Asked by Mary
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
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
Answers
Answered by
bobpursley
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.
You have length, area, inductance, you know mu. Solve for N, the number of turns of wire. The formula is straightforward.