Magnet Myths and Misconceptions

[tinman]

@Tinman
This does not mean that you cannot have your own ideas, nor that you cannot question the established science.
It just makes it simpler to communicate your ideas to others.

I will endevor to do better.

[MileHigh]

Tinman:

You have an interesting conundrum to contemplate.

Okay, we want two coils to dissipate the same amount of power.  Coil #1 (the copper coil) has R resistance and Coil #2 (the iron coil) has 6R resistance (as an example).

So:   I1^2*R = I2^2*6R

        I1^2 =   I2^2*6

        I1 =  Sqrt(6)*I2

        I1 = 2.45*I2

So for example if the iron wire coil has one amp flowing through it and the resistance is 6 ohms, then the power dissipated in the iron wire coil is 6 watts.

Then if the copper wire coil has 2.45 amps flowing though it and the resistance is one ohm, then the power dissipated in the copper wire coil is 6.05 watts.  So the formula checks out.

I recall that you stated that the iron wire coil had the stronger magnetic field.  When we look at the hypothetical example above, the copper wire coil has to have about 2.45 amps running through it compared to only one amp for the iron wire coil.

Plus, we know that some of the iron wire coil's magnetic field is trapped inside the wire itself, suggesting an even weaker external magnetic field than expected.

Since you stated that the iron wire coil had the stronger magnetic field, how can you account for the fact that the copper wire coil actually has more current flowing through it?   That would suggest that it's the copper wire coil that should have the stronger magnetic field.

MileHigh

[MarkE]

Above are equations that calculate the amount of current required to close the relay contacts. What we're dealing with is the "Pull Force" between the electromagnetic coil and the iron armature. The spring exerts a force on the armature of .15 Newtons. Therefore it will require I = 0.138 amps. The flux density will be 0.116 teslas to equal the .15 newtons of spring force.

In view of these mathematical relationships, it's absurd to maintain that "pull force of iron is no measure of field strength" as MileHigh falsely maintains. The "Pull Force" of .15 newtons from the coil on the iron armature is equal to 0.116 teslas.

If one shapes the pole shoes of a magnet, they can change the pull force on an iron test sample sample a lot.  Do you think that shaping the pole shoe changes the energy that is in the magnetic field, or just how that energy is distributed?

[MarkE]

Tinman:

You have an interesting conundrum to contemplate.

Okay, we want two coils to dissipate the same amount of power.  Coil #1 (the copper coil) has R resistance and Coil #2 (the iron coil) has 6R resistance (as an example).

So:   I1^2*R = I2^2*6R

        I1^2 =   I2^2*6

        I1 =  Sqrt(6)*I2

        I1 = 2.45*I2

So for example if the iron wire coil has one amp flowing through it and the resistance is 6 ohms, then the power dissipated in the iron wire coil is 6 watts.

Then if the copper wire coil has 2.45 amps flowing though it and the resistance is one ohm, then the power dissipated in the copper wire coil is 6.05 watts.  So the formula checks out.

I recall that you stated that the iron wire coil had the stronger magnetic field.  When we look at the hypothetical example above, the copper wire coil has to have about 2.45 amps running through it compared to only one amp for the iron wire coil.

Plus, we know that some of the iron wire coil's magnetic field is trapped inside the wire itself, suggesting an even weaker external magnetic field than expected.

Since you stated that the iron wire coil had the stronger magnetic field, how can you account for the fact that the copper wire coil actually has more current flowing through it?   That would suggest that it's the copper wire coil that should have the stronger magnetic field.

MileHigh

Tinman is comparing different weight iron samples that he can lift with one or the other of his electromagnets.  He is measuring force.  Force on a soft iron piece goes with the gradient of the flux density, which is much higher for the iron wire coil.

[tinman]

Tinman:

You have an interesting conundrum to contemplate.

Okay, we want two coils to dissipate the same amount of power.  Coil #1 (the copper coil) has R resistance and Coil #2 (the iron coil) has 6R resistance (as an example).

So:   I1^2*R = I2^2*6R

        I1^2 =   I2^2*6

        I1 =  Sqrt(6)*I2

        I1 = 2.45*I2

So for example if the iron wire coil has one amp flowing through it and the resistance is 6 ohms, then the power dissipated in the iron wire coil is 6 watts.

Then if the copper wire coil has 2.45 amps flowing though it and the resistance is one ohm, then the power dissipated in the copper wire coil is 6.05 watts.  So the formula checks out.

I recall that you stated that the iron wire coil had the stronger magnetic field.  When we look at the hypothetical example above, the copper wire coil has to have about 2.45 amps running through it compared to only one amp for the iron wire coil.

Plus, we know that some of the iron wire coil's magnetic field is trapped inside the wire itself, suggesting an even weaker external magnetic field than expected.

Since you stated that the iron wire coil had the stronger magnetic field, how can you account for the fact that the copper wire coil actually has more current flowing through it?   That would suggest that it's the copper wire coil that should have the stronger magnetic field.

MileHigh

There is no conundrum to contemplate MH. If the P/in is the same in both coils,then the disipated energy in both coils must be the same,as energy can neither be created nor destroyed only transformed. I would suspect that the coil that uses the soft iron wire would be acting or represent a permanent magnet,where as the copper coil would not,as the copper itself is not magnetised.

EDIT: - i forgot to add this in.
As i said before,i would suspect that the copper coil would have a far larger field to that of the iron wire coil,as not only is the iron wire making the field,it becoms part of the field-much the same as having a core.

Now in saying that MH,here is a thought experiment(brain fart).
Lets take two cores that are identical in every way-lets say they are 1/2 inch round x 2 inches long,and well use solid ferrite for this thought experiment. We wind 100 turns of .55mm copper wire on one,and 100 turns of .55 soft iron wire on the other-both enameled coated for insulation.
We apply a P/in of say 5 watts to both(now) electromagnets.
Which do you suppose will have the strongest and largest magnetic field for the same P/in?.
An interesting thought experiment i think,as the one useing the iron wire now has a larger core,but uses the same amount of wire,where as the one useing the copper wire has a smaller core,but still has the same size outside diameter