The inductance of a solenoid that is 16.0 cm long and has a cross-sectional area of 1.00 × $10^{ - 4}$ $m ^ { 2 }$ is $1.00 \mathrm { mH }$ How many turns of wire does this solenoid have? (μ₀ = 4π × 10^-7 T ∙ m/A)

Question

Quizplus · Accepted Answer

The formula for the inductance of a solenoid is:

L = (μ₀N²Aℓ) / ℓ

Where:

L = inductance
N = number of turns
A = cross-sectional area
ℓ = length
μ₀ = vacuum permeability

Rearranging for N, we get:

N = √(Lℓ / (μ₀A))

Plugging in the given values, we get:

N = √((0.001 H)(0.16 m) / (4π × 10^-7 T∙m/A)(1.00 × 10^-4 m^2))

N = √(0.04 / (4π × 10^-3)) = √(10/π) ≈ 3.56

Therefore, the solenoid has approximately 1130 turns of wire (rounded to the nearest integer).

The correct answer is (C) 1130.

The Inductance of a Solenoid That Is 16 10−410^{ - 4}10−4

The Inductance of a Solenoid That Is 16 $10^{ - 4}$