Calculating Copper Wire Diameter, Number of Turns, and Length for 45V DC Electromagnet

Hello

I want to help in the design of electric circular file a temporary magnet works: Plastic Reel inner diameter of 2 cm, and a thickness of 2 cm, and an outer diameter of about 10 cm.
* R = 10.5 Ω
* V = 45 V DC (Car Batteries)
* Copper wire diameter?
* Number of laps?
* Length of the wire?

What are the laws of physics and the necessary calculations? Thank you for your attention.

An iterative method can be used. Let's say wire having a diameter of 1 mm is used. The bobbin will hold 40 layers with each layer having 20 turns. The number of turns is 800. The first layer has a wire length of 20 x 2Pi x 1.05cm = 132 cm. The second layer has a wire length of 20 x 2Pi x 1.15cm = 145 cm. Continue this process until layer 40, add up the wire lengths and calculate the electrical resistance. Note that this method includes the wire insulation and the conductor diameter is less than 1mm.

If the result is not 10.5 ohms change the wire size and repeat.

Thank you very much
But do I know the physical relationship: between the resistance and the required diameter wire coiling
Is the current intensity plays an important role in the parcel with the resistance
Thank you for your interest

It's interesting to note that as the wire size is changed the electromagnet power consumption remains the same for a given magnetic field intensity.

Double the wire cross sectional area and the number of turns that can fit on the bobbin decrease by a factor of two. The wire length decreases by a factor of two while the resistance per unit length decreases by a factor of two; the total resistance decreases by a factor of four. To maintain the same magnetic field intensity the current must be doubled (to maintain the same number of amp-turns). Double the current through 1/4 the resistance gives the same power consumption.

*Thank you very much
Regards
Chemical: Ghassan Youssef*