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?
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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
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
To solve this problem, you want to use the formula for the self-inductance of a solenoid, which is given by L = μ₀ * N² * A / l, where L is the self-inductance, μ₀ is the permeability of free space (approximately 0.000001257 H/m), N is the number of turns of wire, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.
Given:
L = 1.2 mH = 0.0012 H
A = 1.2 * 10^-3 m²
l = 0.048 m
First, rearrange the formula to solve for N:
L = μ₀ * N² * A / l
N² = L * l / (μ₀ * A)
Plug in the known values:
N² = 0.0012 * 0.048 / (0.000001257 * 1.2 * 10^-3)
Calculate:
N² = 0.0000576 / (0.000001257 * 1.2 * 10^-3)
N² = 0.045797851085985
N = √0.045797851085985
N ≈ 0.213858
Therefore, the number of turns of wire needed is approximately 0.213858.