Question

Suppose you wish to make a solenoid whose self-inductance is 1.2 mH. The inductor is to have a cross-sectional area of 1.2 10-3 m2 and a length of 0.048 m. How many turns of wire are needed?

Can you please clarify this for me. When I first post it my reponse was:


# turns= # of turns per meter length x l of solenoid.

I still don't get it please explain.
N=nl

Answer 1

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html#c1

You have length, area, inductance, you know mu. Solve for N, the number of turns of wire. The formula is straightforward.