Magnet Myths and Misconceptions

[TinselKoala]

Once again... you have to match the _current_, that is, amp-turns, not the _power_, because the resistance of the iron is greater than the copper by a fair amount.

Copper has a resistivity coefficient of 1.724 x 10^-8 ohm-meters, and iron is 9.71 x 10^-8 ohm-meters. So all other things being equal the iron coil will have 9.71/1.724 = about 5.6 times the resistance of the copper coil.
http://www.engineeringtoolbox.com/resistivity-conductivity-d_418.html

Say the copper coil has 3 ohms resistance and you are using your 12 volt DC regulated supply. You therefore have I=V/R or 4 amps current, for a supplied power P = I²R = 16x3 = 48 Watts. If the iron coil has 3 ohms x 5.6 = 16.8 ohms, to get the same _4 amps current_, that is the same amp-turns as the copper coil, you need to use V = IR = 4 amps x 16.8 ohms = 67.2 volts, for a supplied power of P = I²R = 16 x 16.8 = 268.8 Watts.

268.8/48 = 5.6 (duh).  So to get the same amp-turns in the iron as in the copper, you need to supply not the _same power_ but 5.6 times _more power_ to the iron. This power is dissipated as heat.

This is why relay, motor, generator, solenoid, etc designers prefer to use copper rather than the -much cheaper- iron wire in their coils.

[synchro1]

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,

The point is not how magnet shape changes the magnetic field but how two magnets of identiclal shape and unequal strength attract an iron armature differently.

[synchro1]

Okay, so I can suggest a follow-up test for Tinman.

You line up the axis of your copper-wire coll on magnetic east-west.   Then you put a compass say 10 inches away from the end of the coil on the line of the coil's axis.   Then you can energize the coil and note the amount of compass needle deflection for a certain amount of power dissipated in the coil, and also for a certain amount of current flowing through the coil.  You can obviously tweak the distances and amount of current flowing through the coil to give you a "nice" deflection of the compass needle, say somewhere between 30 and 45 degrees.

Then repeat the whole thing again for the iron-wire coil and compare results.

This test should give you a nice indication of the relative magnetic field strengths for both types of coils.  (This is not to be confused with the magnetic attraction force measurement.)  You basically have the compass needle aligning itself with the net magnetic field as supplied by the Earth and the coil under test (the two magnetic field sources will be at right angles to each other.)

MileHigh

@Tinman,

Suspend an iron keeper from a long string attached to the ceiling overhead. Secure a laser pen to the string, and mark the spots on the wall where the light beam touches.

[tinman]

This power is dissipated as heat.

Is not all the P/in disipated as heat in an inductor being supplied with a DC current?. I mean,it dosnt require power to maintain a magnetic field,as the field can do no useful work,and a PM dose not require any power input to maintain it's field.The resistance of the coil,and the amps to the coil tells us how much power we are using,so it's all accounted for.This means that it takes no power to create and maintain the actual magnetic field,as all power in is disipated as heat.

[tinman]

In fact,thinking about my last statement,i believe i can prove that a permanent magnet can/is doing useful work.

Want to run with me on this one MarkE?-work together on proving that a PM can/is doing useful work.

P.S-we can do it useing your very well loved physics.