# Electromagnet, ideal turns depending on ohm

More turns -> stronger field
more turns -> longer copper wire
longer copper wire -> more resistance(ohm)

at what turn does the resistance make the electromagnet weaker? - I want to make an ideal electromagnet.

(Sorry for being vague but I'm looking for some kind of formula or an example)

You need to combine some equations, lets list them:

Whe asume that you want the formula in terms of potential diferential, with Ohm's Law, $\Delta V = I R$, and the magnetic field inside one solenoid:

$$B = \frac{N}{L}\mu I = \frac{N}{L}\mu \frac{\Delta V}{R} \quad\quad (1)$$

where $N$ , $L$, $\mu$, stand for the number of turns, lenght of the solenoid and magnetic permeability of you core respectively. Then we also know the dependance of $R$ over the lenght of the wire, $R = \rho\frac{l}{A}$, where $\rho$ is the resistivity, $A$ the cross section of the wire and $l$ the lenght. We can compute the lenght by:

$$l = 2\pi rN$$

so if we subtitute the last equations in $(1)$:

$$B = \mu\frac{N\Delta V}{L}\frac{A}{\rho 2 \pi r N } = \mu\frac{\Delta V A}{2 \pi \rho r L}$$

surprisingly it doesn't care the number of turns!!

Those equasions are good but complex, the important factors are turns and current. If you can get say 100 turns out of the wire you have it's better to halve it and have two lots of 50 turns, same amount of turns but more current.