Suppose you wish to make a solenoid whose self-inductance is 1.8 mH. The inductor is to have a cross-sectional area of 1.60 10-3 m2 and a length of 0.049 m. How many turns of wire are needed?

turns

To calculate the number of turns required for the solenoid, we can use the formula for the self-inductance of a solenoid:

L = (μ₀ * N² * A) / l

Where:
L is the self-inductance of the solenoid
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid
μ₀ is the permeability of free space, approximately equal to 4π × 10⁻⁷ T·m/A

Rearranging the formula to solve for N:

N = √((L * l) / (μ₀ * A))

Let's plug in the given values:

L = 1.8 mH = 1.8 × 10⁻³ H
A = 1.60 × 10⁻³ m²
l = 0.049 m
μ₀ = 4π × 10⁻⁷ T·m/A

Calculating N:

N = √((1.8 × 10⁻³ * 0.049) / (4π × 10⁻⁷ * 1.60 × 10⁻³))

N ≈ 111 turns

Therefore, approximately 111 turns of wire are needed for the solenoid.