Inductance of a Solenoid – Physics Problem
Question:
Tarik winds a small paper tube uniformly with 191 turns of thin wire to form a solenoid. The tube’s diameter is 8.59 mm and its length is 2.29 cm. What is the inductance L, in microhenrys (μH), of Tarik’s solenoid?
Answer with Detailed Explanation:
Step 1: Write down the known values
Number of turns, N = 191
Diameter, D = 8.59 mm = 0.00859 m
Length, l = 2.29 cm = 0.0229 m
Permeability of free space, μ₀ = 4π × 10⁻⁷ H/m
Number of turns, N = 191
Diameter, D = 8.59 mm = 0.00859 m
Length, l = 2.29 cm = 0.0229 m
Permeability of free space, μ₀ = 4π × 10⁻⁷ H/m
Step 2: Use the formula for inductance of a solenoid
L = (μ₀ × N² × A) / l
where A is the cross-sectional area of the solenoid:
A = π × (D / 2)²
Now compute area A:
A = π × (0.00859 / 2)² = π × (0.004295)² ≈ π × 1.844 × 10⁻⁵ m²
A ≈ 5.793 × 10⁻⁵ m²
Now substitute values into the inductance formula:
L = (4π × 10⁻⁷) × (191)² × 5.793 × 10⁻⁵ / 0.0229
L ≈ (4π × 10⁻⁷) × 36481 × 5.793 × 10⁻⁵ / 0.0229
L ≈ 1.159 × 10⁻⁴ H = 115.9 × 10⁻⁶ H
✅ Therefore, the inductance of the solenoid is: L = 115.9 μH