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?
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html#c1
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.
To determine the number of turns of wire needed to create a solenoid with a given self-inductance, we can use the formula:
L = (μ₀ * N² * A) / l
Where:
L is the self-inductance of the solenoid.
μ₀ is the permeability of free space (constant value of 4π × 10⁻⁷ H/m).
N is the number of turns of wire.
A is the cross-sectional area of the solenoid.
l is the length of the solenoid.
In this case, we can rewrite the formula to solve for N:
N = √((L * l) / (μ₀ * A))
Now, let's plug in the given values:
L = 1.2 mH = 1.2 × 10⁻³ H
A = 1.2 × 10⁻³ m²
l = 0.048 m
μ₀ = 4π × 10⁻⁷ H/m
Substituting these values into the formula, we get:
N = √((1.2 × 10⁻³ * 0.048) / (4π × 10⁻⁷ * 1.2 × 10⁻³))
Simplifying this equation, we have:
N = √((0.0576) / (4π × 10⁻¹⁰))
Now, we can calculate this value:
N = √(0.0576) / √(4π × 10⁻¹⁰) ≈ 38.183
Therefore, approximately 38 turns of wire are needed to create the solenoid with a self-inductance of 1.2 mH.