the field. A sphere has perfect symmetry, so rotation will not change the picture in any way. If there is translational movement then all that can happen if the sphere were to move is that the distortion of the field around the original position of the sphere will disappear and the same distortion will be re-established around the new new position; the total system energy will remain unchanged.

OK, instead of the sphere let's try an iron rod. This is different because we've lost symmetry. What happens is that the axis of the rod will be drawn into alignment with the field - like a compass needle. The flux lines prefer the iron to the air because of the higher permeability. Equation EFB has µ on the denominator so the field energy is lower here than in the air, and the further the flux can go through the iron the lower the energy. Think of current flow through a resistor; the current has an easier time going through a low resistance than a high resistance. Flux goes easier through high permeability than through low. When the rod is aligned with the field the flux can go further through a high permeability region. Note that we still don't have a translational force (provided that the field is uniform on the scale of the rod). Think about the famous experiment with iron filings sprinkled onto a piece of cardboard above a bar magnet. The filings tend to line up with the field but don't generally move much because they are so small that the field appears uniform to them.

So for there to be a force on a piece of iron then a displacement of the iron must result in an alteration to the field energy. The electromagnet you are using will have an opinion about changes to the field it generates. It will say that its inductance is changing. This is the basis of one solution to the problem:

F = (I^{2}/2)dL/dx newtons |
Equation EFS |

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- Set a high coil current by turning the power supply up towards 12 volts.
- Push the plunger in until it is retained by the field.
- Reduce the length of the cord until either the balance reads maximum load or the solenoid just retains the plunger. A turnbuckle is handy here.
- Make a note of the force measured by the balance.
- Reduce the current slowly, keeping an eye on the meter reading, until the solenoid lets go, and note the current. The restraining cord is essential unless you want to be hit in the eye by a lump of iron moving at very high speed :-)

Hmmm ... what may be happening is that non-linearities in the permeability of the iron are affecting the field. If I add a 2 mm thick piece of brass on the end of the plunger then I get:

Notice that the retaining force is now much lower even though a higher coil current has been used. Well, this line has a slope of about 2.2 - a bit closer to theory. In an air gap the flux density is exactly proportional to field strength (and thus current). As far as a static magnetic field is concerned brass behaves just the same as air: the permeabilty is a steady µ

It would be nice to extend this experiment by a measurement of coil inductance against force in order to test Equation EFS. The difficulty is that inductance meters use AC test signals. Without laminated iron (which is only found in solenoids designed for AC operation) the reading will be affected by large eddy current losses.

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E-mail: R.Clarke@surrey.ac.uk

Last revised: 2001 March 11th.