Question

Posted by Mary on Saturday, October 6, 2007 at 5:13pm.

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?



For Further Reading


Physics please help! - bobpursley, Saturday, October 6, 2007 at 6:28pm
L= mu*N^2*A/l

you have everything, assume air core for mu, and solve for N


Please tell me where I am going wrong!


l = length of solenoid 0.048m
A = Cross-sectional area 0.0012m^2
n = Number of turns ?
L = self inductance 0.0012H
Mu= 0.000001257H/m


L = (Mu)(n^2)(A) / l

then I solve for n:

(L x l)/(Mu x A)= n^2

(0.0012H/m)(0.048m) / (0.00000125H x (0.0012m^2) = n^2

0.0000576/0.000000002= n^2

28800 = n^2

169.70563 = n

Answer 1

Put your numbers in scientific notation. In the denominator, you are losing accuracy due to digits being terminated.
1,25E-6x1.2E-3=1.5E-9, and you got 2E-9

Big difference.

I get about 197 turns