Electromagnet strength independant of number of turns?

[dorker]

Help me clear up something. I know the magnetic field of an electromagnet's core is given by

,
where N is the number of turns of wire. Now, the resistance of the wire is given by R = ρ*λ/A, where ρ is the resistivity, and λ the length of the wire. Using Ohm's law to combine them, I get:

But, disregarding the uncoiled ends, the length of the wire can be approximated as λ = n*N, where n is the length of one turn of wire. Replacing this in the last equation makes both Ns disappear, thus suggesting that, at least in steady state, the field strength is independant of the number of turns. Is this correct, or did I make a wrong step somewhere?

 

[DrZoidberg]

If the voltage stays constant you are right. If you double the number of turns you get twice the resistance and therefore half the current. (1/2 * I) * (2 * N) = I*N

 

[Enthalpy]

No, if the power is constant. Then you can adapt the number of turns so the resistance of the coil matches your needs, that is, the voltage and current available from the power source.

Because as the wire length doubles, the section is halved due the the available room, so the resistance is multiplied by 4, which matches a source of identical power whose voltage is double but current half.

And with half the current in twice as many turns, you get the same magnetomotive force.

Becare this is true in DC or low frequency. At higher frequencies, you have Kelvin effect which prevents using the whole wire section, and much worse, you have eddy currents in the wires which create horrible losses.

Even at 50Hz, this is a constraint, for instance in turbo-generators, where the bars have to consist of several insulated conductors which exchange their positions regularly so current flows everywhere. These conductors are of rectangular section for better filling, and some conductors are hollow for cooling water or hydrogen... and then just a conductor gets a high-tech object.